Complex Analysis I
All My Students
Version 1
Fall, 2009
Version 1
Fall, 2009
2 of 84
Contents
1 COMPLEX NUMBERS
1.1 Sums and Products . . . . . . . . . . . . .
1.2 Basic Algebraic Properties . . . . . . . . .
1.3 Further Properties . . . . . . . . . . . . .
1.4 Vectors and Moduli . . . . . . . . . . . .
1.5 Complex Conjugates . . . . . . . . . . . .
1.6 Exponential Form . . . . . . . . . . . . .
1.7 Products and Powers in Exponential Form
1.8 Arguments of Products and Quotients . .
1.9 Roots of Complex Numbers . . . . . . . .
1.10 Examples . . . . . . . . . . . . . . . . . .
1.11 Regions in the Complex Plane . . . . . . .
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5
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2 ANALYTIC FUNCTIONS
2.1 Functions of a Complex Variable . . . . .
2.2 Mappings . . . . . . . . . . . . . . . . . .
2.3 Mappings by the Exponential Function . .
2.4 Limits . . . . . . . . . . . . . . . . . . . .
2.5 Theorems on Limits . . . . . . . . . . . .
2.6 Limits Involving the Point at Infinity . . .
2.7 Continuity . . . . . . . . . . . . . . . . .
2.8 Derivatives . . . . . . . . . . . . . . . . .
2.9 Differentiation Formulas . . . . . . . . .
2.10 Cauchy–Riemann Equations . . . . . . . .
2.11 Sufficient Conditions for Differentiability
2.12 Polar Coordinates . . . . . . . . . . . . .
2.13 Analytic Functions . . . . . . . . . . . . .
2.14 Examples . . . . . . . . . . . . . . . . . .
2.15 Harmonic Functions . . . . . . . . . . . .
2.16 Uniquely Determined Analytic Functions
2.17 Reflection Principle . . . . . . . . . . . .
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3 ELEMENTARY FUNCTIONS
3.1 The Exponential Function . . . . . . . .
3.2 The Logarithmic Function . . . . . . . .
3.3 Branches and Derivatives of Logarithms
3.4 Some Identities Involving Logarithms .
3.5 Complex Exponents . . . . . . . . . . .
3.6 Trigonometric Functions . . . . . . . .
3.7 Hyperbolic Functions . . . . . . . . . .
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Version 1
3.8
Fall, 2009
Inverse Trigonometric and Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . .
4 INTEGRALS
4.1 Derivatives of Functions w(t) . . . . . . . . . . . . . . . . . . .
4.2 Definite Integrals of Functions w(t) . . . . . . . . . . . . . . .
4.3 Contours . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Contour Integrals . . . . . . . . . . . . . . . . . . . . . . . . .
4.5 Some Examples . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6 Examples with Branch Cuts . . . . . . . . . . . . . . . . . . . .
4.7 Upper Bounds for Moduli of Contour Integrals . . . . . . . . .
4.8 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.9 Proof of the Theorem . . . . . . . . . . . . . . . . . . . . . . .
4.10 Cauchy–Gorsat Theorem . . . . . . . . . . . . . . . . . . . . .
4.11 Proof of the Theorem . . . . . . . . . . . . . . . . . . . . . . .
4.12 Simply Connected Domains . . . . . . . . . . . . . . . . . . . .
4.13 Multiply Connected Domains . . . . . . . . . . . . . . . . . . .
4.14 Cauchy Integral Formula . . . . . . . . . . . . . . . . . . . . .
4.15 An Extension of the Cauchy Integral Formula . . . . . . . . . .
4.16 Some Consequences of the Extension . . . . . . . . . . . . . . .
4.17 Liouville’s Theorem and the Fundamental Theorem of Algebra
4.18 Maximum Modulus Principle . . . . . . . . . . . . . . . . . . .
4 of 84
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46
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Chapter 1
COMPLEX NUMBERS
§1.1 Sums and Products.
§1.2 Basic Algebraic Properties.
§1.3 Further Properties.
§1.4 Vectors and Moduli.
§1.5 Complex Conjugates.
§1.6 Exponential Form.
§1.7 Products and Powers in Exponential Form.
§1.8 Arguments of Products and Quotients.
§1.9 Roots of Complex Numbers.
§1.10 Examples.
§1.11 Regions in the Complex Plane.
5
Version 1
Fall, 2009
6 of 84
Chapter 2
ANALYTIC FUNCTIONS
§2.1 Functions of a Complex Variable.
If D and R are sets of real numbers, then a function f : D → R is called a real–valued function of a
real variable. It is customary to write such a function as y = f (x). If D and R are sets of complex numbers,
then a function f : D → R is called a complex–valued function of a complex variable. It is customary to
write such a function as w = f (z).
Representation of w = f (z).
1. Using rectangular coordinates on z and w: f (x + iy) = u(x, y) + iv(x, y).
2. Using polar coordinates on z and w: f (reiθ ) = u(r, θ) + iv(r, θ).
Example 2.1.1. If f (z) = z 2 , then
f (x + yi) = (x + yi)2 = x2 − y 2 + 2xyi.
So in the notations above, we have
u(x, y) = x2 − y 2
and
v(x, y) = 2xy.
When polar coordinates are used,
f (reiθ ) = (reiθ )2 = r2 e2iθ = r2 cos(2θ) + ir2 sin(2θ).
Hence,
u(r, θ) = r2 cos(2θ)
and
v(r, θ) = r2 sin(2θ).
We note that if v = 0, then f (z) = u + iv becomes f (z) = u, which is a real–valued function of a
complex variable.
Example 2.1.2. An important real–valued function of a complex variable: for z = x + yi,
f (z) = |z|2 = x2 + y 2 + i0.
Just as in Real Analysis, the function
P (z) = a0 + a1 z + a2 z 2 + · · · + an z n ,
where n = 0 or a positive integer and an ̸= 0 and a0 , a1 , . . . , an are complex numbers, is called a polynomial
of degree n. Domain of P (z) is the whole complex plane. The quotient P (z)/Q(z) of polynomials is called
a rational function and defined at the point z where Q(z) ̸= 0.
Usually, when we say a function, it means a rule assigning one value to one point in the domain.
Explicitly, it is called the single–valued function. We can generalize this rule by assigning more than one
values to one point in the domain. It is called the multiple–valued function. We will study functions of a
complex variable from the viewpoint of the multiple–valued function.
Example 2.1.3. For z ̸= 0, we recall from Section 8. Roots of Complex Numbers,
(
)
Arg(z)
1/2
z = ±|z| exp i
,
2
where −π < Arg(z) ≤ π is the principal value of Arg(z). As we can see, the function f (z) = z 1/2 has two
values so it is a multiple–valued function. However, if we choose the positive value of ±|z| and write
(
)
Arg(z)
f (z) = |z| exp i
,
2
then it becomes the single–valued function.
7
Version 1
Fall, 2009
§2.2 Mappings.
As we approached a complex number from the geometrical viewpoint, we investigate the geometrical
work of a complex function. Since a complex function is defined on the complex plane and maps a point in
the plane to another point in the plane, we need to distinguish those two planes (domain plane and image
plane). Usually we say xy–plane (or z–plane) for the domain and uv–plane (or w–plane) for the image, which should
be clear from the expression f (x + iy) = u + iv introduced in the previous section. So a complex function
becomes a mapping from the xy–plane to the uv–plane. (Recall that for a real function, it is a mapping from
the x–axis to the y–axis.)
Three Geometrical Characteristics
1. Translation: For example, the mapping
w = z + 1 = (x + 1) + iy,
where z = x + iy, can be thought of as a translation of each point z one unit to the right.
2. Rotation: Since i = eiπ/2 , the mapping
π
w = iz = rei(θ+ 2 ) ,
where z = reiθ , rotates the radius vector for each nonzero point z through a right angle about the origin
in the counterclockwise direction.
3. Reflection: The mapping
w = z̄ = x − iy
transforms each point z = x + iy into its reflection in the real axis.
Example 2.2.1. Sketch the image of the mapping w = z 2 in the uv–plane.
Answer. Putting z = x + iy into the function, we have
u = x2 − y 2 ,
v = 2xy.
(Case 1) u = x2 − y 2 = c1 (positive constant): the graph of x2 − y 2 = c1 > 0 in the xy–plane is the
hyperbola cutting the x–axis. But in the uv–plane u = c1 represents a vertical line. That is, hyperbolas
in the xy–plane are mapped to the vertical lines in the uv–plane.
(Case 2) v = 2xy = c2 (positive constant): the graph of 2xy = c2 > 0 in the xy–plane is the hyperbola
located in the first and the third quadrants. But in the uv–plane v = c2 represents a horizontal line.
That is, hyperbolas in the xy–plane are mapped to the horizontal lines in the uv–plane.
Example 2.2.2. Find the image of S = {z = x + iy ∈ C : x > 0, y > 0, xy < 1} under the mapping w = z 2 .
Answer. The mapping w = z 2 implies v(x, y) = 2xy from the previous example. So if 0 < xy < 1, then
0 < v(x, y) = 2xy < 2. Again from the previous example, the mapped image of the hyperbola is a horizontal
line. So xy = 1 in the xy–plane is mapped into a horizontal line v = 2 in the uv–plane and thus S is mapped
to the infinite strip S ′ = { w = u + iv ∈ C | 0 < v < 2 }.
Exercise 2.2.3. Sketch the domain in the xy–plane whose image by the mapping w = z 2 is the square
domain S ′ = {(u, v) : 1 ≤ u ≤ 2, 1 ≤ v ≤ 2}.
Answer. The domain having such an image is
S = {(x, y) : 1 ≤ x2 − y 2 ≤ 2, 1 ≤ 2xy ≤ 2}
Example 2.2.4. The mapping w = z 2 becomes w = r2 ei2θ when z = reiθ . Hence, if w = ρeiϕ , then we have
ρeiϕ = r2 ei2θ , i.e., ρ = r2 and ϕ = 2θ + 2kπ, where k is an integer. So the first quadrant in the xy–plane is
mapped to the upper half–plane of the uv–plane under the mapping w = z 2 .
Exercise 2.2.5. Sketch the image of the sector S = {(r, θ) : r ≤ 1, 0 ≤ θ ≤ π4 } by the mapping (1) w = z 2 ,
(2) w = z 3 , (3) w = z 4 .
§2.3 Mappings by the Exponential Function.
8 of 84
Version 1
Fall, 2009
§2.4 Limits.
The concepts of limits and continuity are similar to that of real variables. In this sense our discussion
can serve as a brief review of many previously understood notions. Consider a function w = f (z) defined
at all points in some neighborhood of z = z0 , except possibly for z0 itself. We say f (z) has the limit w0 if as
z approaches z0 , f (z) approaches w0 (z0 , w0 finite). Mathematically, we say
(2.4.1)
lim f (z) = w0 ,
z → z0
if for every (sufficiently small) ε > 0 there is a δ > 0 such that
∥f (z) − w0 ∥ < ε,
whenever 0 < ∥z − z0 ∥ < δ.
(2.4.2)
This definition is clear when z0 is an interior point of a region R in which f (z) is defined. If z0 is a
boundary point of R, then we require Eq. (2.4.2) to hold only for those z ∈ R.
Geometrically, under the mapping w = f (z), all points interior to the circle ∥z − z0 ∥ = δ with z0
deleted are mapped to points interior to the circle ∥w − w0 ∥ = ε. The limit will exist only in the case when
z approaches z0 (that is, z → z0 ) in an arbitrary direction; then this implies that w → w0 .
Example 2.4.1. Show that
(
lim 2
z→i
z 2 + iz + 2
z−i
)
= 6i.
Proof. We must show that given ε > 0, there is a δ > 0 such that
( 2
(
)
)
z + iz + 2
(z + 2i)(z − i)
2
< ε,
−
6i
−
6i
=
2
z−i
z−i
whenever
0 < ∥z − i∥ < δ.
(2.4.3)
(2.4.4)
(2.4.5)
Since z = i, inequality (2.4.1) implies that 2 ∥z − i∥ < ε. Thus if δ = ε/2, Eq. (2.4.5) ensures that
Eq. (2.4.4) is satisfied. Therefore Eq. (2.4.3) is demonstrated.
Exercise 2.4.2. Compute the limit in the open disk ∥z∥ < 3:
iz − 1
.
z→3
2
lim
When a limit of a function f (z) exists at a point z0 , the limit is unique. That is, if
lim f (z) = w0
and
z → z0
lim f (z) = w1 ,
z → z0
then w0 = w1 .
Proof. For every ε/2 > 0 there is a δ0 > 0 such that
∥f (z) − w0 ∥ < ε,
whenever 0 < ∥z − z0 ∥ < δ0 .
For every ε/2 > 0 there is a δ1 > 0 such that
∥f (z) − w1 ∥ < ε,
whenever 0 < ∥z − z0 ∥ < δ1 .
We observe for 0 < ∥z − z0 ∥ < δ = min {δ0 , δ1 },
0 ≤ ∥w0 − w1 ∥ = ∥(f (z) − w1 ) − (f (z) − w0 )∥ ≤ ∥f (z) − w1 ∥ + ∥f (z) − w0 ∥ < ε/2 + ε/2 = ε.
Since ε > 0 can be chosen arbitrarily small, hence we deduce
∥w0 − w1 ∥ = 0,
i.e.,
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w 0 = w1 .
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We recall from Real Analysis or Calculus, the limit exists only when two sided limit exist and they are
same. However, in Complex Analysis, if z0 is an interior point of the domain of f and the limit is to exist,
the inequality ∥f (z) − w0 ∥ < ε must hold for all points in the deleted neighborhood 0 < ∥z − z0 ∥ < δ.
Thus, the symbol z → z0 implies that z is allowed to approach z0 in arbitrary manner, not just from some
particular direction.
Example 2.4.3. Consider the limit of f (z) =
z
as z approach 0.
z̄
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§2.5 Theorems on Limits.
Theorem 2.5.1. Suppose that
f (z) = u(x, y) + iv(x, y),
z0 = x0 + iy0 ,
w0 = u0 + iv0 .
Then
lim f (z) = w0
z→z0
if and only if
lim
(x,y)→(x0 ,y0 )
lim
u(x, y) = u0 ,
(x,y)→(x0 ,y0 )
v(x, y) = v0 .
Theorem 2.5.2. Suppose that
lim f0 (z) = w0 ,
z→z0
lim F (z) = W0 .
and
z→z0
Then
lim (f (z) + F (z)) = w0 + W0 = lim f (z) + lim F (z),
z→z0
z→z
(
)( 0
)
lim (f (z)F (z)) = w0 W0 = lim f (z)
lim F (z) ,
z→z0
z→z0
z→z0
lim
z→z0
w0
limz→z0 f (z)
f (z)
=
=
,
F (z)
W0
limz→z0 F (z)
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z→z0
(if W0 ̸= 0).
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§2.6 Limits Involving the Point at Infinity.
We introduce an “ideal” point which we call the point at infinity, denoted by ∞. The points in the plane
together with the point at infinity form the extended complex plane. We agree that every straight line shall
pass through the point at infinity. By contrast, no half plane shall contain the ideal point.
How to visualize the point at infinity? Consider a unit sphere S centered at the origin and think of
the complex plane as the plane passing through the equator of the unit sphere S. Let N be the north pole
on the sphere S. (1) To each point z in the complex plane, there corresponds exactly one point P on the
surface of the sphere, which can be found as the intersection point between the sphere and the straight line
connecting z and N . (2) On the other way, to each point P on the sphere, there corresponds exactly one
point z in the complex plane, which can be found as the intersection point between the complex plane and
the straight line connecting P and N . (3) We let the north pole N correspond to the point at infinity, ∞.
From (1) to (3), we obtain a one to one correspondence between the points of the sphere and the points of
the extended complex plane. The sphere is known as the Riemann sphere and the correspondence is called a
stereographic1 projection.
In fact, except (0, 0, 1), to a point (x1 , x2 , x3 ) on the unit sphere S centered at the origin, we can
associate a complex number
x1 + ix2
z=
,
1 − x3
and this correspondence is one to one. Indeed, from the map above, we obtain
(
∥z∥ = z z̄ =
2
x1 + ix2
1 − x3
)(
x1 + ix2
1 − x3
)
=
1 − x23
1 + x3
x1 + ix2 x1 − ix2
x2 + x22
=
= 1
,
2
2 =
1 − x3 1 − x3
1 − x3
(1 − x3 )
(1 − x3 )
where x21 + x22 + x23 = 1 is used. It implies
∥z∥2 − 1
.
x3 =
∥z∥2 + 1
Further computation gives
z + z̄ =
(
)
2x1
= 1 + ∥z∥2 x1 ,
1 − x3
x1 =
z + z̄
,
1 + ∥z∥2
similarly,
z − z̄
).
x2 = (
i 1 + ∥z∥2
The correspondence can be completed by letting the point at infinity correspond to (0, 0, 1), and we can
thus regard the sphere as a representation of the extended plane or of the extended number system. We
note that the hemisphere x3 < 0 corresponds to the disk ∥z∥ < 1 and the hemisphere x3 > 0 to its outside
∥z∥ > 1. Moreover, for each small positive real number ε, those points in the complex plane exterior to the
circle ∥z∥ = 1/ε correspond to the points on the sphere close to N . We thus call the set ∥z∥ > 1/ε an ε
neighborhood, or just neighborhood, of ∞. Unless specifically mentioned, when we refer a point z, it means
a point in the finite plane.
Since we have the point at infinity, ∞, we can have the following theorem.
Theorem 2.6.1. If z0 and w0 are points in the z and w planes, respectively, then
lim f (z) = ∞
if and only if
lim f (z) = w0
if and only if
lim f (z) = ∞
if and only if
z → z0
z→∞
z→∞
1
1
= 0,
z → z0 f (z)
( )
1
lim f
= w0 ,
z→0
z
1
= 0.
lim
z → 0 f (1/z)
lim
stereography: the representation of three–dimensional things on a two–dimensional surface
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Proof. 1. Suppose lim f (z) = ∞ and we prove lim
z → z0
z → z0
exists a δ > 0 such that
∥f (z)∥ >
It implies
1
ε
1
= 0. By the hypothesis, for each ε > 0, there
f (z)
whenever 0 < ∥z − z0 ∥ < δ.
1
f (z) − 0 < ε
whenever 0 < ∥z − z0 ∥ < δ,
1
1
= 0. By the similar argument, we can prove lim
= 0 implies lim f (z) = ∞.
z → z0 f (z)
z → z0
( ) z → z0 f (z)
1
= w0 . By the hypothesis, for each ε > 0, there
2. Suppose lim f (z) = w0 and we prove lim f
z→∞
z→0
z
exists a δ > 0 such that
1
∥f (z) − w0 ∥ < ε whenever ∥z∥ > .
δ
When we replace z by 1/z, the inequalities above become
( )
1
f
< ε whenever 0 < ∥z − 0∥ < δ,
−
w
0
z
( )
1
that is, lim f
= w0 . By the similar argument, we can prove the other implication.
z→0
z
1
3. Suppose lim f (z) = ∞ and we prove lim
= 0. By the hypothesis, for each ε > 0, there
z→∞
z → 0 f (1/z)
exists a δ > 0 such that
1
1
∥f (z)∥ >
whenever ∥z∥ > .
ε
δ
When we replace z by 1/z, the inequalities above become
1
f (1/z) − 0 < ε whenever 0 < ∥z − 0∥ < δ,
that is, lim
1
= 0. By the similar argument, we can prove the other implication.
z → 0 f (1/z)
that is, lim
Example 2.6.2. Compute the limits:
2z + i
,
z→∞ z + 1
z2 − 1
,
2. lim
z→1 z − 1
1. lim
2z 3 − 1
,
z → ∞ z2 + 1
1
lim 2
,
z→1 z − 1
lim
4z 2
,
z → ∞ (z − 1)2
1
lim
,
z → 1 (z − 1)3
lim
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z2 + 1
,
z→∞ z − 1
iz + 3
lim
z → −1 z + 1
lim
lim
z→∞
z
2 + iz
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§2.7 Continuity.
Definition 2.7.1. A function f is said to be continuous at z0 if
lim f (z) = z0 .
z → z0
If f is continuous at each point in a region R, then the function is said to be continuous in the region. If a
complex function f is not continuous at a point z0 , then we say that f is discontinuous at z0 .
The equation above contains the existence of the limit lim f (z) and the value f (z0 ).
z → z0
1
continuous at z = ±i?
1 + z2
(2). Is f (z) = z 2 − iz + 2 continuous at z = 1 − i?
√
(3). Is the principal square root function f (z) = z 1/2 = ∥z∥ ei Arg(z)/2 continuous at z = −1?
Example 2.7.2. (1). Is the function f (z) =
The following results are easily deduced:
1. For a continuous function f and a complex number c, cf is continuous.
2. Sum/Product/Quotient of two continuous functions is also continuous.
3. If f is continuous, then so are Re (f (z)), Im (f (z)), complex conjugate f (z), and ∥f (z)∥. Specifically,
for continuous f (z) = f (x, y) = u(x, y) + iv(x, y), √
the functions Re (f (z)) = u(x, y) and Im (f (z)) =
v(x, y) and f (z) = u(x, y) − iv(x, y) and ∥f (z)∥ = u2 (x, y) + v 2 (x, y) are continuous.
4. A complex polynomial is continuous in the entire complex plane.
More important results are given in the theorems.
Theorem 2.7.3. A composition g ◦ f of two continuous functions f and g is itself continuous.
Since the proof is almost similar to the one for the real–valued functions of real variables, we skip it.
Theorem 2.7.4. If a function f is continuous and nonzero at a point z0 , then there exists a neighborhood of
z0 where f (z) ̸= 0 at any point in that neighborhood.
