Class note on basics of complex analysis
Alok Pan
I.
COMPLEX NUMBER
A complex number z can be written as z = (a + i b), where the real part is a and the imaginary
part is b i.e. a = Re(z),
b = Im(z). Hence, we can write the complex number z as, z = Re(z) +
i Im(z).
• Complex conjugate: z∗ = Re(z) − i Im(z)
• Modulus: |z|2 = z · z∗ = (a + i b)(a − i b) = a2 + b2
A.
Properties
(i) Addition of complex numbers: z1 + z2 = (a + i b) + (c + i d) = (a + c) + i (b + d) = z3
(ii) Equality of complex numbers: z1 = z2 means that Re(z1 ) = Re(z2 ) and Im(z1 ) = Im(z2 )
(iii) Multiplication of complex numbers: z1 · z2 = (a + i b)(c + i d) = (ac − bd) + i (bc + ad) = z3
(iv) Division of complex numbers: zz12 is defined only if z2 , 0.If z2 = 0 then c = 0 and d = 0.
z1 a + i b (a + i b)(c − i d) (ac + bd) + i (bc − ad)
=
=
=
z2 c + i d (c + i d)(c − i d)
|z2 |2
(v) Order of complex numbers: If there are two complex numbers z1 and z2 , then z1 > z2 is
meaningless. The modulus of the complex numbers can be ordered i.e., if z1 is greater than
z2 , then we can write |z1 | > |z2 |.
(vi) Re(z), Im(z) ≤ |z|
∗
∗
(vii) Re(z) = z+z2 , Im(z) = z−z2
(viii) |z1 · z2 | = |z1 | · |z2 |
|z1 · z2 | = |(a + i b)(c + i d)| = |(ac − bd) + i (ad + bc)| =
√
√
= a2 + b2 · c2 + d2
= |z1 | · |z2 |
p
(ac − bd)2 + (ad + bc)2
2
(ix) Triangle Inequalities: |z1 + z2 | ≤ |z1 | + |z2 |
|z1 + z2 |2 = (z1 + z2 ) · (z∗1 + z∗2 )
= |z1 |2 + |z2 |2 + z1 z∗2 + z2 z∗1 = |z1 |2 + |z2 |2 + z1 z∗2 + (z∗2 z1 )∗
= |z1 |2 + |z2 |2 + 2 Re(z1 z∗2 )
≤ |z1 |2 + |z2 |2 + 2 |z1 z∗2 | = |z1 |2 + |z2 |2 + 2 |z1 | |z∗2 | = (|z1 | + |z2 |)2
This implies that |z1 + z2 | ≤ |z1 | + |z2 | General form of Triangle inequality
X
B.
zi ≤
X
|zi |
Geometrical Interpretation
Complex Plane:
z3 ≤ |z1 | + |z2 |
|z1 + z2 | ≤ |z1 | + |z2 |
Complex number in Polar Co-ordinate The complex number z = a + i b is represented by the
point (a, b) in the xy plane. In the polar coordinate, it is given by (r, θ) where
√
r = |z| = a2 + b2
!
b
−1 b
tan θ =
=⇒ θ = tan
a
a
Here,we have a = r cos θ, b = r sin θ, which implies that
z = a + ib = r cos θ + i r sin θ = r(cos θ + i sin θ)
3
Using the series expansion formula, we have
θ2 θ3 θ4 θ5
+ + + +−−−−−−−−−−−
2! 3! 4! 5!
θ2 iθ3 θ4 iθ5
+ +
+−−−−−−−−−−−
eiθ = 1 + iθ − −
2! 3! 4! 5!
θ2 θ4
θ3 θ5
= (1 − + − − − − − −−) + i(θ − + + − − − − − − −)
2! 4!
3! 5!
= cos θ + i sin θ
eθ = 1 + θ +
The complex number is given by z = reiθ , where |z| = r, arg(z) = θ. Also, we have z∗ = re−iθ
Multiplication of complex numbers in polar form:
z1 = r1 eiθ1 ,
z2 = r2 eiθ2 =⇒ z1 z2 = r1 r2 ei(θ1 +θ2 )
Similarly, we find that z1 z∗1 = r12 ei(θ1 −θ1 ) ,
II.
Ex.1 :
|z1 |2 = r12
PLOT IN THE COMPLEX PLANE
|z| = 2, a2 + b2 = 4
FIG. 1: |z| = 2
Ex.2 :
4 ≤ |2z − 6| ≤ 8 =⇒ 1 ≤ |z − 3| ≤ 2
p
=⇒ |a + ib − 3| = |(a − 3) − ib| = (a − 3)2 + b2
1 ≤ (a − 3)2 + b2 ≤ 4
This represents the circle with center (3, 0) and the the modulus of z i.e., r is given by 1 < r < 2.
Hence the required region is expressed as
4
FIG. 2: 4 ≤ |2z − 6| ≤ 8
Convert the complex numbers to rectangular form:
1. Example 1: eiπ
eiπ = cos π + i sin π = 1
√
iπ
2. Example 2: 2 · e 4
√
!
π √
√ π
1
1
2 · e = 2 cos
+ i sin
= 2 √ +i√ =1+i
4
4
2
2
iπ
4
De Moivre’s Theorem: (cos θ + i sin θ)n = cos(nθ) + i sin(nθ)
Proof: First, we show that (cos θ + i sin θ)2 = cos(2θ) + i sin(2θ) is true.
eiθ = cos θ + i sin θ
eiθ · eiθ = = (cos θ + i sin θ) · (cos θ + i sin θ)
(eiθ )2 = cos2 θ − sin2 θ + 2i cos θ sin θ
(cos θ + i sin θ)2 = cos(2θ) + i sin(2θ)
Similarly, we can check that
eiθ · eiθ · eiθ = ei3θ = cos(3θ) + i sin(3θ)
which shows that
(cos θ + i sin θ)3 = cos(3θ) + i sin(3θ)
This finally shows that
einθ = (cos θ + i sin θ)n = cos(nθ) + i sin(nθ)
5
Properties of Arguments:
arg(z1 z2 ) = arg(z1 ) + arg(z2 )
We have z1 z2 = r1 r2 ei(θ1 +θ2 ) . Hence we have
arg(r1 r2 ei(θ1 +θ2 ) ) = θ1 + θ2 = arg(z1 ) + arg(z2 )
Roots of a Complex Number: z = reiθ . We can write
rei(2π+θ) = rei2π eiθ = reiθ (cos(2π) + i sin(2π)) = reiθ
(1)
This shows that z = reiθ = rei(θ+2nπ) . Hence, r = |z|, θ = arg(z). This shows that arg(z) =
FIG. 3: arg(z) = θ + 2nπ
θ + 2nπ,
−π ≤ arg(z) ≤ π.
Example 1: Let’s Plot ”i”
FIG. 4: Plot ”i”
i = cos
π
π
+ i sin
= eiπ/2 = ei(π/2+2nπ)
2
2
For n = 1, we have i = e5π/2 . Similarly, we can show that
−1 = eiπ
eiπ + 1 = 0
6
FIG. 5: Root of complex numbers
Roots of a Complex Number
z = reiθ = rei(θ+2nπ)
z1/m = r1/m ei(θ/m+2nπ/m)
θ
(2)
For n = m, we get z1/m = r1/m ei( m +2π) . Hence here we must have n < m. In general, n =
0, 1, 2, . . . , (m − 1).
Example: For m = 3, n = 0, 1, 2.
First root: For n = 0, we have z10 = r1/3 ei(θ/3) .
Second root: For n = 1, we have z20 = r1/3 ei(θ/3+2π/3)
Third root: For n = 2, we have z30 = r1/3 ei(θ/3+4π/3)
Example: If z = i, then find z1/2 .
Solution: z = i implies that z = eiπ/2 . Using Eq. (2), we can write
π
π
z1/2 = ei( 4 + 2 ) = ei( 4 +nπ) ,
2nπ
n = 0, 1, 2, . . . , (m − 1)
Here n = 0, 1 as m = 2
√ ,
First root: z10 = eiπ/4 = 1+i
2
π
5π
√
Second root: z20 = ei( 4 +π) = cos 5π
+
i
sin
= − 1+i
4
4
2
Example: If z = −1, then find z1/3 .
Solution: z = i implies that z = ei(π+2nπ) Using Eq. (2), we can write
π
z1/3 = ei( 3 + 3 )
2nπ
7
FIG. 6: Graph for z1/2
FIG. 7: Graph for z1/3
Here, m = 3, which implies that n = 0, 1, 2.
√
First root: z10 = eiπ/3 = 12 + i 23
π 2π
Second root: z2 = ei( 3 + 3 ) = eiπ = −1
0
π
√
Third root: z30 = ei( 3 + 3 ) = ei 3 = ei2π − eiπ/3 = 12 − i 23
4π
5π
8
III.
COMPLEX ANALYSIS
Neighbouhood of a point: Let us consider a point z0 in the complex plane. The neighborhood
of the point z0 is defined by
Nδ (z0 ) = {z : |z − z0 | < δ}
Complex function: The funcion f : z → w is a complex function if f (z) = w = u + iv where
z = x + iy.
