CHAPTER
17
In elementary algebra courses
you learned about the existence
and some of the properties of
complex numbers. But in courses
such as calculus, it is likely that
you did not even see a complex
number. Introductory calculus is
basically the study of functions
of a real variable. In advanced
courses, you may have seen
complex numbers occasionally
(see Sections 3.3, 8.8, and 10.2).
However, in the next four
chapters we are going to
introduce you to complex
analysis; that is, the study of
functions of a complex variable.
Although there are many
similarities between complex
analysis and real analysis, there
are many interesting differences
and some surprises.
Functions of a
Complex Variable
CHAPTER CONTENTS
17.1
17.2
17.3
17.4
17.5
17.6
17.7
17.8
Complex Numbers
Powers and Roots
Sets in the Complex Plane
Functions of a Complex Variable
Cauchy–Riemann Equations
Exponential and Logarithmic Functions
Trigonometric and Hyperbolic Functions
Inverse Trigonometric and Hyperbolic Functions
Chapter 17 in Review
INTRODUCTION You have undoubtedly encountered complex numbers in your ear
courses in mathematics. When you first learned to solve a quadratic equation ax2 bx c by the quadratic formula, you saw that the roots of the equation are not real, that is, comp
whenever the discriminant b2 4ac is negative. So, for example, simple equations such
x2 5 0 and x2 x 1 0 have no real solutions. For example, the roots of the last equa
"3
1
"3
1
and . If it is assumed that "3 "3"1, then the ro
are 2
2
2
2
1
"3
1
"3
are written "1 and 2
"1.
2
2
2
2
A Definition Two hundred years ago, around the time that complex numbers were g
ing some respectability in the mathematical community, the symbol i was originally used
disguise for the embarrassing symbol "1. We now simply say that i is the imaginary u
and define it by the property i2 1. Using the imaginary unit, we build a general comp
number out of two real numbers.
Definition 17.1.1
Complex Number
A complex number is any number of the form z a ib where a and b are real numbers
and i is the imaginary unit.
Note: The imaginary part of
z 4 9i is 9, not 9i.
Terminology The number i in Definition 17.1.1 is called the imaginary unit. The
number x in z x iy is called the real part of z; the real number y is called the imagin
part of z. The real and imaginary parts of a complex number z are abbreviated Re(z) and Im
respectively. For example, if z 4 9i, then Re(z) 4 and Im(z) 9. A real constant mult
of the imaginary unit is called a pure imaginary number. For example, z 6i is a pure imagin
number. Two complex numbers are equal if their real and imaginary parts are equal. Since
simple concept is sometimes useful, we formalize the last statement in the next definition.
Definition 17.1.2
Equality
Complex numbers z1 x1 iy1 and z2 x2 iy2 are equal, z1 z2, if
Re(z1) Re(z2) and
Im(z1) Im(z2).
A complex number x iy 0 if x 0 and y 0.
Arithmetic Operations Complex numbers can be added, subtracted, multiplied,
divided. If z1 x1 iy1 and z2 x2 iy2, these operations are defined as follows.
Addition:
z1 z2 (x1 iy1) (x2 iy2) (x1 x2) i( y1 y2)
Subtraction:
z1 z2 (x1 iy1) (x2 iy2) (x1 x2) i( y1 y2)
Multiplication:
z1z2 (x1 iy1)(x2 iy2)
x1x2 y1 y2 i( y1x2 x1 y2)
Division:
x1 iy1
z1
z2
x2 iy2
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CHAPTER 17 Functions of a Complex Variable
x1x2 y1y2
y1x2 2 x1y2
i
x 22 y 22
x 22 y 22
Associative laws:
Distributive law:
e
z1 (z2 z3) (z1 z2) z3
z1(z2z3) (z1z2)z3
z1(z2 z3) z1z2 z1z3
In view of these laws, there is no need to memorize the definitions of addition,
and multiplication. To add (subtract) two complex numbers, we simply add (subtr
responding real and imaginary parts. To multiply two complex numbers, we use the
law and the fact that i 2 1.