Proof. Let f is continuous and nonzero at z0 . Then, by the continuity, for ε =
δ > 0 such that
∥f (z0 )∥
> 0, there exists a
2
∥f (z0 )∥
whenever ∥z − z0 ∥ < δ.
2
So if there is a point z in the neighborhood ∥z − z0 ∥ < δ at which f (z) = 0, then we have the contradiction
∥f (z) − f (z0 )∥ < ε =
∥f (z0 )∥ <
∥f (z0 )∥
.
2
Therefore, we should have f (z) ̸= 0 at any point in the neighborhood ∥z − z0 ∥ < δ.
We recall in real analysis: a function that is continuous in a compact (i.e., closed and bounded) region
R attains both maximum and minimum values on R. This result is also available in complex analysis through
the following theorem, which is very useful in applications.
Theorem 2.7.5. If a function f is continuous throughout a compact (i.e., closed and bounded) region R,
then there exists a nonnegative real number M such that
∥f (z)∥ ≤ M
for all points z in R,
where equality holds for at least one such z.
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√
Proof. We observe ∥f (z)∥ = u2 (x, y) + v 2 (x, y) is continuous throughout R. Since the functions of real
variables, u(x, y) and v(x, y), attain both maximum and minimum values on R from real analysis, so f (z)
should be bounded on R.
Example 2.7.6. Check the continuity of the given function:
z+1
f (z) = 2
,
z + 4z + 5
{
f (z) =
Re(z)
z
0
for z ̸= 0
for z = 0,
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

x2
+ 2i
f (z) =
x2 + y 2

2i
for z ̸= 0
for z = 0.
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§2.8 Derivatives.
Suppose z = x + iy and z0 = x0 + iy0 . Then the change in z0 is the difference △z = z − z0 or
△z = x − x0 + i (y − y0 ) = △x + i△y. If a complex function w = f (z) is defined at z and z0 , then the
corresponding change in the function is the difference △w = f (z0 + △z) − f (z0 ). The derivative of the
function f is defined in terms of a limit of the difference quotient △w/△z as △z → 0.
Definition 2.8.1. Suppose the complex function f is defined in a neighborhood of a point z0 . The derivative
of f at z0 , denoted by f ′ (z0 ), is
f ′ (z0 ) = lim
△z → 0
f (z0 + △z) − f (z0 )
,
△z
provided this limit exists.
If the limit above exists, then the function f is said to be differentiable at z0 . Two other symbols denoting
dw
the derivative of w = f (z) are w′ and
. If the latter notation is used, then the value of a derivative at a
dz
dw .
specified point z0 is written
dz z=z0
Example 2.8.2. Use the definition to find the derivative, if it exists:
f (z) = z 2 ,
f (z) = z 2 − 5z.
Answer. (1). f (z) = z 2 has the derivative, f ′ (z) = 2z. (Check by yourself.)
(2). f (z) = z 2 − 5z has the derivative, f ′ (z) = 2z − 5. (Check by yourself.)
Example 2.8.3. Use the definition to find the derivative of f (z) = z̄, if it exists.
Answer. f (z) = z̄ does not have a derivative at any point, i.e., it is not differentiable anywhere. The reason
is as follows.
△w
z + △z − z̄
△z
z̄ + △z − z̄
=
=
=
.
△z
△z
△z
△z
As △z approaches 0 along the real axis, i.e., △z = (△x, 0) → (0, 0), we have
△z = △x + i0 = △x − i0 = △x + i0 = △z,
and so in this case,
△w
△z
△z
=
=
= 1,
△z
△z
△z
i.e., the limit along the real axis is 1. Similarly, as △z approaches 0 along the imaginary axis, i.e., △z =
(0, △y) → (0, 0), we have
△z = 0 + i△y = 0 − i△y = − (0 + i△y) = −△z,
and so in this case,
−△z
△w
△z
=
=
= −1,
△z
△z
△z
i.e., the limit along the imaginary axis is −1. Since the limit should be unique, it follows dw/dz does not
exist anywhere.
Example 2.8.4. Use the definition to find the derivative of f (z) = ∥z∥2 , if it exists.
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Answer. f (z) = ∥z∥2 has a derivative, f ′ (z) = 0, only at z = 0, i.e., f ′ (0) = 0 and it is not differentiable
except at z = 0. The reason is as follows.
(
)
(z + △z) z̄ + △z − z z̄
∥z + △z∥2 − ∥z∥2
△w
△z
=
=
= z̄ + △z + z
.
△z
△z
△z
△z
As the solution of the example above, the horizontal and vertical approaches of △z toward the origin gives
us, respectively,
△z = △z,
and
△z = −△z.
It implies
△w
= z̄ + △z + z
△z
△w
= z̄ + △z + z
△z
△z
= z̄ + △z + z,
△z
△z
= z̄ − △z − z,
△z
when △z = (△x, 0),
when △z = (0, △y).
Again, by the uniqueness of the limit, as △z approaches 0, we should have
z̄ + z = z̄ − z,
i.e.,
z = 0.
Therefore, dw/dz cannot exist when z ̸= 0.
However, when z = 0, we have
△w
= △z,
△z
△w
= lim △z = 0.
△z → 0 △z
△z → 0
and
lim
That is, dw/dz does exist when z = 0 and its value at z = 0 is 0.
Now, we may raise the following questions:
I. (Relation between Continuity and Differentiability)
I.1 If f is continuous at a point z0 , then is it differentiable at that point?
Answer: No. See the Example 2.8.3 above. f (z) = z̄ is continuous everywhere but it is not differentiable anywhere.
I.2 If f is differentiable at a point z0 , then is it continuous at that point?
Answer: Yes. We have the following fact: if f is differentiable at z0 , then f is continuous at z0 .
Proof. Suppose f ′ (z0 ) exists. Then,
[
]
f (z) − f (z0 )
lim [f (z) − f (z0 )] = lim
(z − z0 )
z → z0
z → z0
z − z0
[
][
]
f (z) − f (z0 )
= lim
lim (z − z0 ) = f ′ (z0 ) · 0 = 0,
z → z0
z → z0
z − z0
which implies
lim f (z) = f (z0 ),
z → z0
i.e., f is continuous at z0 .
II. (Relation between Function and Its Real/Imaginary Parts on Differentiability)
II.1 If f (z) = u(x, y) + iv(x, y) is differentiable at a point z0 = (x0 , y0 ), then are its real and imaginary
parts, u(x, y) and v(x, y), differentiable at the point (x0 , y0 )? Answer: Yes.
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II.2 If both u(x, y) = Re (f (z)) and v(x, y) = Im (f (z)) are differentiable at a point z0 = (x0 , y0 ), then
is the function f (z) differentiable at the point z0 = (x0 , y0 )? Answer: No.
The relations (especially II.2) in the part II above are very important in Complex Analysis and so it will
be discussed in Section 21 Cauchy–Riemann Equations in detail. We may briefly discuss the relations through
the Example 2.8.4 above. The Example 2.8.4 says that a function w = f (z) can be differentiable at a point
but nowhere else in any neighborhood of that point. From f (z) = ∥z∥2 = x2 + y 2 = u(x, y) + iv(x, y), we
have
u(x, y) = x2 + y 2 ,
v(x, y) = 0,
which are differentiable everywhere. So the Example 2.8.4 gives us a function which is not differentiable
at a point but its real part u(x, y) and imaginary part v(x, y) are differentiable at that point. Moreover, if a
function f is differentiable at a point z, then both u(x, y) and v(x, y) should be differentiable there.
Geometric interpretation of a derivative of a complex function is not easy to study and thus it will be
postponed until Chapter 9 Conformal Mapping.
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§2.9 Differentiation Formulas.
By the definition, we can deduce the following formulas: for differentiable functions f and g and a
complex number c,
d
d
1. (Constant Rules)
c = 0;
(cf ) = cf ′
dz
dz
d
d n
2. (Power Rule)
z = nz n−1 , where n ̸= 0 is an integer;
z=1
dz
dz
d
3. (Sum Rule)
(f ± g) = f ′ ± g ′
dz
d
4. (Product Rule)
(f g) = f ′ g + f g ′
dz ( )
f ′g − f g′
d
f
=
5. (Quotient Rule)
, where g ̸= 0.
dz g
g2
d
6. (Chain Rule)
(g (f )) = g ′ (f ) f ′
dz
Proof of Chain Rule. Let a point z0 be fixed at which f ′ (z0 ) exists. We write w0 = f (z0 ) and assume that
g ′ (w0 ) exists. Then there is some ε neighborhood ∥w − w0 ∥ < ε of w0 such that for all points w in that
neighborhood, we can define a function Φ having the values Φ(w0 ) = 0 and
Φ(w) =
g(w) − g(w0 )
− g ′ (w0 ),
w − w0
when w ̸= w0 .
(2.9.1)
Since g ′ (w0 ) exists, by taking the limit on both sides, we can deduce
lim Φ(w) = 0 = Φ(w0 ),
w → w0
i.e., Φ(w) is continuous at w0 .
The equation (2.9.1) can be rewritten by
g(w) − g(w0 ) = [g ′ (w0 ) + Φ(w)] (w − w0 )
(∥w − w0 ∥ < ε)
(2.9.2)
which is also valid when w = w0 . Since f is differentiable at z0 and so continuous there, we can choose a
δ > 0 such that
∥f (z) − f (z0 )∥ < ε whenever 0 ∥z − z0 ∥ < δ.
It allows us to replace w in the equation (2.9.2) by f (z) where z lies in the δ deleted neighborhood of z0 .
Hence, we get
g(f (z)) − g(f (z0 )) = [g ′ (f (z0 )) + Φ(f (z))] (f (z) − f (z0 ))
(0 < ∥z − z0 ∥ < δ)
Dividing the whole equation by z − z0 , which does not vanish in the δ deleted neighborhood, we have
f (z) − f (z0 )
g(f (z)) − g(f (z0 ))
= [g ′ (f (z0 )) + Φ(f (z))]
z − z0
z − z0
(0 < ∥z − z0 ∥ < δ)
Taking the limit as z approaches z0 , we deduce
]
[
][
]
[
d
d
′
(g (f (z)))
= g (f (z0 )) + lim Φ(f (z))
f (z)
z → z0
dz
dz
z=z0
z=z0
Since f is continuous at z0 and Φ is continuous at the point w0 = f (z0 ) and Φ(w0 ) = 0, so we have
lim Φ(f (z)) = Φ(f (z0 )) = Φ(w0 ) = 0.
z → z0
Therefore, finally we conclude
[
]
[
]
d
d
′
(g (f (z)))
= g (f (z0 ))
f (z)
,
dz
dz
z=z0
z=z0
at z = z0 . Since z0 is arbitrary, the equation holds for any z.
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i.e.,
[g(f (z))]′ = g ′ (f (z))f ′ (z)
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Example 2.9.1. Find the derivatives.
f (z) = (2z 2 + i)5
f (z) = sin z
f (z) = cos z
f (z) = ez
f (z) = 3z 4 − 5z 3 + 2z
z2
6. f (z) =
4z + 1
( 2
)5
7. f (z) = iz + 3z
1.
2.
3.
4.
5.
Example 2.9.2 (L’Hopitals Rule). Establish a special
∑ case of l’Hopitals rule. Suppose that f (z) and g(z) have
formal power series about z = a (e.g., f (z) = n an (z − a)n ), and
f (a) = f ′ (a) = f ′′ (a) = · · · = f (k) (a) = 0,
g(a) = g ′ (a) = g ′′ (a) = · · · = g (k) (a) = 0.
If f (k+1) (a) and g (k+1) (a) are not simultaneously zero, show that
f (z)
f (k+1) (a)
= (k+1) .
z → a g(z)
g
(a)
lim
What happens if g (k+1) (a) = 0?
Example 2.9.3. If |g(z)| = M , M > 0 for all z in a neighborhood of z = z0 , show that if limz → z0 f (z) = 0,
then
lim (f (z)g(z)) = 0.
z → z0
Example 2.9.4. Find the set of points where following functions are differentiable.
1. f (z) = sin z
2. f (z) = tan z
z−1
3. f (z) = 2
z +1
4. f (z) = e1/z
5. f (z) = 2z̄
We end the section with nowhere differentiable functions. In Calculus/Real Analysis, is there a function which is not differentiable anywhere, i.e., nowhere differentiable? If so, what is it? If not, why not? We
raise the same question in Complex Analysis. Is there a function which is not differentiable anywhere?
Example 2.9.5. Prove that the following functions are nowhere differentiable.
1. f (z) = x + 4iy
2. f (z) = Re (z)
3. f (z) = Im (z)
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§2.10 Cauchy–Riemann Equations.
We will study one of the most important theorems in Complex Analysis: Theorem with Cauchy–Riemann
Equations. Simply speaking, the theorem gives the relations on differentiability between a complex function
f and its real/imaginary parts Re(f ), Im(f ).
Theorem 2.10.1. The function f (z) = u(x, y) + iv(x, y) is differentiable at a point z = x + iy of a region
in the complex plane if and only if the partial derivatives ux , uy , vx and vy are continuous and satisfy the
Cauchy–Riemann equations
ux = vy ,
uy = −vx
(2.10.1)
at the point z = x + iy.
Proof on Necessary Condition. We start with the definition of the derivative of f .
f (z + △z) − f (z)
.
△z → 0
△z
f ′ (z) = lim
We write the real and imaginary parts of f (z), f (z) = u(x, y) + iv(x, y), and compute the right–hand side
of the equation above for(1) △z = △x real and (2) △z = i△y pure imaginary (i.e., we take the limit along
the real and then along the imaginary axis). Then, for case (1),
(
)
u(x + △x, y) − u(x, y)
v(x + △x, y) − v(x, y)
′
f (z) = lim
+i
= ux (x, y) + ivx (x, y).
(2.10.2)
△x → 0
△x
△x
For case (2),
′
f (z) = lim
△y → 0
(
u(x, y + △y) − u(x, y) i (v(x, y + △y) − v(x, y))
+
i△y
i△y
)
= −iuy (x, y) + vy (x, y). (2.10.3)
Setting equations (2.10.2) and (2.10.3) equal yields
u x = vy ,
uy = −vx ,
which are called the Cauchy–Riemann equations. The proof on sufficient condition is skipped.
Example 2.10.2. Show that the function
f (z) = z 2 = x2 − y 2 + i2xy
is differentiable everywhere and that f ′ (z) = 2z.
Using the necessary condition in the theorem for the differentiability of a function f at a point, we find
the points at which f is not differentiable.
Example 2.10.3. Find the set of points at which the given function is differentiable.
1. f (z) = ∥z∥2
2. f (z) = z̄
3. f (z) = 2x2 + y + i(y 2 − x)
4. Pn (z) = a0 + a1 z + · · · + an z =
n
n
∑
ak z k (a polynomial of degree n)
k=0
Using the sufficient condition, we can find the derivative of f .
Example 2.10.4. Find the derivative of the given function at a given point.
1. f (z) = ez at any point z
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2. f (z) = ∥z∥2 at z0 = 0
Example 2.10.5. Find real constants a, b, c and d so that the given function is differentiable.
1. f (z) = 3x − y + 5 + i (ax + by − 3)
(
)
2. f (z) = x2 + axy + by 2 + i cx2 + dxy + y 2
We end with the following theorem, which is a direct consequence of the Cauchy–Riemann equations.
Theorem 2.10.6 (Constant Functions). Suppose the function f (z) = u(x, y) + iv(x, y) is differentiable in a
region R in the complex plane.
1. If ∥f (z)∥ is constant in R, then so is f (z).
2. If f ′ (z) = 0 in R, then f (z) = c in R, where c is a constant.
§2.11 Sufficient Conditions for Differentiability.
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§2.12 Polar Coordinates.
We can restate the Theorem in the previous section using polar coordinates x = r cos θ and y = r sin θ.
Theorem 2.12.1. The function f (z) = u(r, θ) + iv(r, θ) is differentiable at a point z = reiθ of a region in
the complex plane if and only if the partial derivatives ur , uθ , vr and vθ are continuous and satisfy the polar
coordinates version of Cauchy–Riemann equations
rur = vθ ,
uθ = −rvr
(2.12.1)
at the point z = reiθ .
The derivative of f (z) = u(r, θ) + iv(r, θ) is
f ′ (z) = e−iθ (ur + ivr ) .
Key Idea of Proof. We prove only the key idea, the polar coordinates version of Cauchy–Riemann equations.
From x = r cos θ, y = r sin θ, r2 = x2 + y 2 and tan θ = xy , we get
)
∂ ( 2)
∂ ( 2
r =
x + y2 ,
∂x
∂x
∂ (y)
∂
(tan θ) =
,
∂x
∂x x
)
∂ ( 2)
∂ ( 2
r =
x + y2 ,
∂y
∂y
∂
∂ (y)
(tan θ) =
,
∂y
∂y x
∂r
∂x
∂θ
sec2 θ
∂x
∂r
2r
∂y
∂θ
sec2 θ
∂y
2r
∂r
∂x
∂θ
∂x
∂r
∂y
∂θ
∂y
= 2x,
=−
y
,
x2
= 2y,
=
1
,
x
x
r cos θ
=
= cos θ
r
r
y
r sin θ
sin θ
=− 2 2 =− 2
=
−
x sec θ
r cos2 θ sec2 θ
r
y
r sin θ
= =
= sin θ
r
r
1
1
cos θ
=
=
=
.
2
2
x sec θ
r cos θ sec θ
r
=
Combining the results above with the chain rule for differentiating real–valued functions of two variables
implies the differential relationships:
∂
∂ ∂r
∂ ∂θ
∂
sin θ ∂
=
+
= cos θ −
∂x
∂r ∂x ∂θ ∂x
∂r
r ∂θ
∂
∂ ∂r
∂ ∂θ
∂
cos θ ∂
=
+
= sin θ +
.
∂y
∂r ∂y ∂θ ∂y
∂r
r ∂θ
By the relationships, we get
sin θ
uθ ,
r
sin θ
vx = cos θ vr −
vθ ,
r
cos θ
uθ ,
r
cos θ
vy = sin θ vr +
vθ .
r
ux = cos θ ur −
uy = sin θ ur +
(2.12.2)
(2.12.3)
The Cauchy–Riemann equations says ux = vy and uy = −vx . Applying them above, we deduce
sin θ
cos θ
uθ = ux = vy = sin θ vr +
vθ ,
r
r
sin θ
cos θ
uθ = uy = −vx = − cos θ vr +
vθ .
sin θ ur +
r
r
cos θ ur −
Multiplying the first of these equations by cos θ, the second by sin θ, and adding yields rur = vθ . Similarly,
multiplying the first by sin θ, the second by − cos θ, and adding yields uθ = −rvr . That is,
rur = vθ ,
uθ = −rvr .
Using the relations of (2.12.2) and (2.12.3) in f ′ (z) = ux + ivx with the results above yields
(
)
sin θ
sin θ
′
f (z) = cos θ ur −
uθ + i cos θ vr −
vθ
r
r
= (cos θ − i sin θ) (ur + ivr ) = e−iθ (ur + ivr ) .
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Example 2.12.2. Consider the function
f (z) =
1
1
1
1
= iθ = e−iθ = (cos θ − i sin θ) ,
z
re
r
r
where z ̸= 0. f (z) has the component
u(r, θ) =
and we have
ur = −
cos θ
,
r2
cos θ
,
r
uθ = −
We observe
rur = −
sin θ
,
r
cos θ
= vθ
r
v(r, θ) = −
vr =
uθ = −
sin θ
r
sin θ
,
r2
vθ = −
cos θ
.
r
sin θ
= −rvr
r
at any nonzero point. Therefore f ′ (z) exists when z ̸= 0 and it is in fact
(
)
cos θ
sin θ
f (z) = e (ur + ivr ) = e
− 2 +i 2
r
r
1
1
1
1
= − 2 e−iθ (cos θ − i sin θ) = − 2 e−iθ e−iθ = −
2 = − 2.
iθ
r
r
z
(re )
′
−iθ
−iθ
Example 2.12.3. Consider the function
f (z) = z 1/3 =
√
[ 3]reiθ/3 ,
(r > 0,
α < θ < α + 2π),
where α is a fixed real number. Prove that f is differential everywhere.
Proof. We observe
f (z) =
√
[ 3]re
iθ/3
=
√
(
)
√
√
θ
θ
θ
θ
[ 3]r cos + i sin
= [ 3]r cos + i [ 3]r sin = u(r, θ) + iv(r, θ),
3
3
3
3
i.e.,
u(r, θ) =
√
θ
[ 3]r cos ,
3
v(r, θ) =
√
θ
[ 3]r sin .
3
We compute
r−2/3
θ
ur =
cos ,
3
3
and observe
√
[ 3]r
θ
uθ = −
sin ,
3
3
√
rur =
[ 3]r
θ
cos = vθ
3
3
√
r−2/3
θ
vr =
sin ,
3
3
√
uθ = −
vθ =
[ 3]r
θ
cos ,
3
3
[ 3]r
θ
sin = −rvr
3
3
at any point. Therefore f ′ (z) exists everywhere and it is in fact
(
)
r−2/3
θ
r−2/3
θ
f (z) = e (ur + ivr ) = e
cos + i
sin
3
3
3
3
(
)
−2/3
−2/3
θ
θ
1
r
r
e−iθ cos + i sin
e−iθ eiθ/3 = (√
=
=
3
3
3
3
3
′
−iθ
−iθ
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1
[ 3]reiθ/3
)2 =
1
.
3f 2 (z)
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The example above shows us
d 1/3
1
z =
,
dz
3(z 1/3 )2
which is an example of a Power Rule with non–integer power. Be careful: The power rule in Section 20.
Differentiation Formulas says
d n
z = nz n−1 ,
dz
for integer n.