FIG. 8: Complex function
Example: w = f (z) = z2 + 2z
f (z) = z2 + 2z
= x2 + (iy)2 + 2ixy + 2x + 2iy)
= (x2 − y2 + 2x) + i(2xy + 2y)
= u(x, y) + iv(x, y)
(3)
where u(x, y) = x2 − y2 + 2x, and v(x, y) = 2xy + 2y.
Limit of a Complex function: Let f (z) be a single-valued function, defined in all points of the
neighborhood of z0 . The function f (z) is said to have the limit L as z → z0 along any path, if for
a given ϵ > 0, ∃ a δ > 0 such that
| f (z) − L| < ϵ
∀z : 0 < |z − z0 | < δ
2
2
Example: Let us consider z = x + iy. The function is defined as |z|z2 . Let us check if limz→0 |z|z2
9
exists or not. Let us assume
|z|2
f (z) =
Re(z)2
x2 + y2
lim 2
= ( Case I)
x→0,y→0 x − y2
y2
= −1
x=0,y→0 −y2
x2
= (Case II)
lim 2 = 1
x→0,y=0 x
x2 (1 + m2 ) 1 + m2
=
(when y = mx, x → 0, y → 0) = (Case III)
lim 2
x→0,y→0 x (1 − m2 )
1 − m2
which is different for different values of m
lim
2
(4)
2
|z|
|z|
Since, limz→0 Re(z
2 ) does not have a unique value, the limit does not exist. Hence, limz→0 z2 does
not exist.
Continuity: A complex function is said to be continuous at a point z0 if f (z) is defined in every
neighborhood of z0 and for every ϵ > 0, there exists a δ > 0 such that
| f (z) − f (z0 )| < ϵ
∀z : 0 < |z − z0 | < δ
This implies that
lim f (z) = f (z0 )
z→z0
Multivalued function: Let us consider the function f (z) = z1/2 ,
z = |z|eiθ .
If we consider z → −1, then we get the following
lim z1/2 =⇒ lim |z|eiθ/2
z→−1
|z|→1
(5)
10
Case 1: if θ → π,
then
Case 2: if θ → −π,
then
lim |z|eiθ/2
= eiπ/2 = i
lim |z|eiθ/2
= e−iπ/2 = −i
|z|→1
|z|→1
Derivative Let us assume that the function f (x) is continuous. From the definition of differentiation, we can write
f (x + ∆x) − f (x)
∆x→0
∆x
f ′ (x) = lim
If f (z) is a single-valued complex function, by replacing x by z, we get
f (z + ∆z) − f (z)
,
∆z→0
∆z
f ′ (z) = lim
∆z = ∆x + i∆y
(6)
Clearly, when ∆z → 0, we have ∆x → 0, ∆y → 0. This implies that
′
f (x + ∆x + iy + i∆y) − f (x + iy)
∆z→0
∆x + i∆y
f (z) = lim
Example 1: Let us consider the function f (z) = z2 . Find the derivative of f (z).
Solution: From the definition, we can write
f (z) = z2
f (z + ∆z) − f (z)
∆z→0
∆z
(z + ∆z)2 − z2
= lim
∆z→0
∆z
z2 + 2z∆z + (∆z)2 − z2
= lim
∆z→0
∆z
= 2z
f ′ (z) = lim
(7)
11
Example 2: Let us consider the function f (z) = z∗ . Find the derivative of f (z).
Solution: From the definition, we can write
f (z) = z∗
f (z∗ + ∆z∗ ) − f (z∗ )
∆z→0
∆z
z∗ + ∆z∗ − z∗
= lim
∆z→0
∆z
∗
∆z
= lim
∆z→0 ∆z
∆x − i∆y
= lim
∆z→0 ∆x + i∆y
f ′ (z) = lim
∆x
=1
∆x→0,∆y=0 ∆x
−i∆y
= −1
Case 2: ∆z → 0 =⇒
lim
∆x=0,∆y→0 i∆y
Case 1: ∆z → 0 =⇒
lim
This shows that the limit doesn’t exist, implying that f (z) is not derivable.
(8)
12
IV.
ANALYTIC FUNCTION
Analytic function: If the derivative of the function f (z) exists at the point z0 and every point
of the neighborhood of z0 , then the function f (z) is called an analytic function.
If f (z) is analytic at z0 then f ′ (z) exists for any z such that |z − z0 | < δ.
Theorem 1. If a complex function is analytic, it satisfies the Cauchy-Reimann relation ( necessary
condition for being an analytic function).
Proof: The derivative of the function f (z) is defined as follows:
f (z + ∆z) − f (z)
∆z→0
∆z
f (x + ∆x + iy + i∆y) − f (x + iy)
= lim
∆z→0
∆x + i∆y
′
f (z) = lim
(9)
Since, f (z) = u(x, y) + i v(x, y), we have
f (z + ∆z) = u(x + ∆x, y + ∆y) + i v(x + ∆x, y + ∆y)
Taking ∆z → 0 and∆y = 0, ∆x → 0, we have
f (z + ∆z) = u(x + ∆x, y) + i v(x + ∆x, y)
Putting all these in Eq. (9) we get,
′
u(x + ∆x, y) + i v(x + ∆x, y) − (u(x, y) + i v(x, y))
∆x→0
∆x
v(x + ∆x, y) − v(x, y)
u(x + ∆x, y) − u(x, y)
+i
= lim
∆x→0
∆x
∆x
∂u
∂v
+i
=
∂x
∂x
f (z) = lim
(10)
Now taking ∆z → 0 and ∆x = 0, ∆y → 0 we get,
′
f (x + y + i∆y) − f (x + iy)
∆y→0
i∆y
u(x, y + ∆y) + i v(x, y + ∆y) − (u(x, y)
= lim
∆y→0
i∆y
u(x, y + ∆y) − u(x, y) + i v(x, y)
v(x, y + ∆y) + v(x, y)
= lim
+i
∆y→0
i∆y
i∆y
∂v
∂u
=
−i
∂y
∂x
f (z) = lim
(11)
13
From Eq. (10) and Eq. (11) we get,
∂u
∂v ∂v
∂u
+i
=
−i
∂x
∂x ∂y
∂x
(12)
Comparing real and complex part from Eq. (12) we get,
∂u ∂v
=
=⇒ u x = vy
∂x ∂y
∂v
∂u
=−
=⇒ v x = − uy
∂x
∂x
(13)
Eq. (13) are the Cauchy-Rieman equations. Hence Proved.
Theorem 2. If CR relations are satisfied then the function is an analytic function (sufficient condition).
Proof: The derivative of the function f (z) is
′
f (z + ∆z) − f (z)
∆z→0
∆z
f (z) = lim
(14)
Clearly,
f (z + ∆z) = u(x + ∆x, y + ∆y) + i v(x + ∆x, y + ∆y)
Taylor’s theorem for two variable functions is given by
!
∂g
∂g
g(x + h, y + k) = g(x, y) + h + k
+ (h2 + · · · · · · )
∂x
∂y
This implies that
∂u
+
∆y
u(x + ∆x, y + ∆y) ≈ u(x, y) + ∆x ∂u
∂x
∂y
∂v
∂v
v(x + ∆x, y + ∆y) ≈ v(x, y) + ∆x ∂x + ∆y ∂y
(15)
(16)
Using it, we find that
!
∂u
∂v
∂u
∂v
f (z + ∆z) = u(x, y) + iv(x, y) + ∆x + ∆y + i ∆x + ∆y
∂x
∂y
∂x
∂y
!
!
∂u
∂v
∂u
∂v
= f (z) + ∆x
+i
+ ∆y
+i
∂x
∂x
∂y
∂y
!
!
∂v
∂v
∂u
∂u
= f (z) + ∆x
+i
+ ∆y − + i
(Using CR equations)
∂x
∂x
∂x
∂x
!
!
∂u
∂v
∂u
∂v
= f (z) + ∆x
+i
+ i∆y
+i
∂x
∂x
∂x
∂x
!
∂u
∂v
= f (z) +
+i
(∆x + i∆y)
∂x
∂x
!
∂u
∂v
= f (z) +
+i
∆z
∂x
∂x
(17)
14
This implies that
∂v
f (z + ∆z) − f (z) ∂u
=
+i
∆z→0
∆z
∂x
∂x
(18)
lim
Example 1: ez = e x+iy = e x (cos y + isin y). Show that the function is an analytic function.
Solution: u(x, y) = e x cos y
∂u
= e x cos y,
∂x
v(x, y) = e x sin y.
∂v
= e x cos y,
∂y
∂u
= − e x sin y,
∂y
∂v
= e x sin y
∂x
(19)
This shows that u x = vy and v x = − uy . Thus the CR equations are satisfied. Hence the given
function is analytic.
Example 2: If f (z) = x2 + i y2 , find where the function f is analytic.
Solution: For the given function, we have u = x2 , v = y2 .
The function should satisfy the CR equations to be ananlytic. We have u x = 2x, uy = 0, vy =
2y, v x = 0. To satisfy CR equations, we should have u x = vy . If we plot u x vs vy we will get, This
shows that the function is analytic for x = y.
Example 3: If f (z) = x2 + i y3 , find where the function f is analytic.
Solution: For the given function, we have u = x2 , v = y3 .
The function should satisfy the CR equations to be analytic. We have u x = 2x, uy = 0, vy =
3y2 , v x = 0. To satisfy CR equations, we should have u x = vy . This gives that x = 32 y2 . This shows
that the function is analytic for the points on the parabola x = 23 y2 .