EXAMPLE 1
Addition and Multiplication
If z1 2 4i and z2 3 8i, find (a) z1 z2 and (b) z1z2.
SOLUTION
(a) By adding the real and imaginary parts of the two numbers, w
(2 4i) (3 8i) (2 3) (4 8)i 1 12i.
(b) Using the distributive law, we have
(2 4i)(3 8i) (2 4i)(3) (2 4i)(8i)
6 12i 16i 32i 2
(6 32) (16 12)i 38 4i.
There is also no need to memorize the definition of division, but before discus
need to introduce another concept.
Conjugate If z is a complex number, then the number obtained by changin
its imaginary part is called the complex conjugate or, simply, the conjugate of z. I
then its conjugate is
z x 2 iy.
For example, if z 6 3i, then z 6 3i; if z 5 i, then z 5 i. If z is a
say z 7, then z 7. From the definition of addition it can be readily shown that t
of a sum of two complex numbers is the sum of the conjugates:
z1 z2 z1 z2 .
Moreover, we have the additional three properties
z1 2 z2 z1 2 z2 ,
z1z2 z1z2 ,
z1
z1
a b .
z2
z2
The definitions of addition and multiplication show that the sum and product o
number z and its conjugate z are also real numbers:
z z (x iy) (x iy) 2x
zz (x iy)(x iy) x2 i2y2 x2 y2.
The difference between a complex number z and its conjugate z is a pure imaginar
z z (x iy) (x iy) 2iy.
Since x Re(z) and y Im(z), (1) and (3) yield two useful formulas:
Re(z) zz
2
and
Im(z) z2z
.
2i
17.1 Complex Numbers
Division
EXAMPLE 2
If z1 2 3i and z2 4 6i, find (a)
z1
1
and (b) .
z2
z1
SOLUTION In both parts of this example we shall multiply both numerator and denomin
by the conjugate of the denominator and then use (2).
2 2 3i
2 2 3i 4 2 6i
8 2 12i 2 12i 18i 2
4 6i
4 6i 4 2 6i
16 36
(a)
10 2 24i
5
6
2
i.
52
26
13
1
1 2 3i
2 3i
2
3
i.
2 2 3i
2 2 3i 2 3i
49
13
13
(b)
y
z = x + iy
x
FIGURE 17.1.1 z as a position vector
Geometric Interpretation A complex number z x iy is uniquely determined by
ordered pair of real numbers (x, y). The first and second entries of the ordered pairs correspo
in turn, with the real and imaginary parts of the complex number. For example, the ordered
(2, 3) corresponds to the complex number z 2 3i. Conversely, z 2 3i determines
ordered pair (2, 3). In this manner we are able to associate a complex number z x iy w
a point (x, y) in a coordinate plane. But, as we saw in Section 7.1, an ordered pair of real numb
can be interpreted as the components of a vector. Thus, a complex number z x iy can a
be viewed as a vector whose initial point is the origin and whose terminal point is (x, y). T
coordinate plane illustrated in FIGURE 17.1.1 is called the complex plane or simply the z-plane.
horizontal or x-axis is called the real axis and the vertical or y-axis is called the imaginary a
The length of a vector z, or the distance from the origin to the point (x, y), is clearly "x 2 This real number is given a special name.
Modulus or Absolute Value
Definition 17.1.3
The modulus or absolute value of z x iy, denoted by ZzZ, is the real number
ZzZ "x 2 y 2 "z z.
Modulus of a Complex Number
EXAMPLE 3
If z 2 3i, then Zz Z "22 (3)2 "13.
y
As FIGURE 17.1.2 shows, the sum of the vectors z1 and z2 is the vector z1 z2. For the trian
given in the figure, we know that the length of the side of the triangle corresponding to the ve
z1 z2 cannot be longer than the sum of the remaining two sides. In symbols this is
z1 + z2
z1
z1
Zz1 z2 Z Zz1 Z Zz2 Z.
z2
x
FIGURE 17.1.2 Sum of vectors
The result in (5) is known as the triangle inequality and extends to any finite sum:
Zz1 z2 z3 p zn Z Zz1 Z Zz2 Z Zz3 Z p Zzn Z.