Exercise 2.12.4. Discuss the differentiability of the given function in the indicated domain of definition. If
the derivative exists, then find it.
1. f (z) = z̄ = e−iθ
(C = the whole complex plane)
√ iθ/2
2. f (z) = r e
(r > 0 and α < θ < α + π)
3. f (z) = e−θ cos(ln r) + ie−θ sin(ln r)
(r > 0 and 0 < θ < 2π)
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§2.13 Analytic Functions.
Definition 2.13.1. Let f be a complex function.
1. If f has a derivative at each point in some neighborhood of z, then f is analytic (or regular, holomorphic)
at z.
2. If f is analytic at each point in a domain D or region R, then f is analytic in D or R. (Recall: a domain
D means a connected open set and a region R means a domain with some, none, or all of its boundary
points.) 2
3. If f is analytic in the entire complex plane, then f is an entire function.
4. If f is not analytic at z0 but it is analytic at some point in every neighborhood of z0 , then z0 is a singular
point or singularity of f .
Example 2.13.2.
1. f (z) = ∥z∥2 is nowhere analytic. It is differentiable only at z = 0. So there is no neighborhood of
z = 0 in which f is differentiable. Moreover, f (z) = ∥z∥2 has no singular point.
2. f (z) = 1/z is analytic at each point z ̸= 0 and z = 0 is a singular point of f .
3. A polynomial Pn (z) = a0 + a1 z + · · · + an z n is an entire function. (Are trigonometric functions and
exponential function entire? The answer will be discussed later.)
Using the theorems on previous sections, we can deduce the following results.
Theorem 2.13.3. Let f and g be analytic and c any complex number. Then, cf , f ± g, f g, and f /g with
g ̸= 0, and f ◦ g (composition) are analytic.
The following property of analytic functions is especially useful.
Theorem 2.13.4. If f ′ (z) = 0 everywhere in a domain D, then f (z) must be constant throughout D.
Proof. Step 1: Let f (z) = u(x, y) + iv(x, y) satisfy f ′ (z) = 0 in D. That is,
f ′ (z) = ux + ivx = 0,
ux = 0,
vx = 0.
Since f is differentiable in D, it satisfies the Cauchy–Riemann equations in D and so we have uy = 0 = vy .
That is, at each point in D,
ux = 0 = uy = vx = vy .
Step 2: We show that u(x, y) is constant along any line segment L from a point P to a point P ′ and
lying entirely in D.
Let s denote the distance along L from the point P and let U denote the unit vector along L in the
direction of increasing s. Then from Calculus, the directional derivative of u along L is given by
du
= (∇u) · U,
ds
where ∇u = ux i + uy j is the gradient of u. Since ux = 0 = uy , the gradient vanishes and so the directional
derivative becomes zero along L. It means u is constant along L. By generalizing this argument, we deduce
that u(x, y) = a is constant in D. By the similar argument, constant v(x, y) = b can be obtained. Therefore,
we find that f (z) = u(x, y) + iv(x, y) = a + ib at each point in D.
Before we end this section, we summarize the relation between differentiability and analyticity.
1. If f is analytic at z0 , then f is differentiable at z0 .
2. On the other hand, even if f is differentiable at z0 , f may not be analytic at z0 . (ex: f (z) = ∥z∥2 at
z = 0.) However, if f is differentiable in some neighborhood of z0 , then by definition, f is analytic at
z0 .
2
If f is analytic at a point on the boundary, then it means f is analytic in an open set containing the boundary point.
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§2.14 Examples.
We study various examples.
Example 2.14.1. Since f (z) = z 3 + 4 and g(z) = (z 2 − 3)(z 2 + 1) are √
entire (i.e., analytic everywhere),
so the rational function h = f /g is analytic everywhere except at z = ± 3 and z = ±i. Moreover, those
points are singularities of the rational function h. The derivative of h is obtained by the Quotient Rule.
Since the analyticity is deduced from the differentiability, we use the Cauchy–Riemann equations to
check the differentiability and thus eventually the analyticity.
Example 2.14.2. Trigonometric functions (sin z and cos z) are entire. We prove this for f (z) = sin z.
Proof. Let us take the followings formally (will be discussed later):
f (z) = sin (x + iy) = sin x cos (iy) + cos x sin (iy) = sin x cosh y + i cos x sinh y,
where sinh y and cosh y are hyperbolic trigonometric functions with properties
sinh z = −i sin (iz) ,
cosh z = cos (iz) ,
d
sinh z = cosh z,
dz
d
cosh z = sinh z.
dz
Hence, from f (z) = u(x, y) + iv(x, y), we have
u(x, y) = sin x cosh y,
v(x, y) = cos x sinh y,
which are differentiable for any x and y.
Now we check the Cauchy–Riemann equations:
ux = cos x cosh y = vy ,
uy = sin x sinh y = −vx ,
at any point. Since the component functions of f satisfy the equations at any point, thus f is differentiable
everywhere, i.e., analytic everywhere, i.e., entire.
Example 2.14.3. Suppose f (z) = u(x, y) + iv(x, y) and its conjugate f (z) = u(x, y) − iv(x, y) are both analytic in a domain D. Then f must be constant in D.
Proof. We write f (z) as
f (z) = U (x, y) + iV (x, y),
where
U (x, y) = u(x, y)
and V (x, y) = −v(x, y).
(2.14.1)
Because of the analyticity of f , the Cauchy–Riemann equations
ux = vy ,
uy = −vx
(2.14.2)
hold in D. Also the analyticity of f (z) tells us
Ux = Vy ,
Uy = −Vx .
In view of relations (2.14.1), these last two equations can be written
ux = −vy ,
u y = vx .
(2.14.3)
By adding corresponding sides of the first of equations (2.14.2) and (2.14.3), we find that ux = 0 in D. Similarly, subtraction involving corresponding sides of the second of equations (2.14.2) and (2.14.3) reveals that
vx = 0. Hence,
f ′ (z) = ux + ivx = 0 + i0 = 0,
and by the Theorem in the previous section, f is constant throughout D.
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Example 2.14.4. Let f be analytic in a domain D and let its modulus ∥f ∥ be constant in D. Then f must be
constant in D.
Proof. By the hypothesis, ∥f (z)∥ = c, for all z in D, where c is a real constant. If c = 0, then f = 0 at all
points in D, which implies constant f = 0 in D. Now let c ̸= 0. Taking squared the equation gives
∥f ∥2 = c2 ,
f f¯ = c2 .
Since c ̸= 0, so neither f nor f¯ should be zero, which allows us to have
c2
f¯ =
f
for all points in D. Since f and c2 are analytic in D, the quotient f¯ is also analytic there. The Example 2.14.3
above says that if both g and ḡ are analytic in a domain, then g is constant there. By this result, we conclude
that f is also constant in D.
Combining Examples 2.14.3 and 2.14.4, we can say for an analytic function f in a domain D,
1. If f¯ is analytic, then f is constant.
2. If ∥f ∥ is constant, then f is constant.
Exercise 2.14.5. Find each region on which f (z) = 2xy + i(x2 − y 2 ) is differentiable and analytic, respectively.
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§2.15 Harmonic Functions.
Definition 2.15.1. A real–valued function H(x, y) of real variables is called to be harmonic in a domain D
if
1. first and second derivatives, Hx , Hy , Hxx , Hyy , Hxy and Hyx , are continuous
2. Hxx and Hyy satisfies
△H ≡ Hxx + Hyy = 0
for all (x, y) in D
which is called the Laplace equation.
Example 2.15.2. The function T (x, y) = e−y sin x is harmonic in any domain of the xy–plane, particularly,
in the semi–infinite vertical strip 0 < x < π and y > 0. Laplace equation is satisfied and the followings hold
T (x, 0) = sin x,
lim T (x, y) = 0,
y→∞
T (0, y) = 0 = T (π, y).
Theorem 2.15.3. If f (z) = u(x, y) + iv(x, y) is analytic in a domain D, then its component functions u and
v are harmonic in D.
Proof. We borrow a result from Chapter 4 Integrals:
If a function w = f (z) = u(x, y) + iv(x, y) of a complex variable is analytic at z0 = x0 + iy0 ,
then both u and v have continuous partial derivatives of all orders at (x0 , y0 ).
Assume f is analytic in D. Then its component functions u and v should satisfy the Cauchy–Riemann equations,
u x = vy ,
uy = −vx
at any point in D.
By the Theorem above from Chapter 4 Integrals, we have at any point in D,
uxx = (ux )x = (vy )x = vyx ,
uxy = (ux )y = (vy )y = vyy ,
uyx = (uy )x = (−vx )x = −vxx ,
uyy = (uy )y = (−vx )y = −vxy .
Again by the Theorem above, ux , uy , vx and vy are continuous. From Advanced Calculus,
uxy = uyx ,
vxy = vyx .
Hence, we deduce
uxx + uyy = vyx − vxy = vxy − vxy = 0,
vxx + vyy = −uyx + uxy = −uxy + uxy = 0.
That is, u and v are harmonic in D.
Be careful! The converse of the Theorem is not true.
Example 2.15.4. The functions p(x, y) = 2xy and q(x, y) = x2 − y 2 are harmonic through the plane, because pxx + pyy = 0 and qxx + qyy = 0. However, the function
(
)
f (z) = iz 2 = 2xy + i x2 − y 2 = p(x, y) + iq(x, y)
is not analytic anywhere, because its component functions do not satisfy the Cauchy–Riemann equations in
an open set:
px = 2y = −2y = qy ,
py = 2x = −2x = −qx ,
only at z = (x, y) = 0
That is, f (z) is differentiable only at z = 0 and so it is analytic nowhere.
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Example 2.15.5. Consider f (z) = −ieiz = e−y sin x − ie−y cos x = u(x, y) + iv(x, y). We observe f (z) is
entire, i.e., analytic everywhere, by checking the Cauchy–Riemann equations with u and v. The Theorem above implies that u(x, y) = e−y sin x and v(x, y) = −e−y cos x are harmonic everywhere. We already
checked in Example 2.15.2 that u(x, y) is harmonic everywhere and one can easily check that v is also harmonic everywhere.
Example 2.15.6. Consider f (z) =
i
where z ̸= 0.
z2
i
iz 2
2xy + i (x2 − y 2 )
=
=
= u(x, y) + iv(x, y),
z2
∥z∥4
(x2 + y 2 )2
2xy
x2 − y 2
u(x, y) =
,
v(x, y) =
.
(x2 + y 2 )2
(x2 + y 2 )2
f (z) =
Since u and v satisfy the Cauchy–Riemann equations at any point except at (x, y) = (0, 0), so f (z) is analytic
everywhere except at z = 0 and by the Theorem, u and v are harmonic everywhere except at (x, y) = (0, 0).
Definition 2.15.7. Suppose two real–valued functions p(x, y) and q(x, y) of real variables are harmonic in
a domain D and their first partial derivatives satisfy the Cauchy–Riemann equations through D, i.e.,
px = qy ,
py = −qx .
Then q is said to be a harmonic conjugate of p.
Let us remark three points.
1. (Existence) For a harmonic function p in a domain of certain type, does its harmonic conjugate exist?
Yes, it does. Confer Section 104 Harmonic Conjugates in Chapter 9 Conformal Mapping.
2. (Uniqueness) For a given function, its harmonic conjugate is unique except for an additive constant.
3. (Commutativity) If q is a harmonic conjugate of p, then in general, p may not be a harmonic conjugate
of q. See the Example 2.15.9 following the Theorem below.
Theorem 2.15.8. A function f (z) = u(x, y) + iv(x, y) is analytic in a domain D if and only if v is a harmonic
conjugate of u.
Proof. Suppose v is a harmonic conjugate of u in D. Then first derivatives of u and v satisfy the CauchyRiemann equations in D. By the Theorem in Section 22, f = u + iv is analytic in D.
Suppose f = u + iv is analytic in D. Then the Theorem 2.15.3 implies u and v are harmonic in D. The
Theorem in Section 21 gives that first derivatives of u and v satisfy the Cauchy-Riemann equations. Hence,
v is the harmonic conjugate of u.
Example 2.15.9. Suppose
u(x, y) = x2 − y 2 ,
v(x, y) = 2xy.
Then, we observe:
1. The function u(x, y) + iv(x, y) = x2 − y 2 + 2xyi = z 2 is entire, i.e., analytic everywhere. So by the
Theorem 2.15.8, v is the harmonic conjugate of u.
2. However, v(x, y) + iu(x, y) = 2xy + i (x2 − y 2 ) = iz 2 is not analytic anywhere by the Example 2.15.4.
By the Theorem 2.15.8, u is not a harmonic conjugate of v.
For a harmonic function u, there exists a unique (up to constant) harmonic conjugate of u. This implies
that every harmonic function is the real part of an analytic function.
As the last subject in this section, for a given harmonic function, how can we find its harmonic conjugate and an analytic function of which real part is the given harmonic function?
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Example 2.15.10. The function u(x, y) = y 3 − 3x2 y is harmonic everywhere, because
△u = uxx + uyy = −6 + 6 = 0
at any point (x, y).
Find the harmonic conjugate v of u and the analytic function f such that Re (f ) = u.
Answer. First derivatives of u and its harmonic conjugate v should satisfy the Cauchy–Riemann equations:
u x = vy ,
uy = −vx .
(2.15.1)
From the first equation,
ˆ
)
∂ ( 3
vy =
y − 3x2 y = −6xy,
∂x
v=
(−6xy) dy = −3xy 2 + ϕ(x),
where ϕ(x) is an arbitrary function of x. Using the second equation in (2.15.1), we have
)
)
∂ ( 3
∂ (
y − 3x2 y = uy = −vx = −
−3xy 2 + ϕ(x) = 3y 2 − ϕ′ (x)
∂y
∂x
′
2
3
ϕ (x) = 3x ,
ϕ(x) = x + C,
3y 2 − 3x2 =
where C is a real constant of integration. Therefore, we deduce
v(x, y) = −3xy 2 + x3 + C.
The corresponding analytic function is
(
)
f (z) = u(x, y) + iv(x, y) = y 3 − 3x2 y + i −3xy 2 + x3 + C .
When we express it in terms of z, the function becomes
(
)
(
)
f (z) = y 3 − 3x2 y + i −3xy 2 + x3 + C = i z 3 + C .
Exercises: Solve all the problems on page 81–82 in the book. The followings are challengeable.
• Consider the function u(x, y) = x3 − 3xy 2 − 5y.
1. Verify that u is harmonic in the entire complex plane.
2. Find the harmonic conjugate function v of u.
3. Find the analytic function f having u as its real part.
2 −y 2
• Consider the function u(x, y) = ex
cos (2xy).
1. Verify that u is harmonic in an appropriate domain.
2. Find the harmonic conjugate function v of u.
3. Find the analytic function f having u as its real part and satisfying f (0) = 1.
• Consider the function v(x, y) =
x2
x
.
+ y2
1. Verify that v is harmonic in an appropriate domain.
2. Find the analytic function f having v as its imaginary part.
3. Express f in terms of z.
• Consider the polynomial of form: u(x, y) = ax3 + bx2 y + cxy 2 + dy 3 , where a, b, c and d are constants.
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1. Find the most general harmonic polynomial of the given form.
2. Find the harmonic conjugate function v of u.
3. Find the analytic function f having u as its real part.
• Suppose f (z) = u(r, θ) + iv(r, θ) is analytic. Use the Cauchy–Riemann equations in polar coordinates
form to show that u satisfies the Laplace equation in polar coordinates:
r2 urr + rur + uθθ = 0.
• Prove the proposition:
If u is harmonic and v is its harmonic conjugate, then ϕ = uv is also harmonic.
Hints. Suppose f = u + iv is analytic. Then f 2 is also analytic. Examine f 2 .
§2.16 Uniquely Determined Analytic Functions.
§2.17 Reflection Principle.
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Chapter 3
ELEMENTARY FUNCTIONS
§3.1 The Exponential Function.
Definition 3.1.1. The function ez defined by
exp z = ez = ex eiy ,
or
ez = ex cos y + iex sin y
is called the (complex) exponential function.
When y = 0, i.e., z = x + i0, the definition gives ez = ex . So for z = 1/n + i0 = 1/n with n ∈ Z+ ,
we have e1/n = ex , which is one real number with x = 1/n. That is, the positive nth root e1/n of the Euler
number e is defined as one real number, i.e., e1/n = ex , where x = 1/n with n ∈ Z+ .
Be careful: according to Section 9 Roots of Complex Numbers, the positive nth root z 1/n of z is the set
of n different complex numbers. However, the definition in Section 9 does not hold for e1/n . That is, e1/n is
exceptionally not the set of n different complex numbers.
Now let us see some properties.
Theorem 3.1.2 (Algebraic Properties of ez ). If z1 and z2 are complex numbers, then
1. e0 = 1
2. ez =
̸ 0 for any z
z1 z2
3. e e = ez1 +z2
ez1
4. z2 = ez1 −z2
e
5. (ez1 )n = enz1 , n ∈ Z
6. ez = ez̄
Proof. 1. e0 = e0+i0 = e0 cos 0 + ie0 sin 0 = 1 + i0 = 1.
2. ez = ex cos y + iex sin y implies |ez | = ex ̸= 0. Hence, ez ̸= 0 for any z. (Recall: |z| = 0 if and only
if z = 0.)
3. With z1 = x1 + iy1 and z2 = x2 + iy2 , we have
ez1 ez2 = ex1 +iy1 ex2 +iy2 = ex1 eiy1 ex2 eiy2 = ex1 +x2 ei(y1 +y2 ) = ex1 +x2 +i(y1 +y2 ) = ez1 +z2 .
Another way is as follows.
ez1 ez2 = (ex1 cos x1 + iex1 sin x1 ) (ex2 cos x2 + iex2 sin x2 )
= ex1 ex2 cos x1 cos x2 − ex1 ex2 sin x1 sin x2 + i (ex1 ex2 cos x1 sin x2 + ex1 ex2 sin x1 cos x2 )
= ex1 +x2 (cos x1 cos x2 − sin x1 sin x2 ) + iex1 +x2 (cos x1 sin x2 + sin x1 cos x2 )
= ex1 +x2 cos (x1 + x2 ) + iex1 +x2 sin (x1 + x2 ) = ez1 +z2 .
4. Using the result 3 above, we have ez1 −z2 ez2 = ez1 −z2 +z2 = ez1 . Dividing by ez2 , we get
5. De Moivre’s Formula implies, for n ∈ Z,
e z1
= ez1 −z2 .
e z2
(ez1 )n = (ex cos y + iex sin y)n = [ex (cos y + i sin y)]n = (ex )n (cos y + i sin y)n
= enx (cos (ny) + i sin (ny)) = enx cos (ny) + ienx sin (ny) = enz1 .
6.
ez = ex cos y + iex sin y = ex cos y − iex sin y = ex cos (−y) + iex sin (−y) = ex e−iy = ex−iy = ez̄ .
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Theorem 3.1.3 (Analyticity of ez ). The exponential function ez is entire and its derivative is given by
d z
e = ez .
dz
It can be proved by using the Cauchy–Riemann equations.
(
)
2
Exercise 3.1.4. Find the derivative of each of the following functions: (1) iz 4 z 2 − ez and (2) ez −(1+i)z+3 .
Analogous to real periodic functions, we say that a complex function f is periodic with period T if for
all complex z,
f (z + T ) = f (T ).
Theorem 3.1.5 (Properties Only for Complex Exponential Function). The following properties does not hold
for a real exponential function.
1. ez may be a negative real number.
2. ez is periodic with pure imaginary period 2πi.
3. For a given w0 ̸= 0, there may many values of z such that ez = w0 . It implies that the inverse of the
exponential function can have many values.
Proof. 1. When y = nπ, n ∈ Z, we have
z
x
x
x
{
n
n x
e = e cos y + ie sin y = e (−1) + i0 = (−1) e =
ex
−ex
for n = 2m
for n = 2m + 1,
m ∈ Z.
That is, e(2m+1)πi = −1, and e2mπi = 1, m ∈ Z
2. From 1 above, we have e2mπi = 1, m ∈ Z. It implies
ez+2πi = ez e2πi = ez 1 = ez .
Hence, ez is periodic with the pure imaginary period 2πi.
3. For a given w0 = ∥w0 ∥ ei Arg(w0 ) ̸= 0, the equation
ex eiy = ez = w0 = ∥w0 ∥ ei Arg(w0 )
implies
w0 = e x ,
Arg (w0 ) = y + 2nπi,
i.e.,
x = ln w0 ,
y = Arg (w0 ) + 2nπ,
where n ∈ Z. That is, z = x + iy satisfying x = ln w0 and y = Arg (w0 ) + 2nπ is the solution of the equation
ez = w0 . Since y have infinitely many values with n ∈ Z, there are many solutions of the equation.
From the result 3 in the Theorem above, we may expect the inverse (called a logarithmic function) of
the exponential function is a multiple–valued function, which will be studied in next section.
Example 3.1.6. Find z such that ez = 1 + i.
√
Answer. Since 1 + i has the modulus ∥1 + i∥ = 2 and the principal argument Arg (1 + i) = π/4, so we
have
√
ex eiy = ez = 1 + i = 2 eπi/4 .
It implies
√
e = 2,
x
π
y = + 2nπ,
4
√
ln 2
x = ln 2 =
,
2
i.e.,
(
y=
)
1
+ 2n π,
4
where n ∈ Z. Thus, we conclude the solution of ez = 1 + i is obtained by
(
)
1
ln 2
+i
+ 2n π,
z = x + iy =
2
4
which is not unique, because of n ∈ Z.
z̄
Exercise 3.1.7. Determine the complex function e is analytic.
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§3.2 The Logarithmic Function.