Example 4: Proof f (z) = z∗ is not differentiable.
Solution: Solve it using CR relations.
A curious feature: Let us assume f (z) = u(x, y) + i v(x, y). If ∂∂zf (z)
∗ = 0, then it satisfies CR
equations, provided u x , uy , v x , and vy are continuous over a region.
15
∂f
∂u ∂x ∂u ∂y
∂v ∂x ∂v ∂y
=
+
+i
+
∗
∗
∗
∂z
∂x ∂z
∂y ∂z
∂x ∂z∗ ∂y ∂z∗
∗
!
(20)
∗
We know that, x = z+z2 and y = z−z
. Using it, we get
2i
!
!
!
!!
∂v 1
∂f
∂u 1
∂u −1
∂v −1
=
+
+i
+
∂z∗ ∂x 2
∂y 2i
∂x 2
∂y 2i
v x vy
u x uy
− +i −
=
2 2i
2
2
i
1
=
u x − vy + v x + uy
2
2
(21)
Since we have ∂∂zf (z)
∗ = 0, we get
i
1
u x − vy + v x + uy = 0
2
2
(22)
Equating the real and imaginary parts of both sides, we get
u x − vy = 0 =⇒ u x = vy ,
v x + uy = 0 =⇒ v x = −uy
This implies that the CR equations are satisfied.
Entire function: A function is said to be an Entire function if it is analytic at every point of
the complex plane. Example: ez , z2 , sin z, cos z, zn etc.
1
Note: f (z) = z−a
is not analyitc at z = a.
Harmonic function:
The complex function is given by f (z) = u(x, y) + i v(x, y). The CR equations are satisfied which
implies that u x = vy , uy = −v x . Hence we get
∂2 u
∂u ∂v
∂2 v
=
=⇒
=
∂x ∂y
∂x2 ∂x∂y
∂u
∂v
∂2 u
∂2 v
=−
=⇒
=
−
∂y
∂x
∂y2
∂x∂y
(23)
Adding these, we get,
∂2 u ∂2 u
+
=0
∂x2 ∂y2
(24)
∂2 v ∂2 v
+
=0
∂x2 ∂y2
(25)
Similarly, we can write,
16
This shows that u(x, y) and v(x, y) are Harmonic functions. Hence we can write it for ϕ where
ϕ(u, v).
∇2 ϕ = 0
(26)
Example 1: If z = u(x, y) + iv(x, y) is an analytic function with u(x, y) = x2 − 2xy + 3y. Find z.
Solution: Given that u(x, y) = x2 − 2xy + 3y. This implis that
u x = 2x − 2y,
uy = −2y + 3.
(27)
The function is analytic which implies it satisfies CR equations. This gives that
∂v ∂u
=
= 2x − 2y
∂y ∂x
v = 2xy − y2 + g(x) + c
∂v
′
= 2y + g (x)
∂x
(28)
From CR equations, we get
∂v
∂u
=−
∂x
∂y
′
2y + g (x) = 2x − 3
(29)
′
g (x) = 2x − 2y − 3
g(x) = x2 − 2xy − 3x
Putting the value of g(x) in Eq. (28) we get, v(x, y) = x2 − y2 − 3x + c. Hence, we get z =
(x2 − 2xy + 3y) + i (x2 − y2 − 3x + c). Again for u, we have
∂2 u
∂2 u
∂2 u ∂2 u
=
2,
=
−2
=⇒
+
=0
∂x2
∂y2
∂x2 ∂y2
(30)
∂2 v
∂2 v ∂2 v
=
−2
=⇒
+
=0
∂y2
∂x2 ∂y2
(31)
Again for v, we have
∂2 v
= 2,
∂x2
This shows that u(x, y) and v(x, y) both are harmonic functions.
CR equation in Polar term
Let us consider z = ϵeiα . This implies that
δz = ϵ exp{iα}
= ϵ cos α + iϵ sin α
= δx + iδy
17
where
δx = ϵ cos α,
δy = ϵ sin α
For the function f (z) = u + iv, we get
δ f = δu + iδv
∂u
∂u
δu = ∂x + ∂y
∂x
∂y
∂u
∂u
∂v
∂v
= δx + δx + i( δx + δy)
∂x
∂y
∂x
∂y
= u x δx + uy δuy + i(vy δx + v x δy)
= (u x ϵ cos α + ivy ϵ sin α) + (vy ϵ sin α + iv x ϵ cos α)
z = r exp{iθ} = r cos θ + r sin θ
= u(r, θ) + v(r, θ)
∂u
∂u
du = dx + dy
∂x
∂y
du ∂u ∂x ∂u ∂y
=
+
dθ ∂x ∂θ ∂y ∂θ
= u x (−r sin θ) + uy (r cos θ)
(32)
= r[uy cos θ − u x sin θ]
∂x
∂y
du
= ur = u x + uy = u x cos θ + uy sin θ
dr
∂r
∂r
dv ∂v ∂x ∂v ∂y
vθ =
=
+
dθ ∂x ∂θ ∂y ∂θ
= v x (−r cos θ) + vy (r cos θ)
= r[vy cos θ − v x cos θ] = r[u x cos θ + uy sin θ]
dv
= v x cos θ + vy sin θ
dr
(33)
Comparing ur , vθ and vθ , vr , we get
rur = r[u x cos θ + uy sin θ] = vθ
rvr = −uθ
(34)
18
V.
LINE INTEGRATION OF A COMPLEX VARIBALE
Let us consider the complex number z = x + iy such that dz = dx + idy. The function is given
by f (z) = u + iv. The line integration along a given path C is given by
Z
f (z)dz =
Z
C
(u + iv)(dx + idy) =
C
Z
(udx − vdy) + i(vdx + udy)
C
Example 1: f (z) = z = x + iy.
Here, f (z) = x + iy =⇒ u = x, v = y.
We calculate the line integral from (0, 0) to (1, 1) along three paths. Path I along the lines L1
and L2 , path II along the lines y = x, and Path III along the lines L3 and L4 .
Path IL1 : x = 0 =⇒ dx = 0
Z
f (z)dz =
L1
Z
(u + iv)(dx + idy)
C
=
Z
(udx − vdy) + i(vdx + udy)
L1
=
Z
(xdx − ydy) + i(ydx + xdy)
L1
" 2 #1
1
y
=− .
=
−ydy = −
2 0
2
0
Z 1
L2 : y = 1 =⇒ dy = 0
19
Z
f (z)dz =
L2
=
=
Z
(u + iv)(dx + idy)
ZC
(udx − vdy) + i(vdx + udy)
ZL2
(xdx − ydy) + i(ydx + xdy)
L2
=
Z 1
0
" 2
#1
x
1
xdx + idx =
+ ix = + i.
2
2
0
Hence, considering path I, we get
Z
1 1
f (z)dz = − + + i = i.
2 2
C
Path II
y = x =⇒ dy = dx
Z
f (z)dz =
y=x
Z
(u + iv)(dx + idy)
C
=
Z
(udx − vdy) + i(vdx + udy)
y=x
=
Z
(xdx − ydy) + i(ydx + xdy)
y=x
" 2 #1
x
=
2ixdx = 2i
= i.
2 0
0
Z 1
Hence, considering path II, we get
Z
f (z)dz = i.
C
Path IIIL3 : y = 0 =⇒ dy = 0
Z
f (z)dz =
L3
Z
(u + iv)(dx + idy)
C
=
Z
(udx − vdy) + i(vdx + udy)
L3
=
Z
(xdx − ydy) + i(ydx + xdy)
L3
=
Z 1
0
" 2 #1
1
x
=− .
xdx = −
2 0
2
L4 : x = 1 =⇒ dx = 0
20
Z
f (z)dz =
Z
L4
(u + iv)(dx + idy)
C
=
Z
(udx − vdy) + i(vdx + udy)
L4
=
Z
(xdx − ydy) + i(ydx + xdy)
L4
#1
1
y2
=
(−y + i)dy = − + iy = − + i.
2
2
0
0
Z 1
"
Hence, considering path III, we get
Z
1 1
− + i = i.
2 2
f (z)dz =
C
Example 2: f (z) = z2 = (x2 − y2 ) + 2ixy.
Here, f (z) = (x2 − y2 ) + 2ixy =⇒ u = x2 − y2 , v = 2xy.
We calculate the line integral from (0, 0) to (1, 1) along two paths. Path I along the lines L1 and
L2 , path II along the lines y = x.
Path IL1 : x = 0 =⇒ dx = 0
Then u = −y2 , v = 0.
Z
f (z)dz =
Z
L1
(u + iv)(dx + idy)
L1
=
Z
(udx − vdy) + i(vdx + udy)
L1
" 3 #1
y
i
=
−iy dy = −i
=− .
3 0
3
L1
Z
2
L2 : y = 1 =⇒ dy = 0
21
Then u = x2 − 1, v = 2x.
Z
f (z)dz =
Z
L2
(u + iv)(dx + idy)
L2
=
Z
(udx − vdy) + i(vdx + udy)
L2
=
=
Z
(x2 − 1)dx + 2xidx
L2
Z 1
(x2 + 2xi − 1)dx
0
" 3
#1
1
x
2
2
=
+ ix − x = + i − 1 = − + i.
3
3
3
0
Hence, considering path I, we get
Z
2
i 2
f (z)dz = − − + i = (i − 1).
3 3
3
C
Path II
y = x =⇒ dy = dx
Then u = 0, v = 2ix2 .
Z
f (z)dz =
Z
y=x
(u + iv)(dx + idy)
y=x
=
Z
(udx − vdy) + i(vdx + udy)
y=x
" 3 #1
x
2
=
−2ix dx − 2x dx = 2(i − 1)
= (i − 1).
3 0 3
y=x
Z
2
2
Hence, considering path II, we get
Z
2
f (z)dz = (i − 1).
3
C
Example 3: f (z) = z∗ = x − iy.
Here, f (z) = x + iy =⇒ u = x, v = −y.
We calculate the line integral from (0, 0) to (1, 1) along two paths. Path I along the lines L1 and
L2 , path II along the lines y = x.
Path IL1 : x = 0 =⇒ dx = 0
22
Z
Z
f (z)dz =
L1
(u + iv)(dx + idy)
C
Z
=
(udx − vdy) + i(vdx + udy)
L1
Z
=
(xdx + ydy) + i(−ydx + xdy)
L1
" 2 #1
y
1
=
ydy =
= .
2 0 2
0
Z 1
L2 : y = 1 =⇒ dy = 0
Z
f (z)dz =
L2
Z
(u + iv)(dx + idy)
C
=
Z
(udx − vdy) + i(vdx + udy)
L2
=
Z
(xdx + ydy) + i(−ydx + xdy)
L2
=
Z 1
0
" 2
#1
1
x
xdx − idx =
− ix = − i.
2
2
0
Hence, considering path I, we get
Z
C
f (z)dz =
1 1
+ − i = 1 − i.
2 2
Path II
y = x =⇒ dy = dx
23
Z
f (z)dz =
y=x
Z
(u + iv)(dx + idy)
C
=
Z
(udx − vdy) + i(vdx + udy)
y=x
=
Z
(xdx + ydy) + i(ydx + xdy)
y=x
=
=
Z
(xdx + xdx) + i(−xdx + xdx)
y=x
Z 1
h i1
2xdx = x2 = 1.
0
0
Hence, considering path II, we get
Z
f (z)dz = 1.
C
Since the line integrals are dependent on the path, the given function is not analytic. Similarly,
if we consider f (z) = 1/z, which is not analytic (CR equations are not satisfied), the line integrals
will not be path-independent.
VI.
COMPLEX INTEGRATION FOR A CLOSED REGION
Simply connected region: A region C is said to be a simply connected region if any closed
curve lying inside C can be shrunk to a single point in C. In other words, C does not have any
holes.
Multiply connected region: A region C is multiply connected region if it is not simply connected, implying that every closed curve inside C does not only enclose points of C.
For example, the set of points of the interior of a closed curve is simply connected region but
the annular of two concentric circles is a multiply connected region.
Holomorphic function: A function f is said to be a holomorphic funtion if it is differentiable
at every point of the domain C.
Cauchy’s theorem: If f (z) is analytic in every point of a simply connected region R and C is a
H
closed contour in R, then C f (z)dz = 0.
Extension of Cauchy’s theorem: Let a closed contour C contain another contour C1 and the
function f (t) is analytic in the annular region between C and C1 .
24
I
f (z)dz = 0
Z
Z
= 0
f (z)dz + f (z)dz +
f (z)dz + f (z)dz
ABEFG
Z
ZGH
Z HI J
ZJA
f (z)dz = −
f (z)dz =
f (z)dz =
f (z)dz
Z
Z
C
HI J
JIH
C1
This shows that
Z
f (z)dz =
Z
f (z)dz.
C
C1
Cauchy’s integral formula: If f (z) is a holomorphic function i.e., f ′ (z) is exists at each point
on a closed curve C then for point a inside the closed curve C,
1
f (a) =
2πi
I
f (z)
dz
C z−a
25
For the circle C1 , we have
|z − a| = r < 1
z−a
= reiθ
z = a + reiθ
dz = ireiθ dθ
Since, f is analytic as each point of C and C1 is contained inside it, using the extension of
Cauchy’s theorem, we get
I
f (z)
dz =
C z−a
Z
f (z)
dz
C1 z − a
Now we have
f (a + reiθ ) iθ
ire dθ
iθ
C1 a + re − a
Z
f (a + reiθ ) iθ
=
ire dθ
reiθ
C1
Z
=
i f (a)dθ
C1
Z
= i f (a)
idθ
Z
f (z)
dz =
C1 z − a
Z
(For r → 0)
C1
= 2πi f (a)
Z
f (z)
1
dz
f (a) =
2πi C1 z − a
I
1
f (z)
=
dz.
2πi C z − a
Hence proved.
Cauchy’s integral formula for multiply connected region:
If f (z) is an analytic function in a region enclosed by curve C1 and C2 and a is a point in
between the regions C1 and C2 . Then
1
f (a) =
2πi
Z
f (z)
dz −
C2 z − a
Z
f (z)
dz
C1 z − a
!
H
H
We know C f (z)dz = C f (z)dz
2
1
Now let us consider that there are multiple connected regions contained in C2 :
26
Then we can write,
Z
C2
f (z)dz =
Z
C1
f (z)dz +
Z
d1
f (z)dz +
Z
f (z)dz
e1
If we assume that the point a is contained in the circle d1 then we get the following:
27
Z
Z
f (z)
f (z)
dz +
dz
C1 z − a
d1 z − a
Z
f (z)
=
dz + 2πi f (a)
C1 z − a
!
Z
Z
1
f (z)
f (z)
dz −
dz
f (a) =
2πi C2 z − a
C1 z − a
f (z)
dz =
C2 z − a
Z
Cauchy’s integral formula for n-th derivative:
If a function f (z) is analytic within and on a closed curve C and a is any point within C, then
the derivatives of all order of f (z) exists and it is given by
Z
n!
f (z)
n
dz
f (a) =
2πi C (z − a)n+1
Proof:
From Cauchy’s integral formula, we have
Z
1
f (z)
f (a) =
dz
2πi C z − a
Z
Z
1
− f (z)
1
f (z)
′
f (a) =
(−1)dz
=
dz
2πi C (z − a)2
2πi C (z − a)2
Z
Z
−2 f (z)
f (z)
1
2!
′′
f (a) =
(−1)dz =
dz
3
2πi C (z − a)
2πi C (z − a)3
Z
Z
2!
−3 f (z)
f (z)
3!
′′′
f (a) =
(−1)dz =
dz
4
2πi C (z − a)
2πi C (z − a)4
..
.
Z
n!
f (z)
n
f (a) =
dz
2πi C (z − a)n+1
Hence proved.
R
Example 1: Derive the value of (z2 + 5)dz for i)|z| = 1, ii) |z − 3π| = 10
Solution The given function f (z) = z2 + 5 is analytic in the whole complex plane. Hence it does
not have any singularity in the given regions. Using Cauchy’s theorem, we get
Z
(z2 + 5)dz = 0.
Example 2: Derive the value of
R z2 +5z+6
z−2
dz for
i)|z| = 1 ii)|z| = 3,
Solution: The given function has a singularity at z = 2.
iii)|z| = 2
28
i) |z| = 1 represents the circle with center at z = 0 and radius 1. The singular point z = 2 lies
outside the curve. Hence the curve is analytic within and on |z| = 1. Using Cauchy’s theorem, we
get
Z
z2 + 5z + 6
dz = 0.
z−2
ii) |z| = 3 represents the circle with center at z = 0 and radius 3. The singular point z = 2 lies
inside the curve. If we consider the function f (z) = z2 + 5z + 6, then it is analytic in the complex
plane and hence analytic within and on |z| = 3. Using Cauchy’s integral formula, we get
Z
Z 2
f (z)
z + 5z + 6
dz =
dz = 2πi f (2) = 40πi.
z−2
z−2
iii) |z| = 2 represents the circle with center at z = 0 and radius 2. The singular point z = 2 lies
on the curve. If we consider the function f (z) = z2 + 5z + 6, then it is analytic in the complex plane
and hence analytic within and on |z| = 2. Using Cauchy’s integral formula, we get
Z 2
Z
z + 5z + 6
f (z)
dz =
dz = 2πi f (2) = 40πi.
z−2
z−2
29
Example 3: Derive the value of F1 =
2,
R zez
z−i
dz,
F2 =
R zez
z+i
dz for
i)|z| = 1/2
ii)|z| =
iii)|z| = 1
Solution: Let us consider F1 =
R zez
z−i
dz. It has a singularity at z = i.
i) |z| = 1/2 represents the circle with center at z = 0 and radius 1/2. The singular point z = i lies
outside the curve. Hence the curve is analytic within and on |z| = 1/2. Using Cauchy’s theorem,
we get
zez
dz = 0.
z−i
ii) |z| = 2 represents the circle with center at z = 0 and radius 2. The singular point z = i lies
Z
inside the curve. If we consider the function f (z) = zez , then it is analytic in the complex plane
and hence analytic within and on |z| = 2. Using Cauchy’s integral formula, we get
Z
Z
zez
f (z)
dz =
dz = 2πi f (i) = −2πei .
z−i
z−i
iii) |z| = 1 represents the circle with center at z = 0 and radius 1. The singular point z = i lies on
the curve. If we consider the function f (z) = zez , then it is analytic in the complex plane and hence
analytic within and on |z| = 2. Using Cauchy’s integral formula, we get
Z
Z
zez
f (z)
dz =
dz = 2πi f (i) = −2πei .
z−i
z−i
30
Similarly, we can find
R zez
z+i
dz.