The main goal of this section is to study the inverse of the exponential function, i.e., for a given z ̸= 0,
the solution w of the equation ew = z.
Let us write z = reiΘ , −π < Θ ≤ π and w = u + iv. Then the equation ew = z becomes
eu eiv = eu+iv = z = reiΘ ,
eu = r,
i.e.,
v = Θ + 2nπ,
where n ∈ Z. It implies
u = ln r,
v = Θ + 2nπ,
i.e.,
w = u + iv = ln r + i (Θ + 2nπ) ,
where ln r = loge r. Since ∥z∥ = r and Arg (z) = Θ, thus, we can rewrite the solution w of the equation
ew = z as follows
w = ln ∥z∥ + i (Arg (z) + 2nπ) ,
i.e.,
eln∥z∥+i(Arg(z)+2nπ) = z,
n ∈ Z.
Because of n ∈ Z, there are infinitely many such w’s. If we write log z := ln ∥z∥ + i (Arg (z) + 2nπ), we
have
elog z = eln∥z∥+i(Arg(z)+2nπ) = z,
i.e.,
elog z = z.
For this reason, we have the following definition.
Definition 3.2.1 (Complex Logarithm). The multiple–valued function log z, z ̸= 0, defined by
log z = ln ∥z∥ + i Arg (z) = ln ∥z∥ + i (Arg (z) + 2nπ) ,
n ∈ Z,
is called the complex logarithm or (complex) logarithmic function.
√
Example 3.2.2. Find log z for z = −1 − i 3 .
Answer. We compute ∥z∥ = 2 and Arg (z) = −2π/3. Putting them into the definition, we get
)
(
)
(
2
2π
+ 2nπ = ln 2 + i − + 2n π,
n ∈ Z.
log z = ln 2 + i −
3
3
Exercise 3.2.3. Find all complex solutions to each of the following equations: (1) ew = i, (2) ew = 1 + i and
(3) ew = −2.
Be careful: log (ez ) ̸= z.
Proof. We compute ∥ez = ex+iy ∥ = ex and Arg (ez = ex+iy ) = y + 2nπ. Putting them into the definition,
log (ez ) = ln ∥ez ∥ + i Arg (ez ) = ln ex + i (y + 2nπ) = x + iy + 2nπi = z + 2nπi,
n ∈ Z.
The logarithmic function is multiple–valued. We restrict the argument in the logarithmic function
and deduce a single–valued function.
Definition 3.2.4 (Principal Value of Complex Logarithm). The complex function ln z, z ̸= 0, defined by
ln z = ln ∥z∥ + i Arg (z) ,
−π < Arg (z) ≤ π,
is called the principal value of the complex logarithm log z.
We observe
(1) ln z is a single–valued function.
(2) log z = ln ∥z∥ + i (Arg (z) + 2nπ) = ln z + 2nπi, n ∈ Z.
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Example 3.2.5. A simple computation shows
ln 1 = ln ∥1∥ + i Arg (1) = 0 + i0 = 0.
log 1 = ln 1 + 2nπi,
Thus log 1 = 2nπi, n ∈ Z.
Example 3.2.6. A simple computation shows
log (−1) = ln (−1) + 2nπi,
ln (−1) = ln ∥−1∥ + i Arg (−1) = 0 + iπ = π.
Thus log (−1) = πi + 2nπi = (2n + 1) πi, n ∈ Z.
From those two examples above, we see log (1) ̸= log (−1).
Exercise 3.2.7. Compute the principal value ln z of the complex logarithm for (1) z = i, (2) z = 1 + i and
(3) z = −2.
Since ln z is a single–valued function coming from log z which are the solution of ew = z, so we may
expect ln z to be the inverse of the exponential function ez . In fact, we have the following theorem.
Theorem 3.2.8 (ln z as an Inverse Function of ez ). If the complex exponential f (z) = ez is defined on the
region −∞ < x < ∞, −π < y ≤ π, then f is one–to–one and the inverse function of f is the principal
value of the complex logarithm f −1 (z) = ln z.
By the Theorem, we can have
eln z = z = ln (ez ) ,
which one can check easily.
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§3.3 Branches and Derivatives of Logarithms.
Let us convert log (1 + i) in the form of x + iy.
log (1 + i) = ln ∥1 + i∥ + i Arg (1 + i)
= ln ∥1 + i∥ + i (Arg (1 + i) + 2nπ)
(π
)
√
= ln 2 + i
+ 2nπ
( 4
)
1
ln 2
=
+i
+ 2n π,
n∈Z
2
4
(
)
1
ln 2
=
+i
+0 π
2
4
(
)
ln 2
1
=
+i
±2 π
2
4
(
)
ln 2
1
=
+i
±4 π
2
4
(
)
ln 2
1
+i
±6 π
=
2
4
= ...
If we fix an arbitrary real number α, say α = 7, then can we find only one value of log (1 + i) such that
α < Arg (1 + i) < α + 2π?
(
)
1
The answer is YES! In fact, since α = 7 <
+ 2 π ≈ 7.06858 < 7 + 2π = α + 2π, so the complex
4
number
(
)
ln 2
1
log (1 + i) =
+i
+2 π
(3.3.1)
2
4
satisfies the condition α < Arg (1 + i) < α + 2π.
Now we generalize the argument to any complex number z ̸= 0 instead of 1 + i. We start for z ̸= 0,
log z = ln ∥z∥ + i Arg (z) = ln ∥z∥ + i (Arg (z) + 2nπ) = ln r + i (Arg (z) + 2nπ)
(3.3.2)
where ∥z∥ = r and −π < Arg (z) ≤ π and n ∈ Z.
We recall that Arg (z) is one of the values of Arg (z) such that it belongs to (−π, π]. Just like Arg (z),
we can find one of the values of Arg (z) such that it belong to (α, α + 2π), where α is arbitrarily fixed real
number. Let such a value be θ, i.e., θ is one of the values of Arg (z) and α < θ < α + 2π. Then from
equation (3.3.2), we have
log z = ln r + i (θ + 2mπ) ,
where α < θ < α + 2π and m ∈ Z. Moreover, just like (3.3.1) in the example above, we have
log z = ln r + iθ = u(r, θ) + iv(r, θ),
(r > 0, α < θ < α + 2π) ,
(3.3.3)
where u(r, θ) = ln r and v(r, θ) = θ are single–valued continuous functions in the domain,
}
{
( )
−1 y
< α + 2π .
D = z = x + iy ̸= 0 : α < tan
x
(The domain D is the finite complex plane except the ray θ = α and the origin.)
Remark 3.3.1. The function in (3.3.3) is single–valued and analytic in the domain D and its derivative is
1
d
log z = .
dz
z
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Proof. It is straightforward to see that it is single–valued. To deduce the analyticity, we use the CauchyRiemann equations in the domain D:
1
rur = r = 1 = vθ ,
r
uθ = 0 = −rvr .
Thus, it is analytic in D. Again, by the Theorem,
d
log z = e−iθ (ur + ivr ) = e−iθ
dz
(
1
+0
r
)
=
1
1
= ,
iθ
re
z
where z ∈ D, i.e., ∥z∥ > 0 and α < Arg (z) < α + 2π.
If we choose α = −π, then the function (3.3.3) becomes
log z = ln r + iθ,
(r > 0, −π < θ < π) ,
i.e., we do have the principal value of the complex logarithmic function,
ln z = ln ∥z∥ + i Arg (z) ,
(∥z∥ > 0, −π < Arg (z) < π)
(3.3.4)
d
1
ln z = . (Be careful: There IS a slight difference in the domains of the
dz
z
function (3.3.4) and the principal value of the complex logarithmic function.)
and the derivative of ln z is
Definition 3.3.2. A function F (z) is said to be a branch of a multiple–valued function f (z) in a domain D
if
1. F (z) is single–valued and continuous in D and
2. F (z) has the property that, for each z in D, the value F (z) is one of the values of f (z).
Example 3.3.3. For each fixed α, the single–valued function
log z = ln r + iθ,
(r > 0, α < θ < α + 2π)
(3.3.5)
is a branch of the multiple–valued function log z = ln r + i Arg (z).
Example 3.3.4. With α = −π, the single–valued function
ln z = ln r + iθ,
(r > 0, −π < θ < π)
(3.3.6)
is a branch of the multiple–valued function log z = ln r + i Arg (z). This function is called the principal branch
of the logarithmic function log z. (Again, be careful: Principal value of log z is ln z = ln r + iθ in the domain
r > 0 and −π < θ ≤ π. Principal branch of log z is the same function with the different domain r > 0 and
−π < θ < π. In order for a branch to be analytic in the domain, the domain should be open, not half–open
or half–closed.)
Definition 3.3.5. A branch cut is a portion of a line of curve used to defined a branch F (z) of f (z). A point
that is common to all branch cuts is called the branch point.
We observe that the points on the branch cut for F (z) are singularities or singular points of F (z).
Example 3.3.6. The branches (3.3.5) and (3.3.6) in the Examples above are the rays θ = α and θ = −π,
respectively. The origin is the branch point for branches of the multiple–valued logarithmic function.
Since branches of the logarithmic function have different domains, some identities involving real logarithmic function may not hold for the branches.
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Example 3.3.7.
( π)
( 3)
π
π
ln i = ln (−i) = ln ∥−i∥ + i −
= ln 1 − i = − i,
2
2
2
[
( π )]
[
π ] 3π
3 ln (i) = 3 ln (i) + i
= 3 ln 1 + i =
i.
2
2
2
( )
That is, ln i3 = −πi/2 ̸= 3πi/2 = 3 ln (i).
Appendix
Some textbooks use different notations for the branch for clear and better understanding.
For a fixed real number α, Argα (z) is used to denote the argument between α and α + 2π and the
branch of the logarithmic function log z having the branch cut θ = α is denoted by Lα (z), i.e.,
Lα (z) = ln ∥z∥ + i Argα (z) ,
(∥z∥ > 0, α < Argα (z) < α + 2π) .
For instance, the branch of log (1 + i) having the branch cut θ = 7 is denoted by
(
)
(
)
√
1
ln 2
1
L7 (1 + i) = ln ∥1 + i∥ + i Arg7 (1 + i) = ln 2 + i
+2 π =
+i
+ 2 π,
4
2
4
(
)
1
π
because Arg (1 + i) = and 7 <
+ 2 π ≈ 7.06858 < 7 + 2π.
4
4
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§3.4 Some Identities Involving Logarithms.
In this section, we study only the following identities.
1. log (z1 z2 ) = log z1 + log z2 , where z1 ̸= 0 ̸= z2 .
( )
z1
2. log
= log z1 − log z2 , where z1 ̸= 0 ̸= z2 .
z2
3. z n = en log z , where n is any integer.
1
4. z 1/n = e n log z , where n ̸= 0 is any integer.
Recalling the definition log z = ln ∥z∥ + i Arg (z), we observe that the first and second identities are
based on the identities in Section 8. Arguments of Products and Quotients
( )
z1
Arg (z1 z2 ) = Arg (z1 ) + Arg (z2 ) ,
Arg
= Arg (z1 ) − Arg (z2 ) .
z2
Each identity has three terms. When we know any two terms, the identities give the value of the left one.
Example 3.4.1. For z1 = −1 = z2 , z1 z2 = 1 and
log (z1 z2 ) = log 1 = 2nπi,
log (z1 ) = log (−1) = (2n + 1) πi.
When we choose log (z1 z2 ) = 0 and log (z1 ) = πi, the identity above implies
log (z2 ) = log (z1 z2 ) − log (z1 ) = 0 − πi = −πi.
In fact, by definition,
log (z2 ) = log (−1) = ln ∥−1∥ + i Arg (−1) = ln 1 + i (π + 2nπ) = i (2n + 1) π.
When we choose n = −1, we have log (z2 ) = log (−1) = −πi.
When we replace log z by ln z, in general, the identities may not work.
Example 3.4.2. For z1 = −1 = z2 , z1 z2 = 1 and
ln (z1 z2 ) = ln 1 = 0,
Thus, we get
ln (z2 ) = ln (z1 ) = ln (−1) = πi.
ln (z1 z2 ) = 0 ̸= 2πi = ln (z1 ) + ln (z2 ) .
However, sometimes we can have
ln (z1 z2 ) = ln (z1 ) + ln (z2 ) .
(See Exercise 1 on page 100.)
The third identity is clear to see when n = 1 from the previous section. For the proof, let us use the
exponential form of z, i.e., z = reiθ .
(
)n
z n = reiθ = rn einθ .
(3.4.1)
On the other hand,
)
(
log z = log reiθ = ln reiθ + i Arg reiθ = ln r + iθ,
From (3.4.1) and (3.4.2), we conclude
z n = en log z .
The last identity can be proved similarly.
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en log z = en ln r+inθ = en ln r einθ = rn einθ .
(3.4.2)
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§3.5 Complex Exponents.
From the previous section, we recall that for z ̸= 0,
z n = en log z ,
1
z 1/n = e n log z ,
where n is any integer, n ̸= 0 for the second equality. Now we generalize the integer n to any complex
number.
Definition 3.5.1. Let z ̸= 0 and c be a complex number. Then the function z c is defined by
z c = ec log z .
Example 3.5.2. Express a given complex number i−2i in the form of u(x, y) + iv(x, y).
Answer. We compute log i first.
log i = ln ∥i∥ + i Arg (i) = ln 1 + i
(π
2
(
)
+ 2nπ =
4n + 1
2
)
πi,
where n is an integer, so we get
i−2i = e−2i log i = e−2i(
4n+1
2
) πi = e(4n+1)π ,
which are all real numbers.
Exercise 3.5.3. Express a given complex number (−1)1/π in the form of u(x, y) + iv(x, y).
Answer. (−1)1/π = ei(2n+1) = cos(2n + 1) + i sin(2n + 1).
Property 3.5.4 (Inverse of z c ).
1
1
= c log z = e−c log z = z −c ,
c
z
e
i.e.,
1
= z −c ,
zc
1
= e−z from Section 29. Exponential Function.
ez
When we use the property, for the example above, we have
which is based on the property
1
= i−2i = e(4n+1)π .
i2i
Theorem 3.5.5 (Analyticity & Derivative of z c ). The derivative of z c is given by
d c
z = cz c−1 ,
dz
(|z| > 0, α < Arg(z) < α + 2π)
where α is any real number.
Proof. Let z = reiθ and α be any real number. Then the branch of log z,
log z = ln r + iθ,
(r > 0, α < θ < θ + 2π) ,
is single–valued and analytic in the same domain. So in the same domain, z c = ec log z is also single–valued
and analytic. Using the chain rule, we get
d c log z
c
d c
z =
e
= ec log z .
dz
dz
z
Using the identity z = elog z ,
d c c c log z cec log z
z = e
= log z = ce(c−1) log z = cz c−1 .
dz
z
e
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Now we replace the multiple–valued function log z in z c = ec log z by the principal value of log z, i.e., ln z.
Then, it is called the principal value of z c .
Definition 3.5.6. The Principal Value of z c is defined by
P. V. z c = ec ln z ,
where ln z is the principal value of log z. Similarly, the Principal Branch of z c is defined by ec ln z on the domain
∥z∥ > 0 and −π < Arg (z) < π, where ln z is the principal branch of log z defined on the same domain.
Example 3.5.7. Find the principal value of (−i)i .
Answer. We compute the principal value ln (−i) first.
ln (−i) = ln (−i) + i Arg (−i) = ln 1 −
Hence we deduce
π
π
= − i.
2
2
π
P. V. (−i)i = ei ln(−i) = e 2 ,
which is a real number.
Exercise 3.5.8. Find the principal value of (1 − i)4i .
Answer. P. V. (1 − i)4i = eπ+i2 ln 2 = eπ (cos (2 ln 2) + i sin (2 ln 2)).
Example 3.5.9. Find the principal branch of z 2/3 and the principal value of z 2/3 .
Answer. We compute the principal branch ln z first.
(r = ∥z∥ > 0, −π < Θ = Arg (z) < π)
ln z = ln r + iΘ
Hence we can write the principal branch of z 2/3 as follows
e
2
3
ln z
=e
2
3
ln r+ 23 iΘ
=e
2
3
ln r
e
where r = ∥z∥ > 0 and −π < Θ = Arg (z) < π.
The principal value of z 2/3 follows directly
P. V. z
2/3
=r
2/3
2
3
iΘ
=r
2/3
(
)
2Θ
2Θ
cos
+ i sin
,
3
3
)
(
2Θ
2Θ
+ i sin
,
cos
3
3
of which domain is r = ∥z∥ > 0 and −π < Θ = Arg (z) ≤ π.
In the example above, the principal value of z 2/3 is analytic in the domain of the principal branch, i.e.,
r = ∥z∥ > 0 and −π < Θ = Arg (z) < π.
In general, Laws of Exponents in Calculus do not hold in complex analysis.
Example 3.5.10. Let z1 = 1 + i, z2 = 1 − i and z3 = −1 − i.
Using the principal values,
(z1 z2 )i = 2i = ei ln 2 = ei(ln 2+i0) = ei ln 2 ,
√
z i = ei ln(1+i) = ei(ln 2 +πi/4) = e−π/4 ei(ln 2)/2 ,
1
z2i = ei ln(1−i) = ei(ln
Hence,
(z1 z2 )i = ei ln 2 = z1i z2i ,
√
2 −πi/4)
i.e.,
= eπ/4 ei(ln 2)/2 .
(z1 z2 )i = z1i z2i .
However, again using the principal values, we have
(z2 z3 )i = z2i z3i e−2π ,
i.e.,
(One can check by themselves.)
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(z2 z3 )i ̸= z2i z3i .
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Definition 3.5.11. Let c ̸= 0 be a complex number. Then we define the exponential function with base c by
cz = ez log c .
Since log c is multiple–valued, so cz is also multiple–valued. When c = e, the definition above gives
a multiple–valued function ez , which is not coherent to the fact that a complex exponential function ez is
single–valued. For this reason, when c = e, log c = log e is interpreted as the principal value of e so that
log c = log e = 1.
Property 3.5.12 (Analyticity & Derivative of cz ). When a value of log c is specified, cz is entire. The derivative
of cz is obtained by
d z
d z log c
c =
e
= ez log c log c = cz log c,
dz
dz
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i.e.,
d z
c = cz log c.
dz
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§3.6 Trigonometric Functions.
We recall
eix = cos x + i sin x,
e−ix = cos x − i sin x.
From these two equations, we deduce
sin x =
eix − e−ix
,
2i
cos x =
eix + e−ix
.
2
cos z =
eiz + e−iz
.
2
Definition 3.6.1. For a complex number z, we define
sin z =
eiz − e−iz
,
2i
We have the following properties.
1.
2.
3.
4.
5.
6.
7.
sin z and cos z are entire.
(sin z)′ = cos z, (cos z)′ = − sin z
sin (−z) = − sin z, cos (−z) = cos z
sin2 z + cos2 z = 1
sin (z1 + z2 ) = sin z1 cos z2 + cos z1 sin z2
cos (z1 + z2 ) = cos z1 cos z2 − sin z1 sin z2
sin (2z) = 2 sin z cos z, cos (2z) = cos2 z − sin2 z
(
(
π)
π)
= cos z, sin z −
= − cos z
8. sin z +
2
2
9. sin (z + 2π) = sin z, sin (z + π) = − sin z
10. cos (z + 2π) = cos z, cos (z + π) = − cos z
Definition 3.6.2. Hyperbolic functions are defined by
sinh y =
ey − e−y
,
2
cosh y =
ey + e−y
,
2
where y is any real number.
Using the real hyperbolic functions, we have the following properties.
1.
2.
3.
4.
5.
6.
7.
sinh(iy) = i sinh y, cosh(iy) = cosh y
sinh2 y − cosh2 y = −1
sin z = sin x cosh y + i cos x sinh y, cos z = cos x cosh y − i sin x sinh y
∥sin z∥2 = sin2 x + sinh2 y, ∥cos z∥2 = cos2 x + sinh2 y
Since sinh y is not bounded, from the properties above, sin z and cos z are not bounded, either.
sin z = 0 if and only if z = nπ, where n is an integer.
(
)
(
)
cos z = 0 if and only if z = 12 + n π, where n is an integer, by using the equality cos z = − sin z − π2
and the previous property.
Definition 3.6.3.
tan z =
sin z
,
cos z
cot z =
cos z
,
sin z
sec z =
We have the following properties.
1. 1 + tan2 z = sec2 z, 1 + cot2 z = csc2 z
2. (tan z)′ = sec2 z, (cot z)′ = − csc2 z
3. (sec z)′ = sec z tan z, (csc z)′ = − csc z cot z
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1
,
cos z
csc z =
1
.
sin z
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§3.7 Hyperbolic Functions.
Definition 3.7.1. The hyperbolic sine and cosine functions of a complex variable are defined by
sinh z =
ez − e−z
,
2
cosh z =
ez + e−z
.
2
Hyperbolic functions have the following properties.
Since ez and e−z are entire, sinh z and cosh z are entire.
(sinh z)′ = cosh z, (cosh z)′ = sinh z
−i sinh (iz) = sin z, cosh (iz) = cos z
−i sin (iz) = sinh z, cos (iz) = cosh z
sinh (−z) = − sinh z, cosh (−z) = cosh z
sinh2 z − cosh2 z = −1
sinh (z1 + z2 ) = sinh z1 cosh z2 + cosh z1 sinh z2
cosh (z1 + z2 ) = cosh z1 cosh z2 + sinh z1 sinh z2
sinh z = sinh x cos y + i cosh x sin y
cosh z = cosh x cos y + i sinh x sin y
∥sinh z∥2 = sinh2 x + sin2 y
∥cosh z∥2 = sinh2 x + cos2 y
sinh z and cosh z are periodic with period 2π.
sinh z = 0 if and only if z = nπi, where n is an integer
(
)
1
15. cosh z = 0 if and only if z =
+ n πi, where n is an integer
2
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
Definition 3.7.2.
tanh z =
sinh z
.
cosh z
It is not difficult to seee the following properties.
(tanh z)′ = sech2 z,
(coth z)′ = − csch2 z,
(sech z)′ = − sech z tanh z,
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(csch z)′ = − csch z coth z.
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§3.8 Inverse Trigonometric and Hyperbolic Functions.
Definition 3.8.1. The inverse sine and cosine functions, sin−1 z and cos−1 z, are defined by
[
[
(
)1/2 ]
(
)1/2 ]
sin−1 z = −i log iz + 1 − z 2
,
cosh−1 z = −i log z + i 1 − z 2
.
Let us derive the definition. Let w = sin−1 z. Then, z = sin w, i.e.,
z=
eiw − e−iw
,
2i
(
eiw − e−iw − 2iz = 0,
eiw
)2
− 2izeiw − 1 = 0.
Solving the quadratic equation for eiw , we get
(
)1/2
eiw = iz + 1 − z 2
.
(See Exercise 8 (a) in Section 10. Examples of Roots of Complex Numbers.) Taking logarithms of each side,
we have
[
[
(
)1/2 ]
(
)1/2 ]
iw = log iz + 1 − z 2
,
sin−1 z = −i log iz + 1 − z 2
,
which is clearly a multiple–valued function. For the inverse cosine function, we can argue similarly.
Example 3.8.2. Compute sin−1 (−i).
Answer. By the definition,
−1
sin
]
[
[
√ ]
[
]
(
2 )1/2
1/2
= −i log 1 + 2
= −i log 1 ± 2 .
(−i) = −i log i (−i) + 1 − (−i)
We compute the logarithms.
[
(
(
√ ]
√ √ )
√ )
log 1 + 2 = ln 1 + 2 + i Arg 1 + 2 = ln 1 + 2 + 2nπi,
)
[
(
(√
√ ]
√ √ )
2 − 1 + (2n + 1) πi.
log 1 − 2 = ln 1 − 2 + i Arg 1 − 2 = ln
We observe
ln
It implies
(√
)
2 − 1 = ln
(
√ )
1
√ = − ln 1 + 2 .
1+ 2
[
(
√ ]
√ )
log 1 ± 2 = (−1)n ln 1 + 2 + nπi
and thus
[
[
(
]
(
√ ]
√ )
√ )
sin−1 (−i) = −i log 1 ± 2 = −i (−1)n ln 1 + 2 + nπi = nπ + i (−1)n+1 ln 1 + 2 . Definition 3.8.3. The inverse tangent function, tan−1 z, is defined by
tan−1 z =
i+z
i
log
.
2
i−z
From w = tan−1 z, we have z = tan w. Since
sin w =
eiw − e−iw
,
2i
cos w =
eiw + e−iw
,
2
we get
z = tan w =
1 eiw − e−iw
1 e2iw − 1
sin w
=
=
,
cos w
i eiw + e−iw
i e2iw + 1
Thus,
−2iw = log
i+z
,
i−z
tan−1 z = w = −
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e2iw =
i−z
,
i+z
e−2iw =
1
i+z
i
i+z
log
= log
.
2i
i−z
2
i−z
i+z
.
i−z
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Theorem 3.8.4 (Derivatives). When specific branches of the square roots and logarithmic functions are
used, all three inverse functions become single–valued and analytic having the following derivatives.
(
)′
1. sin−1 z =
2.
3.
4.
5.
6.
1
(1 − z 2 )1/2
( −1 )′
1
cos z = −
(1 − z 2 )1/2
( −1 )′
1
tan z =
1 + z2
(
)′
[
(
)1/2 ]
sinh−1 z = log z + z 2 + 1
(
)′
[
(
)1/2 ]
cosh−1 z = log z + z 2 − 1
(
)′ 1
1+z
tanh−1 z = log
2
1−z
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Chapter 4
INTEGRALS
§4.1 Derivatives of Functions w(t).
Definition 4.1.1. For A complex–valued function w(t) of a real variable t,
w(t) = u(t) + iv(t),
where u and v are real–valued functions of t, its derivative is defined by
w′ (t) =
dw(t)
= u′ (t) + iv ′ (t),
dt
provided each of the derivatives u′ (t) and v ′ (t) exists at t.
Theorem 4.1.2. Using the definition, we deduce the following straightforward results.
1. For a fixed complex constant z0 = x0 + iy0 ,
d
d
[z0 w(t)] = z0 w(t),
dt
dt
i.e.,
[z0 w(t)]′ = z0 w′ (t).
2. For a fixed complex constant z0 = x0 + iy0 ,
d z0 t
e = z0 ez0 t .
dt
3. (Sum) For w1 (t) = u1 (t) + iv1 (t) and w2 (t) = u2 (t) + iv2 (t),
d
[w1 (t) ± w2 (t)] = w1′ (t) ± w2′ (t).
dt
4. (Product) For w1 (t) = u1 (t) + iv1 (t) and w2 (t) = u2 (t) + iv2 (t),
d
[w1 (t)w2 (t)] = w1′ (t)w2 (t) + w1 (t)w2′ (t).
dt
5. (Quotient) For w1 (t) = u1 (t) + iv1 (t) and w2 (t) = u2 (t) + iv2 (t),
[
]
d w1 (t)
w′ (t)w2 (t) − w1 (t)w2′ (t)
= 1
,
dt w2 (t)
w22 (t)
where w2 (t) ̸= 0.
The proofs of the results above are based on the calculus. It allows us to extend a calculus result to the
complex functions. However, the generalization is not always possible. As a counter–example, we recall the
Mean Value Theorem from calculus stating that for a differentiable real–valued function f (x) of a real variable
x on [a, b], there exists a c ∈ (a, b) such that f (b) − f (a) = f ′ (c)(b − a).
Example 4.1.3. Consider w(t) = eit on [0, 2π], which is differentiable at any t. We observe
w(2π) − w(0) = e2πi − e0 = 1 − 1 = 0.
However, w′ (t) = ieit and ∥w′ (t)∥ = |ieit | = 1, i.e., w′ (t) can never be zero. Thus there is no c ∈ (0, 2π) such
that w(2π)−w(0) = 0 = w′ (c)(2π−0). That is, the Mean Value Theorem does not hold for a complex–valued
function.
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§4.2 Definite Integrals of Functions w(t).
Definition 4.2.1. For A complex–valued function w(t) = u(t) + iv(t) on [a, b], its definite integral is defined
by
ˆ b
ˆ b
ˆ b
w(t) dt =
u(t) dt + i
v(t) dt,
a
a
a
provided the individual integrals exist.
From the definition, since u(t) = Re [w(t)] and v(t) = Im [w(t)], so it is easy to observe
ˆ
ˆ
b
b
Re [w(t)] dt + i
w(t) dt =
a
ˆ
b
a
Im [w(t)] dt.
a
Since w(t) is a complex–valued function, its definite integral should be also a complex–valued function,
which implies
[ˆ b
]
[ˆ b
] ˆ b
ˆ b
ˆ b
Re
w(t) dt + i Im
w(t) dt =
w(t) dt =
Re [w(t)] dt + i
Im [w(t)] dt
a
a
a
a
a
[ˆ b
] ˆ b
[ˆ b
] ˆ b
i.e., Re
w(t) dt =
Re [w(t)] dt, and Im
w(t) dt =
Im [w(t)] dt.
a
a
a
a
´
The result above says that the integral operator “ ” and real operators “Re” and “Im” can be changeable,
i.e., the commutativity holds between them.
ˆ 1
Example 4.2.2. Compute the definite integral
(1 + it)2 dt.
0
Answer. Since (1 + it)2 = 1 − t2 + 2ti, so by the definition, we have
ˆ
ˆ
1
2
(1 + it) dt =
0
1
(
1−t
0
2
)
ˆ
dt + 2i
0
1
[
t3
t dt = t −
3
]1
[ ]1
1
2
+ i t2 0 = 1 − + i = + i.
3
3
0
The improper integral of w(t) over unbounded integrals are defined in a similar way. We recall from
calculus that a real–valued function f (x) of a real variable x on [a, b] is said to be piecewise continuous if the
function is continuous except finitely many points where f is discontinuous and the right–hand limit at a
and left–hand limit at b exist.
Definition 4.2.3. When real–valued functions u(t) and v(t) are piecewise continuous on [a, b], w(t) = u(t)+
iv(t) is said to be piecewise continuous on the interval.
From calculus, we have the following generalized results.
Theorem 4.2.4. Let c be a complex constant and w1 (t) = u1 (t) + iv1 (t) and w2 (t) = u2 (t) + iv2 (t) on
[a, b].
ˆ b
ˆ b
1. (Constant Multiple)
cw(t) dt = c
w(t) dt
a
a
ˆ b
ˆ b
ˆ b
2. (Sum)
[w1 (t) ± w2 (t)] dt =
w1 (t) dt ±
w2 (t) dt
a
a
a
ˆ b
ˆ b
3. (Commutativity with Limit) lim
wn (t) dt =
lim wn (t) dt
n→∞ a
a n→∞
ˆ b
ˆ c
ˆ b
4. (Separation)
w(t) dt =
w(t) dt +
w(t) dt
a
a
c
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ˆ
b
5. (Fundamental Theorem of Calculus)
w(t) dt = [W (t)]ba , where W (t) = U (t) + iV (t) and U ′ (t) =
a
u(t) and V ′ (t) = v(t).
ˆ
π/4
eit dt.
Example 4.2.5. Compute the definite integral
0
Answer. We observe
d it
e = ieit ,
dt
d
dt
(
eit
i
)
= eit .
It implies
ˆ
π/4
(π )
]
] 1 [ (π )
1 [ it ]π/4 1 [ iπ/4
e 0 =
e
− e0 =
cos
+ i sin
−1
i
i
i
4
4
[
(
]
)
(
)
1
1
1
1
1
1
1
1
√ +i√ −1 = √ +
√ −1 = √ −i √ −1 .
=
i
i
2
2
2
2
2
2
eit dt =
0
At the end of previous section, we have shown that the Mean Value Theorem for Derivatives in calculus
does not hold for a complex–valued function w(t). The Mean Value Theorem for Integrals in calculus does not
hold for a complex–valued function w(t), either. As a counter–example, see below. To refresh our memory,
we recall the Mean Value Theorem for Integrals in calculus stating that under the appropriate circumstances,
there exists a c ∈ (a, b) such that
ˆ b
f (x) dx = f (c) (b − a) .
a
Example 4.2.6. Consider w(t) = eit on [0, 2π]. We observe
ˆ 2π
ˆ 2π
1 [ it ]2π
e 0
w(t) dt =
eit dt =
i
0
0
] 1
1 [ 2πi
1
=
e − e0 = [cos (2π) + i sin (2π) − 1] = [1 + i(0) − 1] = 0.
i
i
i
However, |w(t)| = |eit | = 1 means w(t) can never be zero. Thus there is no c ∈ (0, 2π) such that
ˆ 2π
w(t) dt = 0 = w(c) (2π − 0) .
0
§4.3 Contours.
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§4.4 Contour Integrals.
We study integrals of complex–valued function f of the complex variable z. Such integral is defined
along a contour C connecting two points in the plane. Since a contour can be viewed as the one generalized
from a line, so the integral over a contour can be discussed from the viewpoint of the line integral. A contour
integral, i.e., integral of f over a contour C is written by
ˆ
f (z) dz,
C
and if the integral does not depend on the choice of the contour C starting a point z1 and ending a point z2 ,
the contour integral can be specifically written by
ˆ z=z2
f (z) dz.
z=z1
Definition 4.4.1 (Contour Integral or Complex Integral). Suppose that
1. C is a contour represented by z = z(t), a ≤ t ≤ b, from z1 = z(a) to z2 = z(b),
2. f (z(t)) is piecewise continuous on C, i.e., piecewise continuous on a ≤ t ≤ b.
Then the contour integral of f along C is defined by
ˆ
ˆ
t=b
f (z) dz =
C
f (z(t))z ′ (t) dt.
t=a
Note that since C is a contour (i.e., piecewise smooth arc), z ′ (t) is piecewise continuous and so the
definite integral does exit. In some books, the definition above is given as a theorem for evaluating a contour
integral.
Remark 4.4.2 (Definition Using Limit–Sum). We can define the contour integral as done by the Riemann
sum in Calculus.
1. Let f be a function of a complex variable z defined at all points on a smooth curve C that lies in some
region of the complex plane. Let C be defined by the parametrization z(t) = x(t) + iy(t), a ≤ t ≤ b.
2. Let P be a partition of the parameter interval [a, b] into n subintervals [tk−1 , tk ] of length △tk =
tk − tk−1 :
a = t0 < t1 < t2 < · · · < tn−1 < tn = b.
The partition P induces a partition of the curve C into n subarcs whose initial and terminal points are
the pairs of numbers
z0 = x(t0 ) + iy(t0 ),
z1 = x(t1 ) + iy(t1 ),
......
zn−1 = x(tn−1 ) + iy(tn−1 ),
z1 = x(t1 ) + iy(t1 )
z2 = x(t2 ) + iy(t2 )
......
zn = x(tn ) + iy(tn )
Let △zk = zk − zk−1 , k = 1, 2, . . . , n.
3. Let ∥P ∥ be the norm of the partition P of [a, b], that is, the length of the longest subinterval.
4. Choose a point zk∗ = x∗k + iyk∗ on each subarc of C.
5. Form n products f (zk∗ )△zk , k = 1, 2, . . . , n, and then sum these products:
n
∑
f (zk∗ )△zk .
k=1
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Then the contour integral is defined by
ˆ
f (z) dz = lim
∥P ∥→0
C
n
∑
f (zk∗ )△zk .
k=1
End of Remark
We point out that the value of the contour integral is invariant under a change in representation of its
contour when the change is made via a strictly increasing function. Specifically, suppose that
1. a contour C is represented by z = z(t), a ≤ t ≤ b,
2. let Zϕ (τ ) = (z ◦ ϕ) (τ ) = z(ϕ(τ )), where z(t) is given in 1 above and ϕ(τ ) is a function satisfying
ϕ′ (τ ) > 0 on α ≤ τ ≤ β and ϕ(α) = a and ϕ(β) = b,
3. let Zψ (µ) = (z ◦ ψ) (µ) = z(ψ(µ)), where z(t) is given in 1 above and ψ(µ) is a function satisfying
ψ ′ (µ) > 0 on γ ≤ µ ≤ δ and ψ(γ) = a and ψ(δ) = b,
4. f (z(t)) is piecewise continuous on C.
Then we have the following equalities,
ˆ
ˆ
t=b
f (z) dz =
ˆ
′
τ =β
f (z(t))z (t) dt =
C
t=a
τ =α
µ=δ
ˆ
=
f (Zϕ (τ ))Zϕ′ (τ ) dτ
ˆ
τ =β
=
f (Zψ (µ))Zψ′ (µ) dµ =
τ =α
µ=δ
ˆ
µ=γ
f (z(ϕ(τ )))z ′ (ϕ(τ ))ϕ′ (τ ) dτ
f (z(ψ(µ)))z ′ (ψ(µ))ψ ′ (µ) dµ.
µ=γ
How do we prove? The simple substitutions t = ϕ(τ ) and t = ψ(µ) into the integrals above give the original
integral expressed in terms of t.
Property 4.4.3. From the definition, it is straightforward to see
ˆ
ˆ
1. (Constant Multiple)
z0 f (z) dz = z0
f (z) dz, where z0 is a complex constant.
C
C
ˆ
ˆ
ˆ
2. (Sum)
(f (z) + g(z)) dz =
f (z) dz +
g(z) dz.
C
C
C
ˆ
ˆ
3. (Reverse Direction)
f (z) dz = − f (z) dz, where −C is a contour passing through exactly same
−C
C
points as C does, but it transverses in the reverse direction of C.
ˆ
ˆ
ˆ
4. (Separation)
f (z) dz =
f (z) dz +
f (z) dz, where C consists of contours C1 and C2 such
C
C1
C2
that C1 starts from the initial point of C and C2 starts from the terminal point of C1 and ends at the
terminal point of C.
Proof. We skip the proof of properties 1 and 2. Let C be a contour represented by z = z(t), a ≤ t ≤ b, with
z1 = z(a) and z2 = z(b).
Reverse Direction: For the contour C, −C has the representation z = z(−t), −b ≤ t ≤ −a. So the
contour integral along −C becomes
ˆ
−C
ˆ
t=−a
f (z) dz = −
′
f (z(−t))z (−t) dt
ˆ
t=−b
τ =b
=−
Subst: τ = −t
f (z(τ ))z ′ (τ ) dτ = −
τ =a
τ =a
=
τ =b
ˆ
f (z) dz.
C
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Separation: Suppose C consists of contours C1 and C2 such that C1 is represented by z = z(t), a ≤ t ≤ c,
and C2 is represented by z = z(t), c ≤ t ≤ b, where a < c < b. Then we get
ˆ
ˆ
t=b
f (z) dz =
C
t=a
t=c
ˆ
=
f (z(t))z ′ (t) dt
′
ˆ
t=b
f (z(t))z (t) dt +
t=a
′
ˆ
f (z(t))z (t) dt =
t=c
ˆ
f (z) dz +
C1
f (z) dz.
C2
For the contours C1 , C2 and C in the property 4 above, we say C is the sum of its legs C1 and C2 and
C is denoted by C = C1 + C2 . Sum of two contours C3 and −C4 is well defined when C3 and C4 have the
same final points, and it is written C3 − C4 .
One of the geometrical interpretation of a definite integral is area in calculus. But in the case of the
contour integrals, such an interpretation is usually not available except in special cases.
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§4.5 Some Examples.
Let us see some examples of the contour integrals.
´
Example 4.5.1. Compute I = C f (z) dz, where f (z) = z̄ and C is the right–hand half z = 2eiθ , −π/2 ≤
θ ≤ π/2, of the circle |z| = 2 from z = −2i to z = 2i. See the figure 4.1.
Answer. Since f (z(θ)) = 2eiθ = 2e−iθ and z ′ (θ) = 2ieiθ , so by definition,
ˆ
I=
ˆ
θ= π2
f (z) dz =
C
ˆ
′
θ= π2
f (z(θ))z (θ) dθ =
θ=− π2
2e
−iθ
ˆ
iθ
θ= π2
2ie dθ =
θ=− π2
4i dθ = 4πi.
θ=− π2
For z on the circle |z| = 2, we have 4 = |z|2 = z z̄ and so z̄ = 4/z, which implies
ˆ
ˆ
ˆ
4
1
4πi =
z̄ dz =
dz,
dz = πi,
C
C z
C z
where C is the contour given in the example 4.5.1.
.y
.2i
.
.C
.O
.−2i
.
.x
.2
.
Figure 4.1: Contour C represented by z = 2eiθ , −π/2 ≤ θ ≤ π/2.
´
Example 4.5.2. Compute I1 = C1 f (z) dz, where f (z) = y − x − i3x2 with z = x + iy and C1 is the
polygonal line composed of the line segments OA and AB. Here O = (0, 0), A = (0, 1) and B = (1, 1). See
the figure 4.2.
Answer. The statement 4 in the proposition 4.4.3 yields
ˆ
ˆ
ˆ
I1 =
f (z) dz =
f (z) dz +
C1
OA
f (z) dz.
AB
We compute
´ each integral.
(1) OA f (z) dz: Along the line segment OA, z = x + iy becomes z = 0 + iy = iy, 0 ≤ y ≤ 1, which
implies that z is a function of y and dz = idy. Along the line segment, f (z) becomes f (z) = y −0−i302 = y,
0 ≤ y ≤ 1. Thus the integral turns to be
ˆ
ˆ
y=1
f (z) dz = i
OA
y=0
[
y2
y dy = i
2
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]y=1
y=0
i
= .
2
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´
(2) AB f (z) dz: Along the line segment AB, z = x + iy becomes z = x + i1 = x + i, 0 ≤ x ≤ 1, which
implies that z is a function of x and dz = dx. Along the line segment, f (z) becomes f (z) = 1 − x − i3x2 ,
0 ≤ x ≤ 1. Thus the integral turns to be
ˆ
ˆ
x=1
f (z) dz =
AB
(
1 − x − i3x
2
[
)
x2
dx = x −
− ix3
2
x=0
Combining those results, we get
ˆ
ˆ
I1 =
f (z) dz =
C1
ˆ
f (z) dz +
f (z) dz =
OA
AB
]x=1
=
x=0
1
− i.
2
i 1
1−i
+ −i=
.
2 2
2
.y
.i
.
.A
.C1
.AB
.B
.
.1 + i
.C2
.OA
.
.
.O
.1
.x
Figure 4.2: Contours C1 and C2 formed by polygonal lines.
´
Example 4.5.3. Compute I2 = C2 f (z) dz, where f (z) = y − x − i3x2 with z = x + iy and C2 is the line
segment OB. Here O = (0, 0) and B = (1, 1). See the figure 4.2.
Answer. Along the line segment OB, z = x+iy becomes z = x+ix = x(1+i), 0 ≤ x ≤ 1, which implies that
z is a function of x and dz = (1 + i)dx. Along the line segment, f (z) becomes f (z) = x − x − i3x2 = −i3x2 ,
0 ≤ x ≤ 1. Thus the integral turns to be
ˆ
ˆ
ˆ x=1
(
)
[ ]x=1
I2 =
f (z) dz =
f (z) dz = (1 + i)
−i3x2 dx = −i (1 + i) x3 x=0 = −i (1 + i) = 1−i. C2
AB
x=0
Remark 4.5.4 (Observation on Examples 4.5.2 and 4.5.3). By those two examples, we can observe
1. although both C1 and C2 have the same initial and terminal points, the integrals of the function f
along each path are different,
2. along the simple closed contour OABO or C1 − C2 , the integral of f has the nonzero value,
ˆ
ˆ
ˆ
−1 + i
1
̸= 0.
f (z) dz =
f (z) dz −
f (z) dz = I1 − I2 = − i − (1 − i) =
2
2
C1 −C2
C1
C2
End of Remark
´
Example 4.5.5. Compute I = C f (z) dz, where f (z) = z and C is a smooth arc represented by z = z(t),
a ≤ t ≤ b, with z1 = z(a) and z2 = z(b).