Example 4: Derive the value of
R
1
dz, for
z−a
i)|z| = a
ii)|z| = a/2,
iii)|z − a| = r
Solution: The given function has a singularity at z = a.
i) |z| = a represents the circle with center at z = 0 and radius a. The singular point z = a lies
on the curve. Hence the curve is not analytic within and on |z| = a. If we consider the function
f (z) = 1, then it is analytic in the complex plane and hence analytic within and on |z| = a. Using
Cauchy’s integral formula, we get
Z
Z
1
f (z)
dz =
dz = 2πi f (a) = 2πi.
z−a
z−a
ii) |z| = a/2 represents the circle with center at z = 0 and radius a/2. The singular point
z = a lies outside the curve. Hence the curve is analytic within and on |z| = a/2. Using Cauchy’s
theorem, we get
Z
1
dz = 0.
z−a
iii) |z − a| = r represents the circle with center at z = a and radius r. The singular point z = a
lies inside the curve. If we consider the function f (z) = 1, then it is analytic in the complex plane
31
and hence analytic within and on |z − a| = r. Using Cauchy’s integral formula, we get
Z
Z
1
f (z)
dz =
dz
z−a
z−a
= 2πi f (a)
= 2πi.
Example 5: Derive the value of
R sin z
z
dz, for
|z| = 1
Solution: The given function has a singularity at z = 0. |z| = 1 represents the circle with center at
z = 0 and radius 1. The singular point z = 0 lies inside the curve. Hence the curve is not analytic
32
within and on |z| = 0. If we consider the function f (z) = sin z, then it is analytic in the complex
plane and hence analytic within and on |z| = 1. Using Cauchy’s integral formula, we get
Z
Z
sin z
f (z)
dz =
dz = 2πi f (0) = 0.
z
z
R
Example 6: Derive the value of cosz z dz, for i) |z| = 1, ii)|z − 2i| = 1
Solution: The given function has a singularity at z = 0. |z| = 1 represents the circle with center at
z = 0 and radius 1. i)The singular point z = 0 lies inside the curve. Hence the curve is not analytic
within and on |z| = 0. If we consider the function f (z) = cos z, then it is analytic in the complex
plane and hence analytic within and on |z| = 1. Using Cauchy’s integral formula, we get
Z
Z
cos z
f (z)
dz =
dz = 2πi f (0) = πi.
z
z
ii) |z − 2i| = 1 represents the circle with center at z = 2i and radius 1. The singular point z = 0
lies outside the curve. Hence the curve is analytic within and on |z − 2i| = 1. Using Cauchy’s
theorem, we get
Z
sin z
dz = 0.
z
33
Example 7: Derive the value of
R
ez
dz, for |z| = 3/2
(z−1)(z+3)2
Solution: The given function has a singularity at z = 1, −3, −3. |z| = 1 represents the circle with
center at z = 0 and radius 1.
The singular point z = 1 lies inside the curve but the singular point z = −3 lies outside the curve.
z
e
Hence the curve is not analytic within and on |z| = 3/2. If we consider the function f (z) = (z+3)
2,
then it is analytic within and on |z| = 3/2. Using Cauchy’s integral formula, we get
Z
Z
e
πie
ez
f (z)
dz = 2πi f (1) = 2πi =
.
dz =
2
(z − 1)(z + 3)
z−1
16
16
R z
Example 8: Derive the value of esin
z −1 dz, for |z + π| = 2
Solution: The given function has a singularity when ez − 1 = 0 i.e., z = 0. |z + π| = 2 represents
the circle with center at z = −π and radius 2.
The singular point z = 0 lies outside the curve. Hence the curve is analytic within and on
|z| = 3/2. Using Cauchy’s theorem, we get
Z
sin z
dz = 0.
ez − 1
34
Example 9: Derive the value of
R
cos z
dz, for |z| = 2
z(z2 +16)
Solution: The given function has singularities at z = 0, z2 + 16 = 0 =⇒ z = 0, ±4i . |z| = 2
represents the circle with center at z = 0 and radius 2.
The singular point z = 0 lies inside the curve but the points z = ±4i lie outside the curve. If
z
we consider the function f (z) = zcos
2 +16 , then it is analytic within and on |z| = 2. Using Cauchy’s
integral formula, we get
Z
Z
cos z
f (z)
1
πi
dz
=
dz
=
2πi
f
(0)
=
2πi
=
.
z(z2 + 16)
z
16
8
R
z
Example 10: Derive the value of z(zsin2 +8)
dz, for |z| = 2
Solution: Follow the method of the previous problem and solve it.
R z2
Example 11: Derive the value of 4−z
2 dz, for |z − 1| = 2
Solution: The given function has singularities at 4 − z2 = 0 =⇒ z = ±2. |z − 1| = 2 represents
the circle with center at z = 1 and radius 2.
The singular point z = 2 lies inside the curve but the point z = −2 lies outside the curve. If
z
we consider the function f (z) = 2+z
, then it is analytic within and on |z − 1| = 2. Using Cauchy’s
35
integral formula, we get
Z
z2
dz =
4 − z2
Z
Example 12: Derive the value of
−
R
f (z)
2
dz = −2πi f (2) = −2πi · = −πi.
z−2
4
z2 +2
dz, for |z| = 2
(z−1)(z−2)
Solution: The given function has singularities at z = 1, 2. |z| = 2 represents the circle with center
at z = 0 and radius 2.
The singular point z = 1 lies inside the curve and z = 2 lies on the curve. We can rewrite the
given integration as follows:
Z
z2 + 2
dz =
(z − 1)(z − 2)
Z
(z2 + 2)[(z − 1) − (z − 2)]
dz =
(z − 1)(z − 2)
Z
z2 + 2
dz −
z−2
Z
z2 + 2
dz
z−1
If we consider the function f (z) = z2 + 2, then it is analytic within and on |z| = 2. Using Cauchy’s
integral formula, we get
Z
Z
Z
z2 + 2
f (z)
f (z)
dz =
dz −
dz = 2πi f (2) − 2πi f (1) = 2πi(6) − 2πi(3) = 6πi.
(z − 1)(z − 2)
z−2
z−1
R
Example 13: Derive the value of z2z+4 dz, for |z| = 3
Solution: The given function has singularities at z = ±2i. |z| = 3 represents the circle with center
at z = 0 and radius 3. The singular points z = 2i, −2i lies inside the curve. We can rewrite the
given integration as follows:
Z
Z
Z
Z
z
z[(z + 2i) − (z − 2i)]
1
z
1
z
dz =
dz =
dz −
dz
2
z +4
4i(z − 2i)(z + 2i)
4i
z − 2i
4i
z + 2i
If we consider the function f (z) = z, then it is analytic within and on |z| = 3. Using Cauchy’s
integral formula, we get
Z
Z
Z
z
1
f (z)
f (z)
1
2πi
(2i + 2i) = 2πi.
dz =
dz −
dz = (2πi f (2i) − 2πi f (−2i)) =
2
z +4
4i
z − 2i
z + 2i
4i
4i
36
Example 14: Derive the value of
R
z2 +1
dz, for |z| = 3
z(2z+1)
R
ez
dz, for |z| = 2
(z+1)4
Solution: Solve it.
Example 15: Derive the value of
Solution: The given function has singularities at z = −1. |z| = 2 represents the circle with center
at z = 0 and radius 2. The singular point z = −1 lies inside the curve. If we consider the function
f (z) = e2z , then it is analytic within and on |z| = 2. Using Cauchy’s integral formula of n-th
derivative, we get
Z
ez
2πi ′′′
πi
=
f (−1) =
· 8e2(−1)
(Since f ′′′ (z) = 8e2z )
4
(z + 1)
3!
3
8πie−2
=
3
R z cos z
Example 16: Derive the value of (z− π )2 dz, for |z − 1| = 1
2
Solution: The given function has singularity at z = π2 . |z − 1| = 1 represents the circle with center
at z = 1 and radius 1. The singular point z = π2 lies inside the curve. If we consider the function
f (z) = z cos z, then it is analytic within and on |z − 1| = 1.
f (z) = z cos z =⇒ f ′ (z) = cos z − z sin z
37
Using Cauchy’s integral formula of n-th derivative, we get
Z
2πi ′ π
π
z cos z
f ( ) = 2πi(− ) = −π2 i.
π 2 dz =
(z − 2 )
1!
2
2
R cos5 z
Example 17: Derive the value of (z−
π 3 dz, for |z − 1| = 1
)
3
Solution: Solve it.
VII.
SERIES OF COMPLEX NUMBERS
Let us consider the series
∞
X
zk
k=1
Ratio test:
zk+1
=L
k→∞ zk
lim
Root test:
√
lim k zk = L
k→∞
• Convergent: L < 1
• Divergent: L > 1
• Inconclusive: L = 1
Power series: The general form of the power series is given by
f (z) =
∞
X
ak (z − z0 )k
k=0
= a0 + a1 (z − z0 ) + a2 (z − z0 )2 + a3 (z − z0 )3 + · · · + an (z − z0 )n
(35)
38
The Cauchy’s integral formula gives
Z
1
f (z)
f (a) =
dz,
2πi
z−a
n!
f (a) =
2πi
n
Z
f (z)
dz,
(z − a)n+1
Concept of the radius of convergence: Let us consider the series
∞
X
ak (z − z0 )k ,
k=0
where limk→∞ aak+1
= L. If we perform the ratio test,
k
ak+1 (z − z0 )k+1
k→∞
ak (z − z0 )k
lim
(36)
It is convergent if
lim
k→∞
ak+1
1
|z − z0 | < 1 =⇒ RL < 1 =⇒ R <
ak
L
Hence, the radius of convergence is L1 which implies that for any z such that |z − z0 | ≤ L1 , the
series is convergent.
Example 1:
∞
P
k=0
zk
.
2k +1
Using Ratio test, we get
1 + 21k
ak+1
2k + 1
1
lim
= lim k+1
= <1
= lim
1
k→∞ ak
k→∞ 2
k→∞
2
+1
2 + 2k
Hence, the series is convergent and the radius of convergence is R = 2.
∞
P
(z−3i)k
Example 2:
.
(1−i)k+1
k=0
Using the Ratio test, we get
ak+1
(1 − i)k+1
1
1+i
1
= lim
= lim
=
= √ <1
k+2
k→∞ ak
k→∞ (1 − i)
k→∞ 1 − i
2
2
√
Hence, the series is convergent and the radius of convergence is R = 2. The center of the circle
lim
of convergence is z = 3i.
39
Taylor’s theorem: If a function f (z) is analytic within a circle C with its center z = a and
radius R, then at every point z inside C, then
f (z) =
∞
X
k=0
ak (z − a) ,
k
f k (a)
ak =
k!
1
Proof: From Cauchy’s integral formula, we have f (a) = 2πi
R f (t)
1
dt.
2πi
t−z
R f (z)
z−a
dz,. This implies that f (z) =
1
1
1
=
=
t − z (t − a) − (z − a) (t − a) 1 − z−a
t−a
1 z − a −1
1−
=
(t − a)
t−a
"
#
1
z − a z − a 2
=
1+
+
+ ···
(t − a)
t−a
t−a
#
Z "
1
z − a z − a 2
f (z) =
1+
+
+ · · · f (t)dt
2πi
t−a
t−a
f ′′ (a)
= f (a) + f ′ (a)(z − a) +
(z − a)2 + · · ·
2!
∞
k
X
f
(a)
=
ak (z − a)k , ak =
k!
k=0
Hence proved.
40
Example 1:
f (z) =
1
1
=
1 − z (1 − i) − (z − i)
1 z − i −1
=
1−
1−i"
1−i
#
1
z − i z − i 2
=
1+
+
+ ···
1−i
1−i
1−i
∞
X
1
=
(z − i)k
k+1
(1
−
i)
k=0
∞
X
ak (z − i)k
=
(37)
k=0
1
ak+1
1
1
(1−i)k+2
lim
= lim
= √
= lim
1
k→∞ ak
k→∞
k→∞ (1 − i)
2
(1−i)k+1
If the radius of convergence is R then R <
√
2.
Hence, the given function represents a circle with center z = i and radius
√
2.
This implies that x2 + 1 = 2 =⇒ x = ±1 i.e., it cuts x axis at (±1, 0). Example 2:
f (z) =
1
1
=
4 − z (4 − 2i) − (z − 2i)
!−1
1
z − 2i
=
1−
4 − 2i
4 − 2i


!2

1 
z − 2i
z − 2i
1 +
=
+
+ · · · 
4 − 2i
4 − 2i
4 − 2i
∞
X
1
=
(z − 2i)k
k+1
(4
−
2i)
k=0
∞
X
=
ak (z − 2i)k
k=0
41
1
ak+1
1
1
(4−2i)k+2
lim
= lim
= lim
= √
1
k→∞ ak
k→∞ (4 − 2i)
k→∞
20
(4−2i)k+1
If the radius of convergence is R then
R<
√
20
Hence, the given function represents a circle with center z = 2i and radius
√
20.
This implies that x2 + 4 = 20 =⇒ x = ±4 i.e., it cuts x axis at (±4, 0).
z−2
Example 3: f (z) = 5−z
,a = 2
z−2
z−2
z−2
z−2
f (z) =
=
=
1−
5 − z (5 − 2) − (z − 2)
3
3
!−1


!2

z−2
z − 2 
z−2
1 +
=
+
+ · · · 
3
3
3
∞
∞
X
X
1
k
(z
−
2)
=
ak (z − 2)k
=
k
3
k=0
k=1
1
ak+1
1
k+1
lim
= lim 3 1 =
k→∞ ak
k→∞
3
3k
If the radius of convergence is R then R < 3.
1
Example 4: f (z) = 1−z
, a = 1/2

−1
!−1
(z − 21 ) 

1
1
1
f (z) =
=
= 2 1 − 1  = 2 1 − 2(z − )
1 − z 12 − (z − 12 )
2
2
!
!2
1
1
=2+4 z−
+8 z−
+ ···
2
2
!k X
!k
∞
∞
X
1
1
k+1
=
2
z−
=
ak z −
2
2
k=0
k=0
ak+1
2k+1
= lim k = 2
k→∞ ak
k→∞ 2
lim
42
If the radius of convergence is R then R < 21 .
Laurent’s theorem: Suppose a function f (z) is analytic in the closed ring bounded by two cocentric circles C1 and C2 of center a and radii R1 and R2 (R2 < R1 ). If z is any point in the annulus,
then
f (z) =
∞
X
an (z − a) +
n=0
n
∞
X
1
an =
2πi
bn (z − a) ,
−n
n=1
Z
f (t)
dt,
n
C1 (t − a)
1
bn =
2πi
Z
Proof: Cauchy’s integral formula gives
1
f (a) =
2πi
Z
1
f (z)
dz −
2πi
C1 z − a
Z
f (z)
dz.
C2 z − a
Hence we get
Z
Z
1
f (t)
1
f (t)
f (z) =
dt −
dt
2πi C1 t − z
2πi C2 t − z
Z
Z
1
f (t)
1
f (t)
=
dt +
dt
2πi C1 t − z
2πi C2 z − t
Z
Z
1
f (t)
1
f (t)
=
dt +
dt
2πi C1 (t − a) − (z − a)
2πi C2 (z − a) − (t − a)
Z
Z
1
f (t)
1
f (t)
dt +
dt
=
2πi C1 (t − a) 1 − z−a
2πi C2 (z − a) 1 − t−a
t−a
= I1 + I2 .
z−a
f (t)
dt.
−n+1
C2 (t − a)
43
1
I1 =
2πi
Z
f (t)
dt
z−a
C1 (t − a) 1 −
t−a
Z
1
1 z − a −1
=
1−
f (t)dt
2πi C1 (t − a)
t−a
"
#
Z
z − a z − a 2 z − a 3
1
1
1+
+
+
+ · · · f (t)dt
=
2πi C1 (t − a)
t−a
t−a
t−a
Z
Z
Z
1
1
f (t)dt
f (t)dt
f (t)dt
2 1
=
+ (z − a)
+ (z − a)
+ ···
2
2πi C1 (t − a)
2πi C1 (t − a)
2πi C1 (t − a)3
Z
∞
X
f (t)dt
n 1
(z − a)
=
2πi C1 (t − a)n+1
n=0
Z
∞
X
1
f (t)dt
n
an (z − a) , an =
=
.
2πi C1 (t − a)n+1
n=0
Z
1
f (t)
dt
I2 =
2πi C2 (z − a) 1 − t−a
z−a
Z
1
t − a −1
1
=
1−
f (t)dt
2πi C2 (z − a)
z−a
#
"
Z
t − a t − a 2 t − a 3
1
1
1+
+
+
+ · · · f (t)dt
=
2πi C2 (z − a)
z−a
z−a
z−a
Z
Z
Z
1
1
1
1
1
1
=
f (t)dt +
(t − a) f (t)dt +
(t − a)2 f (t)dt + · · ·
2πi (z − a) C2
2πi (z − a)2 C2
2πi (z − a)3 C2
Z
∞
X
1
1
(t − a)n−1 f (t)dt
=
·
n 2πi
(z
−
a)
C
2
n=1
Z
∞
X
1
f (t)
−n
=
bn (z − a) , bn =
dt
−n+1
2πi
(t
−
a)
C
2
n=1
Hence proved.
1
Example 1: f (z) = z(z−2)
, 1 < |z − 3| < 3.
Solution:
1 < |z − 3| < 3 =⇒
1
z−1
< 1,
< 1.
z−3
3
"
#
"
#
1
1 1
1
1
1
1
f (z) =
=
−
=
−
z(z − 2) 2 z − 2 z
2 z−3+1 z−3+3



1 
1
1

= 
−
1
2 (z − 3)(1 + z−3
) 3(1 + z−3
)
3

!−1
!−1 
1  1
1
1
z − 3 

= 
1+
− 1+
2 (z − 3)
z−3
3
3
Example 2: f (z) = z2z−1
3 −z2 , 0 < |z| < 1.