Answer. We recall that
d
dt
[
]
(z(t))2
= z(t)z ′ (t).
2
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It implies
ˆ
ˆ
ˆ
t=b
f (z) dz =
z dz =
z(t)z ′ (t) dt
C
C
t=a
[
]
[
]t=b
ˆ t=b
2
(z(t))2
z 2 (b) − z 2 (a)
z 2 − z12
d (z(t))
=
dt =
=
= 2
.
2
2
2
2
t=a dt
t=a
I=
Since the value of this integral does not depend on the arc C, so we can rewrite the integral by specifying
the initial and terminal points,
ˆ z=z2
z 2 − z22
.
z dz = 2
2
z=z1
In the example above, even if we change the smooth arc into a contour, we can have the similar result.
´
Example 4.5.6. Compute I = C f (z) dz, where f (z) = z and C is a contour (i.e., piecewise smooth arc)
represented by z = z(t), a ≤ t ≤ b, starting from z(a) and ending at z(b).
Answer. Suppose a given contour C is composed of finitely many smooth arcs Ck , k = 1, 2, . . . , n, such that
C1 starts from the initial point z1 = z(a) of C, C2 starts from the terminal point z2 of C1 , and Ck starts from
the terminal point zk of Ck−1 and Cn ends at the terminal point zn+1 = z(b) of C. Then we have
C=
n
∑
ˆ
ˆ
and
Ck ,
f (z) dz =
∑n
C
k=1
k=1
f (z) dz =
Ck
n ˆ
∑
k=1
f (z) dz.
Ck
For the integral along each Ck , the example 4.5.5 above implies
ˆ
ˆ
z 2 − zk2
f (z) dz =
z dz = k+1
,
2
Ck
Ck
where Ck has the initial point zk and the terminal point zk+1 by the setup above. Hence, we get
ˆ
f (z) dz =
C
n ˆ
∑
k=1
Ck
f (z) dz =
n ˆ
∑
k=1
z dz =
Ck
n
2
∑
zk+1
− zk2
k=1
2
=
2
zn+1
− z12
,
2
where the last summation has telescoped. Just as the example 4.5.5, the value of the integral depends on
only the initial point z1 and the terminal point zn+1 of the contour C.
We observe from the examples 4.5.5 and 4.5.6 that if C is closed, then the initial point should coincide
with the terminal point and thus the integral over the contour must be zero, i.e.,
ˆ
z dz = 0,
C
where C is a closed curve. However, according to the remark 4.5.4 following the examples 4.5.2 and 4.5.3,
the integral along the closed contour C1 − C2 is not zero, i.e.,
ˆ
−1 + i
,
f (z) dz =
2
C1 −C2
where C1 , C2 and f (z) are given in those
´ examples. Then we may raise a question: what are the conditions
on f and the contour for the integral C f (z) dz to have the value zero? We will start to discuss it from the
Section on Antiderivatives later.
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§4.6 Examples with Branch Cuts.
So far the contour integrals are discussed for the piecewise continuous function f (z) along a contour.
Suppose f is not defined at finite number of points on a given contour, particularly, at the end points of the
contour. Can we still compute the contour integral? Let us see the examples involved with the branch cut.
We consider examples of the contour integrals in which the contour has the end point(s) on the branch cut
of a multiple–valued function f and f is the integrand of the integral.
´
Example 4.6.1. Compute the integral I = C f (z) dz, where C is a contour parameterized by z = 3eiθ ,
1
0 ≤ θ ≤ π and f (z) = z 1/2 = e 2 log z on |z| > 0, 0 < Arg z < 2π is a branch of z 1/2 . See the figure 4.3.
Answer. Although the integrand f (z) is not defined at the point z = 3 (θ = 0 in z = 3eiθ ), which is the
initial point of the given contour
(z) is still piecewise continuous on C and so the integral does exist.
´ x=1C, f√
(One can think of the integral x=0 1/ x dx = 2 in Calculus.) Specifically, on 0 < θ ≤ π,
1
1
iθ
f (z(θ)) = e 2 log(3e ) = e 2 (ln 3+iθ) =
√
f (z(θ))z ′ (θ) =
θ
3 e 2 i,
√ θ i iθ
√ 3θ
3 e 2 3ie = i3 3 e 2 i .
So when we take the right–hand side limit,
√
√ 3θ
lim+ f (z(θ))z ′ (θ) = lim+ i3 3 e 2 i = i3 3 ,
θ→0
θ→0
′
i.e.,
√the limit exists. Hence, f (z(θ))z (θ) is continuous on 0 ≤ θ ≤ π when its value at θ = 0 is defined as
i3 3 . We compute the integral,
]θ=π
[
ˆ
ˆ θ=π
√
√
√ ˆ θ=π 3θ i
2 3θ i
′
2
2
I=
f (z) dz =
e
= −2 3 (1 + i) . f (z(θ))z (θ) dθ = i3 3
e dθ = i3 3
3i
C
θ=0
θ=0
θ=0
.y
.C
.3i
.
.−3
.O
.
.
.3
.x
Figure 4.3: Upper–half circle C from 3 to −3 and branch cut, positive x–axis.
´
Example 4.6.2. Compute the integral I = C f (z) dz, where C is a circle centered at the origin with radius
R from −Ri to Ri, i.e., parameterized by z = Reiθ , −π ≤ θ ≤ π and f (z) = z a−1 = e(a−1) ln z on |z| > 0,
−π < Arg z < π is the principal branch of z a−1 with any nonzero real constant a. See the figure 4.4.
Answer. Similarly as in the example 4.6.1, although the integrand f (z) is not defined at the point z = −R
(θ = −π or θ = π in z = Reiθ ), which are the initial and terminal points of the given contour C, f (z) is still
piecewise continuous on C and so the integral does exist. Specifically, on −π < θ < π,
iθ
f (z(θ)) = e(a−1) ln(Re ) = e(a−1)(ln R+iθ) = Ra−1 e(a−1)θi ,
f (z(θ))z ′ (θ) = Ra−1 e(a−1)θi Rieiθ = iRa eaθi .
So when we take the right–hand side limit of −π and left–hand side limit of π,
lim + f (z(θ))z ′ (θ) = lim + iRa eaθi = iRa e−aπi ,
θ→−π
θ→−π
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lim− f (z(θ))z ′ (θ) = lim− iRa eaθi = iRa eaπi ,
θ→π
θ→π
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i.e., both limits exist. Hence, f (z(θ))z ′ (θ) is continuous on −π ≤ θ ≤ π when its values at θ = −π and
θ = π are defined as iRa e−aπi and iRa eaπi , respectively. We compute the integral,
ˆ
I=
ˆ
θ=π
f (z) dz =
C
= iR
ˆ
f (z(θ))z ′ (θ) dθ
θ=−π
[
θ=π
a
e
aθi
dθ = iR
a
θ=−π
1 aθi
e
ai
]θ=π
=i
θ=−π
2Ra eaπi − e−aπi
2Ra
=i
sin (aπ) .
a
2i
a
.y
.Ri
.C
.
.−R
.O
.
.x
.R
.−Ri
Figure 4.4: Circle C parameterized by z = Reiθ , −π ≤ θ ≤ π, and branch cut, negative x–axis.
Remark 4.6.3. In the example 4.5.6 above,
1. if a = n, n = ±1, ±2, . . . , then the value of the integral becomes
ˆ
ˆ
2Rn
I=
f (z) dz =
z n−1 dz = i
sin (nπ) = 0.
n
C
C
2. If a = 0, then
ˆ
ˆ
I=
f (z) dz =
C
C
1
dz =
z
ˆ
θ=π
θ=−π
1
Rieiθ dθ = i
Reiθ
ˆ
θ=π
1 dθ = 2πi.
θ=−π
End of Remark
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§4.7 Upper Bounds for Moduli of Contour Integrals.
The main result, theorem 4.7.2, in this section is extremely important in applications such as Cauchy
integral formula, Liouville’s theorem, fundamental theorem of calculus, maximum modulus principle and
others.
Lemma 4.7.1. Suppose that w(t) is a piecewise continuous complex–valued function of a real variable t on
a ≤ t ≤ b. Then,
ˆ t=b
ˆ t=b
w(t) dt
∥w(t)∥ dt.
≤
t=a
t=a
´b
Proof. The integral a w(t) dt is a complex number. If the integral is zero, then the inequality is clearly true.
So let us assume the integral is not zero. Then we can express the integral in the exponential form,
ˆ
b
w(t) dt = r0 eiθ0 ,
a
for certain r0 > 0 and θ0 . We observe
−iθ0
ˆ
r0 = e
ˆ
b
b
w(t) dt =
a
e
a
−iθ0
ˆ b
ˆ b
−iθ
0
w(t) dt
w(t) dt, and 0 < r0 = ∥r0 ∥ = w(t) dt
.
e
=
a
a
Since the real part of a real number is the number itself, we find
(ˆ
b
)
−iθ0
e
w(t) dt
r0 = Re (r0 ) = Re
a
ˆ b
(
)
=
Re e−iθ0 w(t) dt.
a
But a simple computation shows
(
) Re e−iθ0 w(t) ≤ e−iθ0 w(t) = ∥w(t)∥ ,
which implies
ˆ b
ˆ b
(
)
w(t) dt = r0 =
Re e−iθ0 w(t) dt
a
a
ˆ b
≤
∥w(t)∥ dt.
a
Therefore, the lemma is proved.
Theorem 4.7.2 (Upper Bounds for Moduli of Contour Integrals). Suppose that
1. C is a contour of total length L,
2. f is a piecewise continuous function on C,
3. M ≥ 0 is a constant such that |f (z)| ≤ M for all z ∈ C.
Then we have
ˆ
f (z) dz ≤ M L.
C
That is, the integral is bounded by the product of the maximum value of f on C and the length of C.
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Proof. Let C be represented by z = z(t), a ≤ t ≤ b. Then
ˆ
ˆ b
ˆ
ˆ b
by Lemma b
′
′
f (z) dz = f (z(t))z (t) dt ≤
∥f (z(t))z (t)∥ dt =
∥f (z(t))∥ ∥z ′ (t)∥ dt
C
a
a
a
ˆ b
ˆ b
Condition 3
′
≤
M ∥z (t)∥ dt = M
∥z ′ (t)∥ dt = M L.
a
a
In the theorem, the condition 3 is essential because the bound M of |f (z)| is used to bound the integral
of f . We recall from Calculus/Real Analysis that a continuous function ϕ(x) on a compact interval (closed
and bounded) a ≤ x ≤ b has the maximum and minimum values on the interval [a, b]. (The corresponding
statement in the theory of the complex variables will be studied in the last section Maximum modulus principle of this course.) In fact, the same is true for a piecewise continuous function f on a piecewise smooth
arc C, i.e., the piecewise continuous function |f (z(t))| on a compact interval a ≤ t ≤ b has the maximum
value on [a, b].
ˆ
6π
z
+
4
Example 4.7.3. Prove that dz
z 3 − 1 ≤ 7 , where C is the arc of the circle |z| = 2 from z = 2 to
C
z = 2i that lies in the first quadrant.
Proof. We parameterize the arc C by z = 2eiθ , 0 ≤ θ ≤ π/2. It is straightforward to see the total length L
of C is given by
1
1
L = (total circumference of the circle |z| = 2) = 2π(2) = π.
4
4
3
Now we find the bound M of the integrand (z + 4)/(z − 1) on the arc C. For z on C, we find
z+4 |z + 4|
6
3
3
3
|z + 4| ≤ |z| + 4 ≤ 2 + 4 = 6,
|z − 1| ≥ ||z| − 1| = |2 − 1| = 7,
z 3 − 1 = |z 3 − 1| ≤ 7 = M.
Therefore, by the theorem, we conclude
ˆ
z
+
4
6
dz
z 3 − 1 ≤ M L = 7 π.
C
ˆ
Example 4.7.4. Prove that lim
f (z) dz = 0. Here CR is the upper–half circle centered at the origin
R→∞
CR
with radius R > 1, parameterized by z = Reiθ , 0 ≤ θ ≤ π, and f (z) = z 1/2 /(z 2 + 1) and z 1/2 is the branch
1
z 1/2 = e 2 log z , |z| > 0, −π/2 < θ < 3π/2.
Answer. It is straightforward to see the total length L of the upper–half circle CR of radius R is given by
LR = πR. We find the bound MR of the integrand f (z) = z 1/2 /(z 2 + 1) on the arc CR . For z on CR , we find
√
1/2 1/2
√
z
|z
|
R
1/2
2
2
2
2
|z | = R ,
|z + 1| ≥ ||z| − 1| = |R − 1| = R − 1,
z 2 + 1 = |z 2 + 1| ≤ R2 − 1 = MR .
Therefore, by the theorem, we conclude
ˆ
ˆ
f (z) dz =
CR
CR
√
z 1/2
πR
R
≤ MR LR = 2
dz .
2
z +1
R −1
When we take the limit to the result above, we have
( √ )
ˆ
ˆ
ˆ
1/2
z
πR R
z 1/2
z 1/2
≤
lim
lim dz
dz
=
0,
dz = 0. =
0,
lim
R→∞ R2 − 1
2
2
R→∞
R→∞ C z 2 + 1
CR z + 1
whole plane z + 1
R
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§4.8 Antiderivatives.
There are certain functions whose integrals from z1 to z2 have values that are independent of path
and in this case, the values of integrals around closed paths are zero. We raise two questions and find the
answers. When is the integration independent of path? When does an integral around a closed path have
value zero?
The answers are involved with the extension of the fundamental theorem of calculus, which deals
with the concept on an antiderivative of a continuous function f on a domain D. Here an antiderivative of
f means a function F such that F ′ (z) = f (z) for all z in D.
Note that
1. if f has an antiderivative F in a domain D, then F is differentiable and moreover analytic in D
2. an antiderivative is unique except for an additive constant.
Proof of 2. Let F and G be antiderivatives of f in a domain D. Then (F − G)′ = F ′ − G′ = f − f = 0, i.e.,
(F − G)′ = 0 in D. Since F − G is analytic and its derivative is zero in D, so F − G is constant in D.
Now we present a theorem giving the relationship between an antiderivative, extension of fundamental theorem of calculus and the integral around a closed curve.
Theorem 4.8.1. Suppose a function f is continuous on a domain D. The following statements are equivalent,
i.e., if one of them is true, the rest of them are also true.
1. f has an antiderivative F throughout D.
2. The integrals of f along contours lying in D and extending from any fixed point z1 to any fixed point
z2 all have the same value, i.e.,
ˆ z2
] z2
f (z) dz = F (z) = F (z2 ) − F (z1 ),
z1
z1
where F is the antiderivative of f .
3. The integrals of f around closed contours lying entirely in D all have value zero, i.e.,
ˆ
f (z) dz = 0,
C
where C is any closed contour in D.
Be careful: The theorem does not say that for a given f , any of them is true. It says only either that all
of them are true or that none of them is true.
Example 4.8.2. Consider f (z) = z 2 . We observe that f has an antiderivative F (z) = z 3 /3 throughout the
plane. By the statement 2 in the theorem, for any contour C from z = 0 to z = 1 + i,
[ 3 ]z=1+i
ˆ
ˆ z=1+i
2
z
(1 + i)3
2
2
= (−1 + i) .
z dz =
z dz =
=
3 z=0
3
3
C
z=0
The statement 3 in the theorem implies that for any closed contour C,
ˆ
z 2 dz = 0.
C
Example 4.8.3. Consider f (z) = 1/z 2 . We observe f is continuous and has an antiderivative F (z) = −1/z
in the domain |z| > 0, i.e., the entire plane with the origin deleted.
The statement 3 in the theorem implies that for any closed contour C in the domain,
ˆ
1
dz = 0.
2
C z
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Example 4.8.4. Consider f (z) = 1/z. We observe f is continuous and analytic in the domain |z| > 0, i.e.,
the entire plane with the origin deleted. We recall that any branch of log z has the derivative 1/z, that is, an
antiderivative of f is a branch F of log z.
Suppose the ray θ = α from the origin is a branch cut used to define the branch F . Then, the branch F
is not differentiable, not even defined along the branch cut θ = α. See the figure 4.5 with a contour z = 2eiθ .
So for a contour C intersecting the branch cut θ = α, the integral
ˆ
f (z) dz
C
cannot be computed using the statement 2 in the theorem. How can we overcome this kind of difficulty in
integration? Let us see the next example with explicitly given contour.
.y
.θ = α
.2i
.
.C
.α
.O
.
.x
.
.−2i
Figure 4.5: At intersection of branch cut and contour (red dot), F is not differentiable.
Example 4.8.5. Compute
´
C
1/z dz, where C is a circle centered at the origin with radius 2. See the figure 4.5.
Answer. We represent the circle C as follows:
(
)
π
3π
− ≤θ≤
.
2
2
iθ
C : z = 2e ,
We separate C into two pieces C1 and C2 , see the figure 4.6,
( π
π)
C1 : z = 2eiθ ,
− ≤θ≤
,
2)
( 2
3π
π
≤θ≤
C2 : z = 2eiθ ,
.
2
2
Then C = C1 + C2 and thus
ˆ
C
1
dz =
z
ˆ
C1
1
dz +
z
ˆ
C2
1
dz.
z
Part I. Integral over C1 Since an antiderivative of 1/z is a branch of log z, we choose a branch cut
θ = −π so that the branch cut does not intersect the curve C1 . The branch of log z having the branch cut
θ = −π is the principal branch, i.e.,
ln z = ln r + iΘ,
(r > 0, −π < Θ < π) .
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.y
.2i
.Branch θ = −π
.O
.−2i
.y
.
.
.C1
.C2
.x
.
.2i
.
.O
.
.
.Branch θ = 0
.−2i
Figure 4.6: Separation of Contour equipped with Branches.
Now we have found an antiderivative of 1/z and so we can use the theorem,
ˆ
ˆ z=2i
[
]z=2i
1
π)
1
π (
= πi.
dz =
dz = ln z
= ln (2i) − ln (−2i) = ln 2 + i − ln 2 − i
2
2
z=−2i
C1 z
z=−2i z
Part II. Integral over C2 Since an antiderivative of 1/z is a branch of log z, we choose a branch cut
θ = 0 so that the branch cut does not intersect the curve C2 . The branch of log z having the branch cut
θ = 0 is given by
log z = ln r + iθ,
(r > 0, 0 < θ < 2π) .
Now we have found an antiderivative of 1/z and so we can use the theorem,
ˆ
ˆ z=−2i
[
]z=−2i
1
1
π)
3π (
dz =
dz = log z
− ln 2 + i
= πi.
= log (−2i) − log (2i) = ln 2 + i
z
2
2
z=2i
C2 z
z=2i
Combining the results from Part I and II, we conclude
ˆ
ˆ
ˆ
1
1
1
dz =
dz +
dz = πi + πi = 2πi.
C z
C1 z
C2 z
´
Example 4.8.6. Compute C1 z 1/2 dz where the integrand is the branch
√
f (z) = z 1/2 = e1/2 log z = r eiθ/2 ,
(r > 0, 0 < θ < 2π)
and C1 is any contour from z = −3 to z = 3 that lies above the x–axis except for both end points.
Answer. We observe that z 1/2 is piecewise continuous on C1 and the integral exists over C1 , but the given
branch f (z) is not defined on the ray θ = 0, particularly at z = 3. It prevents us from using its antiderivative
for the integral. So we introduce another branch
(
)
√ iθ/2
3π
π
r > 0, − < θ <
f1 (z) = r e ,
2
2
which is defined and continuous everywhere on C1 . Then f (z) = f1 (z) at all z on C1 except z = 3. We
replace the integrand f by f1 and f1 has the antiderivative
(
)
π
3π
2 3/2 2 √ i3θ/2
,
r > 0, − < θ <
F1 (z) = z = r r e
.
3
3
2
2
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So the given integral in the problem can be computed by
ˆ
ˆ
ˆ z=3
[
]z=3
√
)
2 √ (
f (z) dz =
f1 (z) dz =
f1 (z) dz = F1 (z)
= 3 3 e0i − ei3π/2 = 2 3 (1 + i) . 3
z=−3
C1
C1
z=−3
We have used the branch f1 with −π/2 < θ < 3π/2. In fact, one can get the same result by using any
other branch with α < θ < α + 2π, α ̸= 0, −π, etc., which contains both z = −3 and z = 3 and the contour
C1 .
We consider another contour in the previous example.
´
Example 4.8.7. Compute C2 z 1/2 dz where the integrand is the branch
f (z) = z 1/2 = e1/2 log z =
√
r eiθ/2 ,
(r > 0, 0 < θ < 2π)
and C2 is any contour from z = −3 to z = 3 that lies below the x–axis except for both end points.
Answer. We observe that z 1/2 is piecewise continuous on C2 and the integral exists over C2 , but the given
branch f (z) is not defined on the ray θ = 0, particularly at z = 3. We introduce another branch
(
)
√ iθ/2
π
5π
f2 (z) = r e ,
r > 0, < θ <
2
2
which is defined and continuous everywhere on C2 . Then f (z) = f2 (z) at all z on C2 except z = 3. We
replace the integrand f by f2 and f2 has the antiderivative
(
)
2 3/2 2 √ i3θ/2
π
5π
F2 (z) = z = r r e
,
r > 0, < θ <
.