(38)
44
Solution: Since we have|z| < 1, we can write f (z) as
f (z) =
z+z−1
1
1
1
1 1
1 1
=
+
=
−
+
=
− − (1 − z)−1
z2 (z − 1) z(z − 1) z2 z − 1 z z2 z2 z
1 1
= 2 − − 1 − z − z2 − z3 − · · ·
z
z
Example 3: f (z) = 1z , 0 < |z − 1| < 1.
Solution: Since we have|z − 1| < 1, we can write f (z) as
∞
X
1
−1
2
3
f (z) =
(−1)k (z − 1)k
= (1 + (z − 1)) = 1 − (z − 1) + (z − 1) − (z − 1) + · · · =
z−1+1
k=0
Note:
∞
X
1
d 1
f (z) = 2 = − ( ) =
k(−1)k (z − 1)k−1
z
dz z
k=1
VIII.
SINGULARITY OF COMPLEX NUMBERS
Singular points: If a function fails to be analytic at a point z0 but is analytic in the neighborhood of z0 , then it z0 is called the singular point of f (z). Otherwise, it is a regular point.
Laurent’s series:
f (z) =
k
X
an (z − z0 )n
n=−k
=
∞
X
n=0
where
∞
P
an (z − z0 )n +
∞
X
bn
(z − z0 )n
n=1
an (z − z0 )n is the analytic part (AP) with no singularities and
n=0
(39)
∞
P
n=1
bn
is the principle
(z−z0 )n
part (PP) with singularities. This is the Laurent series expansion of f around the point z0 .
• Isolated singularities:
45
1. Removable: The Laurent series has no principle part i.e., each bn is 0.
i) lim f (z) exists and finite. ii) lim f (z) is the analytic part.
z→z0
z→z0
Example:
sin z
z"
#
1
z3 z5
=
z − − − ···
z
3! 5!
2
z
z4
= 1 − − − ···
3! 5!
f (z) =
Hence z = 0 is the removable singular point.
2. Pole: If Laurent’s series contains a finite number of terms in PP, then z = z0 is a
pole.
f (z) =
∞
X
an (z − z0 )n +
n=0
b2
b3
b1
+
+
+ ···
1
2
(z − z0 )
(z − z0 )
(z − z0 )3
The order of the pole is the highest power of bn i.e., if bm<n = 0 and bn = 0 then the
order of the pole is m. For example, if the last term of Laurent’s series of f (z) is (z−zb30 )3 ,
then the order of the pole z = z0 is 3.
Example 1:
#
"
z3 z5
1
z
z3
sin z
1
f (z) = 2 = 2 z − − − · · · = − − − · · ·
z
z
3! 5!
z 3! 5!
Here, the order of the pole z = 0 is 1.
Example 2:
"
#
ez
1
z2 z3
1
1
1
1
f (z) = 3 = 3 1 + z + + + · · · = 3 + 2 +
+ + ···
z
z
2! 3!
z
z
2z 3!
Example 3:
e2z
e2t+2 e2 (e2t )
=
=
(z − 1)3
t3
t3
2 e
= 3 1 + 2t + 4t2 + 8t3 + · · ·
t
e2 2e2 4e2
= 3 + 2 +
+8
t
t
t
f (z) =
Here, the order of the pole z = 1 is 3.
46
3. Essential singularity: If Laurent’s series contains an infinite number of terms in
PP, then z = z0 is an essential singularity. In this case, lim f (z) does not exist.
z→z0
Example 1:
f (z) = e1/z = 1 +
1
1
1
+
+
+ ···
2
z 2! · z
3! · z3
lim f (z) does not exist, implying that z → 0 is an essential singularity.
z→0
Example 2:
1
f (z) = sin
z
!
Example 3:
!
!
!
1
1
1
f (z) = (z − 3) sin
= (z − 2) sin
− sin
z−2
z−2
z−2
lim f (z) does not exist, implying that z → 2 is an essential singularity.
z→2
• Non-isolated singularities: If for a known singular point, there exists one or more singular
points in the neighborhood of the point then they are called non-isolated singularities.
Example:
sin
π
1
= 0 =⇒ πz = nπ =⇒ z = , n = ±1, ±2, · · ·
z
n
Also, for n → ∞, z → 0 which shows that z = 0 is the singular point and in every neighborhood of it, there are other singular points z = n1 .
Residue: If the Laurent’s expansion of any function f (z) is given by
f (z) =
∞
X
an (z − z0 )n +
n=0
b1
b2
b3
+
+
+ ···
(z − z0 )1 (z − z0 )2 (z − z0 )3
then for the isolated singularity z = z0 , the coefficient of (z − z0 )−1 i.e., b1 is defined as the residue
of the f (z) at z = z0 .
Let us write the Laurent’s expansion as follows:
f (z) =
∞
X
an (z − z0 ) +
n=0
n
∞
X
bn (z − z0 )−n
n=1
where we have
1
an =
2πi
Z
f (ξ)
dξ,
(ξ − z0 )n+1
1
bn =
2πi
Z
f (ξ)
dξ
(ξ − z0 )−(n+1)
47
i) f (z) is analytic at z = z0 . Then b1 = 0 which implies that Res( f (z), z0 ) = 0.
ii) f (z) has a simple pole at z = z0 i.e., z0 is a pole of order 1.
Then Laurent’s expansion can be written as follows:
b1
z − z0
2
3
(z − z0 ) f (z) = a0 (z − z0 ) + a1 (z − z0 ) + a2 (z − z0 ) + · · · + b1
f (z) = a0 + a1 (z − z0 ) + a2 (z − z0 )2 + · · · +
This shows that
lim (z − z0 ) f (z) = b1 .
z→z0
iii) f (z) has a pole at z = z0 of order n.
Then Laurent’s expansion can be written as follows:
f (z) =
bn
bn−1
b1
+
+ ··· +
+ a0 + a1 (z − z0 ) + a2 (z − z0 )2 + · · ·
n
n−1
(z − z0 )
(z − z0 )
(z − z0 )
(z − z0 )n f (z) = bn + (z − z0 )bn−1 + · · · + b1 (z − z0 )n−1 + a0 (z − z0 )n + a1 (z − z0 )n+1 + a2 (z − z0 )n+2
+···
dn−1 n
(z − z0 ) f (z) = (n − 1)! · b1 + a0 · n!(z − z0 ) + · · ·
dzn−1
This shows that
1
dn−1 n
lim
(z − z0 ) f (z) = b1 .
(n − 1)! z→z0 dzn−1
Example 1: Find the residue for each pole of the given function:
f (z) =
1
(z − 1)2 (z − 3)
Soluion: The given function f (z) has pole z = 1 of order n = 2 and z = 3 of order n = 1.
We know
Res( f (z), z0 ) =
1
dn−1 lim n−1 (z − z0 )n f (z) .
(n − 1)! z→z0 dz
For the pole z = 1 where n = 2, we calculate the residue as follows:
d
1
d 1 1
2
(z − 1)
= lim
=− .
Res( f (z), 1) = lim
2
z→1 dz
z→1 dz (z − 3)
(z − 1) (z − 3)
4
(40)
48
For the pole z = 3 where n = 1, we calculate the residue as follows:
Res( f (z), 3) = lim (z − 3)
z→1
1 1
1
=
lim
= .
z→3 (z − 1)2
(z − 1)2 (z − 3)
4
Example 2: Find the residue for each pole of the given function:
f (z) =
5z2 − 4z + 3
(z + 1)(z + 2)(z + 3)
Example 3: Find the residue for each pole of the given function:
f (z) =
ez
z3
Soluion: Using the series expansion of ez , we get
ez
z3
!
1
z2 z3 z4
= 3 1 + z + + + ···
z
2! 3! 4!
1
z
1
1
1
+ + ···
= 3+ 2+
z
z
2! · z 3! 4!
f (z) =
(41)
1
1
1
Here, the analytic part (AP) is ( 3!1 + z−0
+ · · · ) and the priniple part (PP) is ( (z−0)
3 + (z−0)2 + 2!·(z−0) ).
4!
1
Clearly, the coefficient of (z−0)
is 12 which implies that the residue is 21 .
z
Also, since f (z) = ze3 , z = 0 is apole of order 3 i.e., n = 3. Hence, we get
z
d2 d2 z 1
1
1
3e
Res( f (z), 0) = lim 2 (z − 0) 3 = lim 2 e = − .
2! z→0 dz
z
2 z→0 dz
2
Example 4: Find the residue for each pole of the given function:
f (z) =
z2 − 2z
(z + 1)2 (z2 + 4)
Soluion: The given function f (z) can be written as
f (z) =
z2 − 2z
(z + 1)2 (z + 2i)(z − 2i)
It has pole z = −1 of order n = 2, z = ±2i of order n = 1.
We know
1
dn−1 n
lim
(z − z0 ) f (z) .
Res( f (z), z0 ) =
(n − 1)! z→z0 dzn−1
49
For the pole z = −1 where n = 2, we calculate the residue as follows:
d
d z2 − 2z z2 − 2z
2
Res( f (z), −1) = lim
(z + 1)
= lim
=
z→−1 dz
z→−1 dz (z2 + 4)
(z + 1)2 (z2 + 4)
(z2 + 4)(2z − 2) − (z2 − 2z)2z −20 + 6
14
lim
=
=− .