3
3
2
2
So the given integral in the problem can be computed by
ˆ
ˆ
ˆ z=3
[
]z=3
√
)
2 √ (
f (z) dz =
f2 (z) dz =
f2 (z) dz = F2 (z)
= 3 3 ei3π − ei3π/2 = 2 3 (−1 + i) .
3
z=−3
C2
C2
z=−3
We have used the branch f2 with π/2 < θ < 5π/2. In fact, one can get the same result by using any
other branch with α < θ < α + 2π, α ̸= 0, −π, etc., which contains both z = −3 and z = 3 and the contour
C2 . Explicitly with α > 0, the branch
√
fα (z) = r eiθ/2 ,
(r > 0, α < θ < α + 2π)
has the domain containing both z = −3 (θ = π) and z = 3 (θ = 2π) and the contour C2 . Its antiderivative is
Fα (z) =
and the given integral becomes
ˆ
ˆ
ˆ
f (z) dz =
fα (z) dz =
C2
C2
2 √ i3θ/2
r re
,
3
(r > 0, α < θ < α + 2π)
[
]z=3
√
)
2 √ (
fα (z) dz = Fα (z)
= 3 3 ei3π − ei3π/2 = 2 3 (−1 + i) .
3
z=−3
z=−3
z=3
§4.9 Proof of the Theorem.
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§4.10 Cauchy–Gorsat Theorem.
From the Section 44 Antiderivatives, we recall that for a continuous function f having an antiderivative,
ˆ
f (z) dz = 0
C
over any closed contour C. In this section, we raise a question and look for the answer: for what kind of
function f , can we have
ˆ
f (z) dz = 0
C
over any simple closed contour C? If f is continuous and has an antiderivative, then by Section 44, the integral
is clearly zero. Obviously we look for other condition(s). The result is central to the theory of functions of
complex variables.
Theorem 4.10.1 (Cauchy Theorem). We assume
1. C is a simple closed contour,
2. f is analytic at each point interior to and on C,
3. f ′ is continuous at each point interior to and on C.
Then we have
ˆ
f (z) dz = 0.
C
Theorem 4.10.2 (Cauchy–Gorsat Theorem). We assume
1. C is a simple closed contour,
2. f is analytic at each point interior to and on C.
Then we have
ˆ
f (z) dz = 0.
C
Observe the difference between two theorems. Gorsat proved that the condition of continuity of f ′ can
be omitted. In fact, later in Section 52 (Theorem 1 on page 168), we will study that if f is analytic, then f ′ is
also analytic (so f ′ is clearly continuous). Moreover, all derivatives of f are analytic (so all derivatives are
continuous).
´
3
Example 4.10.3. Compute C ez dz, where C is any simple closed contour.
3
3
Answer. Since ez is a composite function of analytic functions ez and z 3 , so ez is also analytic. By CauchyGorsat theorem, the integral is zero, i.e.,
ˆ
3
ez dz = 0.
C
Proof of Cauchy Theorem. Let C be represented by z = z(t), a ≤ t ≤ b, in positive sense. With f (z) =
u(x, y) + iv(x, y) and z(t) = x(t) + iy(t),
ˆ
ˆ
t=b
f (z) dz =
C
′
ˆ
f (z(t))z (t) dt =
ˆ
t=a
b
b
[u(x(t), y(t)) + iv(x(t), y(t))] [x′ (t) + iy ′ (t)] dt
a
[u(x(t), y(t))x′ (t) − v(x(t), y(t))y ′ (t)] dt
=
a
ˆ
+i
b
[v(x(t), y(t))x′ (t) + u(x(t), y(t))y ′ (t)] dt
a
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ˆ
ˆ
u dx − v dy + i
=
v dx + u dy.
C
C
Since f ′ is continuous, we can use Green’s theorem to get
ˆ
ˆ
ˆ
¨
¨
f (z) dz =
u dx − v dy + i v dx + u dy =
(−vx − uy ) dA + i
(ux − vy ) dA,
C
C
C
R
R
where R is the region consisting of all points interior to and on C. Since f is analytic, f satisfies the CauchyRiemann equations, ux = vy , uy = −vx , and thus the integral becomes
ˆ
¨
¨
f (z) dz =
(−vx − uy ) dA + i
(ux − vy ) dA
C
¨R
¨ R
¨
=
(uy − uy ) dA + i
(ux − ux ) dA =
0 dA = 0.
R
R
R
The orientation of C is not important, because
ˆ
ˆ
0=
f (z) dz = −
−C
C
§4.11 Proof of the Theorem.
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f (z) dz.
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§4.12 Simply Connected Domains.
We recall the Cauchy–Goursat theorem stating the integral of an analytic function is zero over a simple
closed contour. We want to get the same result (integral zero) over a closed contour. For this purpose, a domain
which the contour lies on will be studied.
Definition 4.12.1. A simply connected domain D is a domain (=connected open set) such that every simple
closed contour within it encloses only points of D.
Example 4.12.2. A set of points interior to a simple closed contour is a simply connected domain.
An annular domain (=donut shape) is not simply connected.
Now on the simple connected domain, the Cauchy–Goursat theorem can be written as follows.
Theorem 4.12.3 (Cauchy–Goursat Theorem (on Simple Connected Domain)). We assume
1. D is a simply connected domain,
2. f is analytic throughout D.
Then we have
ˆ
f (z) dz = 0
C
for every closed contour C lying in D.
Rough Proof. We prove only two cases: when C is a simple closed contour and when C is a closed contour
that intersects itself finitely many times.
Case 1. C is a simple closed contour. Since C is a simple and lies in D, the function f is analytic at each
point interior to and on C. The Cauchy–Goursat theorem in Section 46 implies
ˆ
f (z) dz = 0.
C
Case 2. C is a closed contour that intersects itself finitely∑
many times. We can separate C into finitely many simple
closed contours, Ck , k = 1, . . . , n, such that C = nk=1 Ck . By the Cauchy–Goursat theorem in Section 46,
the integral of f over each simple closed contour Ck is zero. Thus it implies
ˆ
n ˆ
n
∑
∑
0 = 0.
f (z) dz =
f (z) dz =
C
k=1
Ck
k=1
In fact, even if a closed contour intersects itself infinitely many times, the theorem still holds.
ˆ
zez
Example 4.12.4. Compute
dz, where C is a closed contour lying in the open disk |z| < 2.
2
5
C (z + 9)
Answer. The open disk |z| < 2 is a simply connected domain. The integrand f (z) = zez /(z 2 + 9)5 has
singularities at z = ±3i, which are not in the open disk |z| < 2. So f is analytic in the given disk. Thus by
the Cauchy–Goursat theorem above, the integral is zero, i.e.,
ˆ
zez
dz = 0.
2
5
C (z + 9)
Corollary 4.12.5. A function f is analytic throughout a simply connected domain D must have an antiderivative everywhere in D.
Proof. Since the analyticity implies the continuity, an analytic function
f in a simply connected domain D
´
is continuous there. The Cauchy–Goursat theorem above implies C f (z) dz = 0 for any closed contour C
lying in D. According to the theorem in Section 44 Antiderivatives, f has an antiderivative in D.
Since a finite plane is a simply connected domain, by the corollary, any entire function must have an
antiderivative everywhere.
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§4.13 Multiply Connected Domains.
A domain that is not simply connected is called to multiply connected domain. We can generalize the
Cauchy–Goursat theorem on a simply connected domain to the multiply connected domain.
Theorem 4.13.1 (Cauchy–Goursat (on Multiply Connected Domain)). Suppose that
1. C is a simple closed contour, described in the counterclockwise direction,
2. Ck , k = 1, 2, . . . , n, are simply closed contours interior to C, all described in the clockwise direction,
that are disjoint and whose interiors have no points in common. See the figure 4.7.
If a function f is analytic on all of these contours and throughout the multiply connected domain consisting
of the points inside C and exterior to each Ck , then
ˆ
n ˆ
∑
f (z) dz +
C
f (z) dz = 0.
Ck
k=1
.C
.y
.Γ1
.L1
.−L1
.L2
.C1
.−L2
.C2
.L3
.−L3
.Γ2
.O
.
.x
Figure 4.7: Separation of Multiply Connected Domain.
Note that in the equation above, the direction of each path of integration is such that the multiply
connected domain lies to the left of the path.
Proof. Without loss of generality, we consider the multiply connected domain in the figure 4.7, specifically,
region interior to C and exterior to C1 and C2 .
We separate the domain by introducing some polygonal paths L1 , L2 and L3 as shown in the figure.
Then we can form two simple closed contour Γ1 and Γ2 as follows:
Γ1 = C u + L1 + C1u + L2 + C2u + L3 ,
Γ2 = C l − L1 + C1l − L2 + C2l − L3 ,
where C u , C1u and C2u represent the upper part of C, C1 and C2 at the level of the polygonal paths, while C l ,
C1l and C2l represent the lower part of C, C1 and C2 at the level of the polygonal paths so that C = C u + C l ,
C1 = C1u + C1l and C2 = C2u + C2l . (Better figure than 4.7 is given in the book on page 159.)
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Each region interior to and on the simple closed contours Γ1 and Γ2 , respectively, is simply closed
domain. Under the assumption, f is analytic on such a simply closed domain. The Cauchy–Goursat theorem
on a simply connected domain implies
ˆ
ˆ
ˆ
ˆ
f (z) dz = 0 =
f (z) dz
or
f (z) dz +
f (z) dz = 0.
Γ1
Γ2
Γ1
Γ2
When we rewrite the equation, we get
ˆ
ˆ
ˆ
ˆ
0=
f (z) dz +
f (z) dz =
f (z) dz +
f (z) dz
Γ1
Γ2
C u +L1 +C1u +L2 +C2u +L3
C l −L1 +C1l −L2 +C2l −L3
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
f (z) dz +
f (z) dz +
f (z) dz +
f (z) dz +
f (z) dz +
f (z) dz
=
L1
C1u
L2
C2u
L3
Cu
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
f (z) dz −
f (z) dz +
f (z) dz −
f (z) dz +
f (z) dz −
f (z) dz
+
L1
C1l
L2
C2l
L3
Cl
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
f (z) dz =
f (z) dz +
f (z) dz +
f (z) dz
=
f (z) dz +
f (z) dz +
C u +C l
ˆ
i.e., 0 =
f (z) dz +
C
2 ˆ
∑
k=1
C1u +C1l
C2u +C2l
C
C1
C2
f (z) dz.
Ck
A similar argument can be applied for the general situation having more than two simple closed contours
interior to C.
Corollary 4.13.2 (Principle of Deformation of Paths). Let C1 and C2 denote positively oriented simple closed
contours where C1 is interior to C2 . See the figure 4.8. If a function f is analytic in the closed region consisting of those contours and all points between them, then
ˆ
ˆ
f (z) dz =
f (z) dz.
C1
C2
.C2
.y
.C1
.O
.
.x
Figure 4.8: Region between C1 and C2 (Annulus).
The corollary tells us that if C1 is continuously deformed into C2 , always passing through points at
which f is analytic, then the value of the integral of f over C1 never changes.
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Proof. By the theorem, we have
ˆ
ˆ
f (z) dz +
f (z) dz = 0,
C2
−C1
Example 4.13.3. Compute
the origin.
ˆ
ˆ
f (z) dz = −
i.e.,
C2
´
C
ˆ
f (z) dz =
−C1
f (z) dz.
C1
1/z dz, where C is any positively oriented simple closed contour surrounding
Answer. For a fixed C, we can choose a positively oriented circle C0 centered at the origin and lying entirely
interior to C. Then C0 is clearly a simple closed contour. Since 1/z is analytic everywhere except at z = 0,
so 1/z is analytic in the closed region consisting of those contours C and C0 and all points between them.
Thus by the corollary,
ˆ
ˆ
1
1
dz =
dz.
C z
C0 z
´
We recall from Example 2 in Section 42 of the textbook, page 134 that C0 1/z dz = 2πi which gives
ˆ
C
1
dz =
z
ˆ
C0
1
dz = 2πi.
z
In the solution of the example, we can choose C0 to be bigger than C so that C lies entirely inside C0 .
The corollary gives the same result.
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§4.14 Cauchy Integral Formula.
Theorem 4.14.1 (Cauchy Integral Formula). Suppose that
1. C is a simple closed contour taken in the positive sense.
2. f is analytic everywhere inside and on C.
If z0 is any point interior to C, then
1
f (z0 ) =
2πi
ˆ
C
f (z)
dz.
z − z0
The theorem tells us that if f is to be analytic within and on a simple closed contour C, then the values
of f interior to C are completely determined by the values of f on C. The equation can be written as
ˆ
f (z)
dz = 2πif (z0 ),
C z − z0
which is useful in computing certain integrals.
Example 4.14.2. For f (z) = z/(9 − z 2 ), compute
|z| = 2.
´
C
f (z)/(z + i) dz, where C is a positively oriented circle
Answer. Since f (z) has singularities at z = ±3 which are exterior to the given contour C, so f is analytic
inside and on C. z = −i is a point interior to C. The Cauchy integral formula implies
ˆ
ˆ
f (z)
f (z)
−i
π
dz =
dz = 2πif (−i) = 2πi
= .
2
9 − (−i)
5
C z +i
C z − (−i)
Proof of Cauchy Integral Formula. Let Cρ be a positively oriented circle |z −z0 | = ρ, where ρ is small enough
that Cρ is interior to C. Since f (z)/(z−z0 ) is analytic between and on the contours Cρ and C, by the principle
of deformation paths (i.e., the corollary of the previous section), we have
ˆ
ˆ
f (z)
f (z)
dz =
dz.
C z − z0
C ρ z − z0
When we add f (z0 )
ˆ
C
´
Cρ
1/(z − z0 ) dz to both sides, we obtain
f (z)
dz − f (z0 )
z − z0
ˆ
Cρ
1
dz =
z − z0
ˆ
Cρ
f (z)
dz − f (z0 )
z − z0
ˆ
Cρ
1
dz =
z − z0
ˆ
Cρ
f (z) − f (z0 )
dz.
z − z0
But Exercise 10 (b), Section 42 in the textbook, page 136 says
ˆ
1
dz = 2πi.
Cρ z − z0
Applying this result, the equation above becomes
ˆ
ˆ
f (z)
f (z) − f (z0 )
dz − 2πif (z0 ) =
dz.
z − z0
C z − z0
Cρ
Now since f is analytic and so continuous at z0 , for any ε > 0, there exists a δ > 0 such that
|f (z) − f (z0 )| < ε
whenever |z − z0 | < δ.
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We choose the radius ρ of the circle Cρ to be smaller than δ. (That is, we chose Cρ to lie entirely inside the
disk |z − z0 | < δ.) Then, for z on Cρ , |z − z0 | = ρ < δ. Thus, by the continuity,
|f (z) − f (z0 )| < ε
for z ∈ Cρ ⊂ { z : |z − z0 | < δ }.
The theorem in Section 43 gives the upper bounds for the moduli of contour integrals,
ˆ
f
(z)
−
f
(z
)
0
dz Cρ
z − z0
≤ (Maximum of Integrand on Contour) · (Total Length of Contour)
ε
< 2πρ = 2πε.
ρ
Therefore, from the equation above, we have
ˆ
ˆ
f
(z)
−
f
(z
)
f
(z)
0
=
0<
dz
−
2πif
(z
)
dz
< 2πε.
0 z − z0
z
−
z
0
Cρ
C
Since ε > 0 can be arbitrary chosen, finally we conclude
ˆ
f (z)
= 0,
dz
−
2πif
(z
)
i.e.,
0
z − z0
C
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ˆ
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§4.15 An Extension of the Cauchy Integral Formula.
In this section, we find an integral form of a derivative of an analytic function.
Theorem 4.15.1 (Cauchy Integral Formula for Derivatives). Suppose that the conditions in the Cauchy integral formula hold. If w is any point interior to C, then the nth derivative of f at w is given by
f
(n)
n!
(z) =
2πi
ˆ
C
f (s)
ds,
(s − z)n+1
where n = 0, 1, 2, . . . and n factorial means n! = n · (n − 1) · (n − 2) · · · 4 · 3 · 2 · 1 with 0! = 1. The equation
can be written as
ˆ
f (s)
2πi (n)
f (z).
n+1 ds =
n!
C (s − w)
Formally, we can see the followings:
f
(0)
(z) =
f (1) (z) =
f (2) (z) =
f (3) (z) =
ˆ
1
f (s)
ds
2πi C s − z
[
]
[
]
ˆ
ˆ
ˆ
d
1
f (s)
1
d f (s)
1
f (s)
ds =
ds = 1
ds
dz 2πi C s − z
2πi C dz s − z
2πi C (s − z)2
[
]
[
]
ˆ
ˆ
ˆ
f (s)
d
f (s)
f (s)
d
1
1
1
1
ds = 1
ds = 2 · 1
ds
2
2
dz
2πi C (s − z)
2πi C dz (s − z)
2πi C (s − z)3
[
]
[
]
ˆ
ˆ
ˆ
f (s)
d
f (s)
f (s)
d
1
1
1
2·1
ds = 2 · 1
ds = 3 · 2 · 1
ds
3
3
dz
2πi C (s − z)
2πi C dz (s − z)
2πi C (s − z)4
......
f
(n)
[
]
[
]
ˆ
ˆ
d
f (s)
1
f (s)
1
d
(z) =
(n − 1) · · · 3 · 2 · 1
ds = (n − 1) · · · 3 · 2 · 1
ds
dz
2πi C (s − z)n
2πi C dz (s − z)n
ˆ
ˆ
1
f (s)
n!
f (s)
= n · (n − 1) · · · 3 · 2 · 1
ds =
ds.
n+1
2πi C (s − z)
2πi C (s − z)n+1
Proof for First Derivative. Let us fix a point z inside C and let d = min{ |s − z| : s ∈ C }. See the figure 4.9.
The Cauchy integral formula says
ˆ
1
f (s)
f (z) =
ds.
2πi C s − z
Using this, we can get
]
[ˆ
ˆ
1 1
f (s)
f (z + △z) − f (z)
f (s)
=
ds −
ds
△z
2πi △z C s − z − △z
C s−z
[
]
ˆ
1
1
f (s)
1
=
−
ds
2πi C △z s − z − △z s − z
ˆ
1
f (s)
ds
=
2πi C (s − z − △z)(s − z)
1
where 0 < |△z| < d. We add − 2πi
1
f (z + △z) − f (z)
−
△z
2πi
ˆ
C
´
f (s)
C (s−z)2
ds to both sides,
ˆ
ˆ
1
1
f (s)
f (s)
f (s)
ds =
ds −
ds
2
(s − z)
2πi C (s − z − △z)(s − z)
2πi C (s − z)2
ˆ
1
△zf (s)
=
ds.
2πi C (s − z − △z)(s − z)2
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Now we use the theorem in Section 43 Upper Bounds for Moduli of Contour Integrals. For this, let M =
max{ |f (s)| : s ∈ C }. Then we observe since |s − z| ≥ d and |△z| < d, so
|s − z − △z| = |(s − z) − △z| ≥ ||s − z| − |△z|| ≥ d − |△z| > 0.
Therefore,
ˆ
ˆ
f (z + △z) − f (z)
1
1
f
(s)
△zf
(s)
|△z|M
=
≤
−
ds
ds
L,
2
2
△z
2πi C (s − z)
2πi C (s − z − △z)(s − z)
(d − |△z|)d2
where L is the total length of C. Take the limit on both sides as △z → 0, we conclude
ˆ
f (z + △z) − f (z)
|△z|M
1
△zf (s) ≤ lim
0 ≤ lim ds
L = 0,
−
2
△z→0 (d − |△z|)d2
△z→0
△z
2πi C (s − z)
i.e., we deduce
1
f (z) −
2πi
′
ˆ
C
f (s)
ds = 0,
(s − z)2
1
f (z) =
2πi
′
ˆ
f (s)
ds.
(s − z)2
C
.C
.y
.
.d
.|s − z|
.|△z|
.s
.
.
.z
.
.z + △z
.O
.
.x
Figure 4.9: Points in Region.
Example 4.15.2. Compute
´
C
e2z /z 4 dz, where C is the positively oriented unit circle |z| = 1.
Answer. Let f (z) = e2z . Then it is easy to see that f is an entire function, i.e., analytic everywhere in the
plane. Since C is a simple closed contour and w = 0 is inside C, so by the Cauchy integral theorem for
derivatives, we have
ˆ
ˆ
ˆ 2z
2πi (3)
f (z)
f (z)
e
dz =
dz =
dz =
f (0).
4
4
4
3!
C z
C (z − 0)
C z
Since f (3) (z) = 23 e2z , so f (3) (0) = 8 and
ˆ 2z
e
2πi (3)
2πi
8πi
dz
=
f
(0)
=
8
=
.
4
3!
6
3
C z
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´
´
Example 4.15.3. Compute the integrals C 1/(z − z0 ) dz and C 1/(z − z0 )n+1 dz, n = 0, 1, 2, . . . , where C
is a positively oriented simple closed contour and z0 is any point interior to C.
Answer. By the Cauchy integral formula for derivatives with f (z) = 1, we have
ˆ
ˆ
1
1
dz = 2πi,
dz = (2πi)(n) = 0,
n = 0, 1, 2, . . . .
n+1
C z − z0
C (z − z0 )
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§4.16 Some Consequences of the Extension.
We study some important consequences of the Cauchy integral formula in the previous section.
Theorem 4.16.1. If a function f is analytic at a given point, then its derivatives of all orders are analytic
there too.
Proof. We claim that if f is analytic at z0 , then f ′ is also analytic at z0 . Since f is analytic at z0 , there is a
neighborhood |z − z0 | < ε of z0 throughout which f is analytic. We consider a positively oriented circle C0
centered at z0 with radius ε/2 such that f is analytic inside and on C0 . Then the Cauchy integral formula for
derivatives implies
ˆ
1
f (s)
′′
f (z) =
ds
πi C0 (s − z)3
´
at each point z interior to C0 . It means that both the integral C0 f (s)/(s − z)3 ds and f ′′ exist and they are
same throughout the neighborhood |z − z0 | < ε/2. Since f ′′ exists in the neighborhood, so f ′ should be
analytic in that neighborhood, particularly at z0 . Therefore, the claim has been proved. We can apply the
similar argument to deduce the analyticity of f ′′ from that of f ′ .
Corollary 4.16.2. If a function f (z) = u(x, y)+iv(x, y) is analytic at a point z = (x, y), then the component
functions u and v have continuous partial derivatives of all orders at that point.
Proof. For analytic f (z) = u(x, y) + iv(x, y), the theorem above implies that f ′ is analytic, and so differentiable and satisfies the Cauchy–Riemann equations ux = vy and uy = −vx and can be expressed by
f ′ = ux + ivx = vy − iuy .
The differentiability of f ′ implies the continuity of f ′ and so all first–order partial derivatives ux , uy , vx and
vy are continuous.
Again, since f ′ = ux + ivx = vy − iuy is analytic, f ′′ is also analytic by the theorem above. So f ′′ is
differentiable and satisfies the Cauchy–Riemann equations
(Re (f ′ ))x = (Im (f ′ ))y ,
(Re (f ′ ))y = − (Im (f ′ ))x ,
uxy = −vxx ,
vyy = uyx ,
i.e., uxx = vxy ,
and vyx = −uyy ,
(from f ′ = ux + ivx )
(from f ′ = vy − iuy )
and f ′′ can be expressed by
f ′′ = (ux + ivx )x = (vy − iuy )x = uxx + ivxx = vyx − iuyx = vxy − iuxy = −uyy − ivyy .
The differentiability of f ′′ implies the continuity of ′′ and so all second–order partial derivatives uxx , …, vyy
are continuous. We can continue on in this way and prove the statement in the corollary.
The proof of the following theorem is due to E. Morera.
´
Theorem 4.16.3. Let f be continuous on a domain D. If C f (z) dz = 0 for every closed contour C in D,
then f is analytic throughout D.
We recall the theorem in Section
48 Simply Connected Domains that if f is analytic throughout a
´
simply connected domain D, then C f (z) dz = 0 for any closed contour C in D. So assuming that f is
continuous, the theorem above gives the converse statement of the one given on the simply connected
domain. In a nutshell, combining those two theorem, we can say
´
Suppose f is continuous on a simply connected domain D. Then f is analytic in D if and only if C f (z) dz = 0
for any closed contour C in D.
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Proof. We recall from Section 44 Antiderivatives that when f is continuous on D, C f (z) dz = 0 for every
closed contour C in D if and only if f has an antiderivative in D. So there exists a function F such that
F ′ = f in D. That is, F is differentiable everywhere in the open set D, so F is analytic in D. By the theorem
above, F ′ is also analytic in D. Thus f is analytic in D.
Using the Cauchy integral theorem for derivatives, we can get the bounds of the derivatives.
Theorem 4.16.4 (Cauchy’s Inequality). Suppose that
1. CR is a positively oriented circle centered at z0 with radius R.
2. f is analytic inside and on CR .
3. MR is the maximum value of |f (z)| on CR , i.e., MR = max { |f (z)| : z ∈ CR }.
Then, we have the bounds of f (n) (z0 ),
(n)
f (z0 ) ≤ n!MR
Rn
(n = 0, 1, 2, . . .) .
Proof. By the Cauchy integral formula for derivatives,
ˆ
ˆ
(n)
n! f (s)
f (s)
f (z0 ) = n!
ds = ds
2πi
n+1
n+1
2πi
C (s − z0 )
CR (s − z0 )
ˆ R
f (s)
n!
ds
≤
2π CR (s − z0 )n+1 n!
· (Maximum of Integrand) · (Total Length of CR )
≤
2π
n!
≤
· (Maximum of Integrand) · (2πR) = n!R · (Maximum of Integrand).
2π
For s ∈ CR , |s − z0 | = R and so |(s − z0 )n+1 | = Rn+1 , which implies for s ∈ CR ,
f
(s)
MR
(s − z0 )n+1 ≤ Rn+1 .
Thus finally we get
(n)
f (z0 ) ≤ n!R · (Maximum of Integrand) ≤ n!R · MR = n!MR .
Rn+1
Rn
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§4.17 Liouville’s Theorem and the Fundamental Theorem of Algebra.
Cauchy’s inequality in the previous section implies the following theorem known as Liouville’s theorem.
Theorem 4.17.1 (Liouville’s Theorem). If a function f is entire and bounded in the complex plane, then f
is constant throughout the plane.
By the theorem, if f is entire, but not constant, then it should be not bounded. For example, f (z) =
sin z is entire (because f ′ (z) = cos z exists at any z) and clearly not constant, and so it should be unbounded.
Proof. Let f be entire and bounded by M in the complex plane, i.e., |f (z)| ≤ M for any z. Let us fix a point
z0 and consider a circle CR centered at z0 with radius R. Then by Cauchy’s inequality, we have
∥f ′ (z0 )∥ ≤
M
.
R
This inequality holds for any choice of CR , i.e., any radius R. It implies ∥f ′ (z0 )∥ = 0. Again since z0 is
arbitrary chosen, so ∥f ′ (z)∥ = 0 for any z. By the theorem in Section 24 Analytic Functions, f is a constant
function in the complex plane.
Using Liouville’s theorem, we can deduce the following theorem known as the fundamental theorem of
algebra.
∑
Theorem 4.17.2. Any polynomial P (z) = a0 + a1 z + a2 z 2 + · · · + an z n = nk=0 ak z k , an ̸= 0, of degree
n ≥ 1 has at least one zero. That is, there exists at least one point z0 such that P (z0 ) = 0.
The theorem says that a polynomial of degree n has no more than n distinct complex zeros/roots. By
the theorem, a polynomial P (z) of degree n has a zero, say z1 . Then we have P (z) = (z − z1 )Q(z), where
Q(z) is a polynomial of degree n − 1. Applying the theorem to Q(z), Q(z) has a zero, say z2 . So we have
Q(z) = (z − z2 )R(z), where R(z) is a polynomial of degree n − 2. Combining all together, we get
P (z) = (z − z1 )Q(z) = (z − z1 )(z − z2 )R(z),
where R(z) is a polynomial of degree n − 2. Continuing in this way, we arrive at
P (z) = c(z − z1 )(z − z2 ) · · · (z − zn ),
where c ̸= 0 is a constant. Thus, P (z) has no more than n distinct complex zeros/roots, z1 , z2 , . . . , zn .
Proof. Assume P (z) is not zero at any z and we deduce a contradiction. Since P (z) ̸= 0 for any z, we can
define f by f (z) = 1/P (z), which is clearly entire and bounded in the complex plane. We introduce
w=
a0
a1
a2
an−1
P (z)
− an = n + n−1 + n−2 + · · · +
n
z
z
z
z
z
so that P (z) = (an + w)z n . We choose R to be sufficiently large such that ∥ak ∥ ≤ ∥an ∥ Rn−k /(2n), k =
0, 1, 2, . . . , n − 1. Then we get for |z| > R,
a Rn−k ∥a ∥
Rn−k ∥an ∥
∥an ∥
k n
<
=
n−k ≤
n−k
n−k
z
2n ∥z ∥
2nR
2n
(k = 0, 1, 2, . . . , n − 1) .
It implies for |z| > R,
a ∥a ∥ ∥a ∥
a a a ∥an ∥
∥an ∥
n−1 0 1 2 n
n
+
+ ··· +
=
.
∥w∥ ≤ n + n−1 + n−2 + · · · + <
z
z
z
z
2n
2n
2n
2
The triangle inequality implies for |z| > R,
∥an + w∥ ≥ ∥∥an ∥ − ∥w∥∥ >
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and thus for |z| > R,
∥P (z)∥ = ∥an + w∥ ∥z n ∥ >
∥an ∥ n
R ,
2
∥f (z)∥ =
1
2
<
.
∥P (z)∥
∥an ∥ Rn
It means that the entire function f is bounded in the region exterior to the disk |z| ≤ R. Clearly f is
continuous in the closed disk |z| ≤ R, so f is bounded in that disk |z| ≤ R. (Recall from Section 18 Continuity
(Theorem 3 on page 54) that a continuous function on a compact set, i.e., closed and bounded set, is bounded
there.) Thus, f is bounded in the complex plane (exterior to |z| ≤ R and inside |z| ≤ R). Liouville’s theorem
implies that the bounded and entire f should be constant in the plane. However, since a polynomial cannot
be bounded in the plane, so f = 1/P cannot be bounded in the plane and we deduce a contraction.
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§4.18 Maximum Modulus Principle.
In Calculus, the absolute extrema on a closed interval can occur either at an interior point or at the
boundary points. However, in the theory of complex variables, under appropriate conditions, the absolute
extrema should occur only at the boundary points, otherwise the function should be constant.
Lemma 4.18.1. Suppose f is analytic in a neighborhood N (z0 , ε) = |z − z0 | < ε of z0 . If |f (z)| ≤ |f (z0 )|
for all z in N , then f is a constant function with value f (z0 ) in N , i.e, f (z) = f (z0 ) for all z in N .
Proof. Let f satisfy the conditions. Let us choose a point z1 ̸= z0 inside the neighborhood N . We claim
∥f (z1 )∥ = ∥f (z0 )∥ so that f is a constant function with value f (z0 ) in N .
To prove the claim, let ρ = |z0 − z1 |. Then since 0 < ρ < ε, the neighborhood N contains the circle
Cρ centered at z0 with radius ρ and the circle Cρ passes through z1 . The Cauchy integral formula says
ˆ
1
f (z)
dz.
f (z0 ) =
2πi Cρ z − z0
Applying the parametric representation z = z0 + ρeiθ , 0 ≤ θ ≤ 2π, for Cρ , the integral becomes
ˆ θ=2π
ˆ θ=2π
(
)
1
f (z0 + ρeiθ )
1
iθ
iθ
f (z0 ) =
ρie
dθ
=
f
z
+
ρe
dθ.
0
2πi θ=0 z0 + ρeiθ − z0
2π θ=0
• Remark. Gauss’s Mean Value Theorem: From Calculus, the average value fave of f on [a, b] is given by
ˆ b
ˆ
1
1
f (Point on Interval) dx.
fave =
f (x) dx =
b−a a
(Length of Interval) (Interval)
The last result above can be interpreted as
ˆ θ=2π
(
)
1
f (z0 ) =
f z0 + ρeiθ dθ
2π θ=0
ˆ
1
f (Point on Circle) dθ
=
(Total Length of Circle) (Circle)
= Mean/Average of f on the Circle.
That is, when a function is analytic within and on a given circle, its value at the center is the arithmetic
mean/average of its values on the circle. This result is called Gauss’s Mean Value Theorem. End of Remark
The last result before the remark gives
ˆ θ=2π
(
)
1
f z0 + ρeiθ dθ.
∥f (z0 )∥ ≤
2π θ=0
(
)
The assumption |f (z)| ≤ |f (z0 )| for all z ∈ N implies f z0 + ρeiθ ≤ ∥f (z0 )∥ and so
1
∥f (z0 )∥ ≤
2π
ˆ
θ=2π
θ=0
(
)
f z0 + ρeiθ dθ ≤ 1
2π
ˆ
θ=2π
θ=0
1
∥f (z0 )∥ dθ = ∥f (z0 )∥
2π
ˆ
θ=2π
1 dθ = ∥f (z0 )∥ .
θ=0
That is, we deduce
ˆ
(
)
f z0 + ρeiθ dθ,
θ=0
ˆ θ=2π
(
)
1
f z0 + ρeiθ dθ
0 = ∥f (z0 )∥ −
2π θ=0
ˆ θ=2π
ˆ θ=2π
(
)
1
1
f z0 + ρeiθ dθ
=
∥f (z0 )∥ dθ −
2π θ=0
2π θ=0
1
∥f (z0 )∥ =
2π
θ=2π
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1
=
2π
ˆ
θ=2π
(
)]
[
∥f (z0 )∥ − f z0 + ρeiθ dθ.
θ=0
´
From Calculus/Real Analysis, we recall a simple fact that if f (x) ≥ 0 on an interval
f (x) dx = 0,
I
( I andiθ )
then f (x) = 0 on the interval I. By the given condition, we have ∥f (z0 )∥ − f z0 + ρe
≥ 0 on the
circle Cρ and its integral is 0. Thus we conclude
(
(
)
)
∥f (z0 )∥ − f z0 + ρeiθ = 0, ∥f (z0 )∥ = f z0 + ρeiθ (0 ≤ θ ≤ 2π) .
It says ∥f (z0 )∥ = ∥f (z)∥ for all points z on the circle Cρ (|z − z0 | = ρ). Since z1 is arbitrary chosen,
∥f (z0 )∥ = ∥f (z)∥ for all points z in the neighborhood N . That is, the modulus |f (z)| is constant with value
|f (z0 )| in N . By the example 4 in Section 25, when the modulus of an analytic function is constant in a
domain, the function itself is constant there. Therefore, f (z) is a constant function in N with the value
f (z0 ).
Theorem 4.18.2 (Maximum Modulus Principle). If a function f is analytic and not constant in a given domain D, then |f (z)| has no maximum value in D. That is, there is no point z0 in the domain D such that
|f (z)| ≤ |f (z0 )| for all points z in D.
• Aside. Logic: Let p, q, r represent, respectively,
1. p := f is analytic in D,
2. q := f is constant in D,
3. r := |f (z)| has a maximum value in D.
Then the given theorem says, in logic notation, p∧ ∼ q =⇒∼ r. In logic, it has the following equivalent
relation,
p∧ ∼ q =⇒∼ r ⇐⇒ p ∧ r =⇒ q,
where the latter means that if f is analytic and |f (z)| has a maximum value in D, then f is constant in D.
We prove this latter equivalent statement.
End of Aside
Proof. Step 1. Setup: Suppose that f is analytic in D and |f (z)| has a maximum value at z0 in D. Let P be an
arbitrarily chosen point in D. Since D is a domain, i.e., a connected open set, we can find a polygonal path L
connecting z0 and P and entirely lying in D. Let d = min{ |z − w| : z ∈ ∂D, w ∈ L }, where ∂D means the
boundary of D. That is, d represents the shortest distance between the points on the boundary of D and the
points on the polygonal path L. (When D is the entire plane, d can be chosen to be any positive number.)
Now we choose points zk , k = 1, 2, . . . , n, on the path L such that |zk − zk−1 | < d, k = 1, 2, . . . , n, and
zn = P . Using the points, we form the neighborhood Nk of zk , k = 0, 1, . . . , n, such that Nk has the center
zk with radius d. Then we observe
1. each Nk entirely lies in D and so f is analytic on each Nk ,
2. the center zk of each neighborhood Nk lies in the neighborhood Nk−1 , i.e., zk ∈ Nk ∩ Nk−1 , k =
1, 2, . . . , n. (Remember Nk has the radius d and |zk − zk−1 | < d.)
Step 2. Applying Lemma:
(1) Since |f (z)| was assumed to have the maximum value in D at z0 , i.e., |f (z)| ≤ |f (z0 )| for all z ∈ D,
so clearly, |f (z)| ≤ |f (z0 )| for all z ∈ N0 . The lemma says f (z) = f (z0 ) throughout N0 . Since z1 ∈ N0 , so
f (z1 ) = f (z0 ).
(2) Since |f (z)| was assumed to have the maximum value in D at z0 , i.e., |f (z)| ≤ |f (z0 )| = |f (z1 )| for
all z ∈ D, so clearly, |f (z)| ≤ |f (z1 )| for all z ∈ N1 . The lemma says f (z) = f (z1 ) throughout N1 . Since
z2 ∈ N1 , so f (z2 ) = f (z1 ) and with the result in (1), we have f (z2 ) = f (z1 ) = f (z0 ).
(3) Since |f (z)| was assumed to have the maximum value in D at z0 , i.e., |f (z)| ≤ |f (z0 )| = |f (z1 )| =
|f (z2 )| for all z ∈ D, so clearly, |f (z)| ≤ |f (z2 )| for all z ∈ N2 . The lemma says f (z) = f (z2 ) throughout N2 .
Since z3 ∈ N2 , so f (z3 ) = f (z2 ) and with the results in (1) and (2), we have f (z3 ) = f (z2 ) = f (z1 ) = f (z0 ).
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Continuing the process, we deduce f (zn ) = f (zn−1 ) = · · · = f (z2 ) = f (z1 ) = f (z0 ), i.e., f has the
same value at z0 and zn = P . Since P is arbitrary chosen in D, thus we conclude that f is a constant function
with the value f (z0 ) in D and the proof is completed.
Corollary 4.18.3. Suppose that
1. f is continuous on a closed bounded region R.
2. f is analytic in the interior of R.
3. f is not constant in the interior of R.
Then |f (z)| has the maximum value in R, which is always reached and occurs on the boundary of R. The
maximum value can never occur at the interior point to R.
Proof. Suppose f is continuous on a closed bounded region R and analytic in the interior of R (i.e., assume
conditions 1 and 2 are satisfied). Then by the theorem in Section 18 Continuity (theorem 3 on page 54), |f (z)|
has a maximum value somewhere in R, i.e., there exists a M ≥ 0 such that |f (z)| ≤ M for all points z ∈ R
and the equality holds at least for one such z. The latter part means that there exists at least one point in
R, say z0 , such that |f (z0 )| = M , i.e., |f (z)| ≤ |f (z0 )| for all z ∈ R. However equipped with the condition
3, the Maximum modulus theorem implies that z0 cannot be an interior point of R. Hence, z0 should be on
the boundary of R and the maximum value occurs on the boundary of R.
Example 4.18.4. Let R denote the rectangular region 0 ≤ x ≤ π, 0 ≤ y ≤ 1. Find the point in R at which
the modulus |f (z)| of the function f (z) = sin z has the maximum value.
Answer. We recall
√ from Section 34 Trigonometric Functions on page 106 that f (z) = sin z = sin (x + iy)
and |f (z)| = sin2 x + sinh2 y .
Method 1. Using Corollary: The corollary says that the maximum value of |f (z)| occurs on the boundary
of R, i.e.,
∂R = { (0, y), (π, y), (x, 0), (x, 1) : 0 ≤ x ≤ π, 0 ≤ y ≤ 1 } .
Putting each point on the boundary to |f (z)|, we get for (x, y) ∈ ∂R,
(π )
|f (0, y)| = | sinh y| = |f (π, y)| ≤ sinh 1 ≈ 1.1752,
|f (x, 0)| = | sin x| ≤ sin
= 1,
2
√
( )
√
√
2
2
2 π
+ sinh2 1 = 1 + sinh2 1 ≈ 1.47486.
|f (x, 1)| = sin x + sinh 1 ≤ sin
2
That is, |f (z)| has the maximum value |f (π/2, 1)| ≈ 1.47486√which occurs at the point (π/2, 1) on the
boundary of R. Method 2. Direct Computation: Since |f (z)| = sin2 x + sinh2 y is a real–valued function
of two real variables x and y, we find the maximum value using Calculus. In fact, since the square root is
an increasing function, so both |f (z)| and |f (z)|2 = sin2 x + sinh2 y have the maximum value at the same
point (x, y). We look for the maximum value of |f (z)|2 . It is easy to see that sin2 x has the maximum value
sin2 (π/2) = 1 at x = π/2 ∈ [0, π] and sinh2 y has the maximum value sinh2 1 at y = 1 ∈ [0, 1]. Therefore,
2
2
|f (z)|2 has the maximum
√ value sin (π/2) + sinh 1 ≈ 2.1752 at (x, y) = (π/2, 1) ∈ ∂R and thus |f (z)| has
the maximum value
see the figure 4.10.
sin2 (π/2) + sinh2 1 ≈ 1.47486 at (x, y) = (π/2, 1) ∈ ∂R. For the graph of sinh x,
Corollary 4.18.5. Let f (z) = u(x, y) + iv(x, y) satisfy all the conditions of the corollary above and R be
the same region as given in the corollary above. Then the maximum value of |u(x, y)| in R, which is always
reached, occurs on the boundary of R. The maximum value can never occur at the interior point to R.
We recall that for analytic f , u = Re (f ) is harmonic at the interior point of the domain R. The
statement above says that the maximum value of such a harmonic function occurs on the boundary.
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.y
.y = sin x
.1
.
.−1
.x
.1
.−1
.y = sinh x
Figure 4.10: Graphs of y = sin x and y = sinh x.
Proof. Consider a composite function g = ef . Since the exponential function is entire, so g is also analytic
in R and satisfies all the same conditions as f does. Because the exponential function is always increasing,
|g(z)| = |eu+iv | = eu |eiv | = eu implies that both |g(z)| and u have the maximum value at the same point.
Applying the corollary above to g(z), the modulus |g(z)| has the maximum value on the boundary of R and
therefore, u also has the maximum value on the boundary of R.
Theorem 4.18.6 (Minimum Modulus Principle). Suppose that
1. f is continuous on a closed bounded region R.
2. f is analytic and not constant throughout the interior of R.
3. f (z) =
̸ 0 anywhere in R.
Then |f (z)| has minimum value in R which occurs on the boundary of R. The minimum value can never
occur at the interior point to R.
Proof. We skip the proof. See the exercise 3 on page 179.
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