2
2
2
z→−1
(z + 4)
5
25
For the pole z = 2i where n = 1, we calculate the residue as follows:
z2 − 2z
z2 − 2z
=
lim
z→2i (z + 1)2 (z + 2i)
z→2i
(z + 1)2 (z + 2i)(z − 2i)
−4 − 4i
−4(1 + i)
7+i
=
=
=
.
2
2
(2i + 1) (4i) (2i + 1) (4i)
25
Res( f (z), 2i) = lim (z − 2i)
Substituting i by −i, we get
Res( f (z), −2i) =
7−i
.
25
Theorem 3. Cauchy’s residue theorem:
If f (z) is analytic in the region bounded by the simple curve C except the points z1 , z2 , z3 , · · · zi
then
I
f (z)dz =
I
I
f (z)dz +
C1
f (z)dz + · · · +
C2
I
f (z)dz =
Cn
Cauchy’s residue theorem says that
I
f (z)dz = 2πi
n
X
Res( f (z), zi )
i=1
Example 1:
H z+6
z2 +4
dz, C : |z − i| = 2
Soluion: The given function f (z) can be written as
f (z) =
z+6
(z + 2i)(z − 2i)
n I
X
i=1
Ci
f (z)dz.
50
The given region is the circle with a center at i and a radius of 2. The given function f (z)
has isolated singularities z = ±2i of order 1. In the above diagram, we can see that the isolated
singularity z = 2i lies inside the circle but the isolated singularity z = −2i lies outside the circle.
Hence, using Cauchy’s residue theorem, we can write
I
Example 2:
H
z+6
dz = 2πi Res( f (z), 2i)
z2 + 4
z+6
= 2πi lim (z − 2i)
z→2i
(z + 2i)(z − 2i)
z+6
= 2πi lim
= π(i + 3)
z→2i z + 2i
1
dz, C : |z − 3i| = 3
z2 +4z+13
Soluion: Solving the equation
z2 + 4z + 13 = 0,
we get z = −2 ± 3i. The given function f (z) can be written as
f (z) =
1
z − (−2 + 3i) z − (−2 − 3i)
51
The given region is the circle with a center at 3i and a radius of 3. The given function f (z) has
isolated singularities z = −2 ± 3i of order 1.
The complex number z = −2 ± 3i represents the points (−2, ±3) in the Cartesian coordinate
system. In the above diagram, we can see that the isolated singularity z = −2 + 3i ≡ (−2, 3) lies
inside the circle but the isolated singularity z = −2 − 3i ≡ (−2, −3) lies outside the circle. Hence,
using Cauchy’s residue theorem, we can write
I
1
dz = 2πi · Res( f (z), −2 + 3i)
2
z + 4z + 3
= 2πi lim z − (−2 + 3i)
z→−2+3i
Example 3:
H
π
1
=
3
z − (−2 + 3i) z − (−2 − 3i)
1
dz, C : |z| = 3
(z−1)(z+2)2
Soluion: The given function f (z) is
f (z) =
1
(z − 1)(z + 2)2
The given region is the circle with a center at z = 0 and a radius of 3. The given function f (z)
has isolated singularities z = −2 of order 2 and z = 1 of order 1. In the above diagram, we can see
that the isolated singularities z = −2 and z = 1 lie inside the circle. Hence, using Cauchy’s residue
theorem, we can write
I
1
dz = 2πi · Res( f (z), −2) + Res( f (z), 1)
(z − 1)(z + 2)2
Now we calculate the residues as follows:
d
1
d 1 1
−1
Res( f (z), −2) = lim
=
lim
=
−
(z + 2)2
=
lim
z→−2 dz
z→−2 dz (z − 1)
z→−2 (z − 1)2
(z − 1)(z + 2)2
9
Res( f (z), 1) = lim (z − 1)
z→1
1 1
1
= lim
=
z→1 (z + 2)2
(z − 1)(z + 2)2
9
52
This implies that
I
Example 4: I =
H
z3
2
e1/z
1 1
1
dz
=
2πi
·
− +
= 0.
(z − 1)(z + 2)2
9 9
dz
Soluion: The given function can be written as
z3
e1/z2
2
= z3 · e−1/z
#
"
1
1
1
3
−
=z · 1− 2 + 4
z
z · 2! z6 · 3!
1
1
= z3 − z +
− 3
+ ···
z · 2! z · 3!
1
Here, the analytic part (AP) is (z3 − z) and the priniple part (PP) is z·2!
− z31·3! + · · · . Here, the only
singularity is z = 0. Clearly, the coefficient of 1z is 12 which implies that the residue of f (z) is 21 .
Hence, using Cauchy’s residue theorem, we can write
I
z3
1
dz = 2πi · Res( f (z), 0) = 2πi · = πi.
2
1/z
2
e
H tan z
Example 5: I =
dz, C : |z − 1| = 2
z
Soluion: The given function tanz z has singular points z = 0, π2 , − π2 .
The given region is the circle with a center at z = 1 and a radius of 3. The given function f (z)
has singularities z = ± π2 and z = 0 of order 1. In the above diagram, we can see that the isolated
singularities z = π2 and z = 0 lie inside the circle. Hence, using Cauchy’s residue theorem, we can
write
I=
I
tan z
π
dz = 2πi · Res f (z), + Res( f (z), 0)
z
2
Now we calculate the residues as follows:
(z − π2 ) cos z + sin z
π
1
2
π tan z π sin z Res f (z),
= limπ (z − )
= limπ (z − )
= limπ
= π =−
z→ 2
z→ 2
z→ 2
2
2 z
2 z cos z
cos z − z sin z
−2
π
53
tan z Res( f (z), 0) = lim z ·
=0
z→0
z
Hence, we get
I=
I
π
2
tan z
dz = 2πi · Res f (z), + Res( f (z), 0) = −2πi · = −4i.
z
2
π
A real function can be converted to a complex function to calculate the integral. For example,
R∞
f (x)dx can be converted to a suitable complex function.
−∞
R 2π
Let 0 F(cos θ, sin θ)dθ can be converted as follows:
eiθ + e−iθ z + z
cos θ =
=
,
2
2
1
1
eiθ − e−iθ z − z
sin θ =
=
2
2
where z = eiθ , |z| = 1, θ ∈ [0, 2π]. This implies that dz = ieiθ dθ = izdθ. Hence, we have dθ = dz
.
iz
R 2π dθ
Example 1: 0 5+4 sin θ
Solution: We convert the given integration to a complex integration as follows:
Z 2π
Z
Z 2π
Z 2π
dz
1 2π
dz
dθ
dz
iz
1 =
=
=
2
2
z+
5 + 4 sin θ
5zi + 2z + 2 2 0 z + 5z2 i − 1
0
0
0
5 + 4 2z
(42)
We find the singularities of the function z2 +dz
. Thus we get
5z
i−1
2
5z
z + i − 1 = 0 =⇒ z =
2
2
− 52 i ±
q
(− 52 i)2 + 4
2
=
− 52 i ±
q
− 25
+4
4
2
1
= − i, −2i
2
Hence, the two singularities are − 21 i, −2i.
In the above diagram, we can see that the isolated singularity z = − 12 i lies inside the circle but
the isolated singularity z = −2i lies outside the circle. Hence, using Cauchy’s residue theorem, we
can write
Z 2π
0
!
dz
1
= 2πi Res f (z), − i
2
2(z2 + 5z2 i − 1)
2π
1
1
=
= 2πi lim (z + i)
2 2(z + 2i)(z + 12 i)
3
z→− 12 i
54
Example 2:
R 2π
Example 3:
R 2π cos θ
0
0
dθ
1+ sin2 θ
3+sin θ
dθ
Let us consider the following integral
Z ∞
Z R
F(x)dx = I = lim
F(x)dx
R→∞
−∞
−R
RR
where limR→∞ −R F(x)dx is convergent.
I
F(z)dz =
Z
F(z)dz +
C1
Example 1:
Z
C2 (real)
f (x)dx = 2πi
n
X
k=1
Z ∞
I
1
1
dx =
dz
2
2
z − 2z + 2
−∞ x − 2x + 2
dz
. Thus we get
We find the singularities of the function z2 −2z+2
z2 − 2z + 2 = 0 √
2± 4−8
z=
= 1 ± i.
2
Res( f (z), zk )
55
Hence, the two singularities are 1±i. In the above diagram, we can see that the singularity z = 1+i
lies inside the semi-circle but the singularity z = 1−i lies outside the circle. Hence, using Cauchy’s
residue theorem, we can write
Z 2π
1
dz = 2πi Res( f (z), 1 + i)
z2 − 2z + 2
0
1
= 2πi lim (z − (1 + i))
z→1+i
(z − (1 + i))(z − (1 − i))
1
1
= 2πi lim
= 2πi · = π
z→1+i z − 1 + i
2i
I
Z
Z R
f (z)dz =
f (z)dz +
f (x)dx = π
C1
−R
where C f (z)dz = I1 . Here, z = Reiθ which gives dz = iReiθ dθ
R
1
iReiθ dθ
I1 = 2 2iθ
R e − 2Reiθ + 2
Z
|iReiθ dθ|
≤
|R2 e2iθ − 2Reiθ + 2|
=0
X
ai ≤
X
|ai |