Using (5) on z1 z2 (z2), we obtain another important inequality:
Zz1 z2 Z Zz1 Z Zz2 Z.
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(
CHAPTER 17 Functions of a Complex Variable
some remarkable differences as well. For example, we cannot compare two compl
z1 x1 iy1, y1 0, and z2 x2 iy2, y2 0, by means of inequalities. In other w
ments such as z1 z2 and z2 z1 have no meaning except in the case when the tw
z1 and z2 are real. We can, however, compare the absolute values of two comple
Thus, if z1 3 4i and z2 5 i, then |z1| 5 and |z2| !26, and consequentl
This last inequality means that the point (3, 4) is closer to the origin than is the po
Exercises
17.1
Answers to selected odd-numbered problems begin on page ANS-39.
In Problems 1–26, write the given number in the form a ib.
3
1. 2i 3i 5i
2. 3i 5 i 4 7i 3 10i 2 9
3. i 8
4. i 11
5. (5 9i) (2 4i)
6. 3(4 i) 3(5 2i)
7. i (5 7i)
8. i (4 i) 4i(1 2i)
9. (2 3i)(4 i)
11. (2 3i)2
2
i
2 2 4i
15.
3 5i
17.
10. ( 12 14 i)( 23 53 i)
12. (1 i)3
18.
24. (2 3i) a
25. a
22i 2
b
1 2i
i
1
b a
b
32i
2 3i
(1 i) (1 2 2i)
(2 i) (4 2 3i)
z2 in the complex plane.
Zz 6 8iZ 12.
Discussion Problems
43. For n a nonnegative integer, in can be one of four v
1
(1 i) (1 2 2i) (1 3i)
17.2
34. z 2z 7 6i
36. z 2 4z
42. Show for all complex numbers z on the circle x 2 20.
26.
33. 2z i(2 9i)
35. z 2 i
39. 10 8i, 11 6i
40. 12 14 i, 23 16 i
41. Prove that |z1 z2| is the distance between the
(4 5i) 2i 3
(2 i)2
22. (1 i )2 (1 i)3
1
23. (3 6i) (4 i )(3 5i ) 22i
(5 2 4i) 2 (3 7i)
(4 2i) (2 2 3i)
21. i (1 i)(2 i)(2 6i)
19.
In Problems 33–38, use Definition 17.1.2 to find a c
number z satisfying the given equation.
22i
z
38.
3 4i
1 3i
1z
In Problems 39 and 40, determine which complex nu
closer to the origin.
14.
(3 2 i) (2 3i)
1i
28. Re(z 2)
30. Im( z 2 z 2)
32. Zz 5z Z
37. z 2z i
1i
10 2 5i
16.
6 2i
13.
In Problems 27–32, let z x iy. Find the indicated e
27. Re(1/z)
29. Im(2z 4z 4i)
31. Zz 1 3iZ
2
i, and 1. In each of the following four cases expre
exponent n in terms of the symbol k, where k 0
(a) in i (b) in 1 (c) in i (d) in
44. (a) Without doing any significant work such as
out or using the binomial theorem, think of
of evaluating (1 i)8.
(b) Use your method in part (a) to evaluate (1 Powers and Roots
INTRODUCTION Recall from calculus that a point (x, y) in rectangular coordin
be expressed in terms of polar coordinates (r, u). We shall see in this section that
express a complex number z in terms of r and u greatly facilitates finding powers a
Polar Form Rectangular coordinates (x, y) and polar coordinates (r, u) are r
equations x r cos u and y r sin u (see Section 14.1). Thus a nonzero comp
z x iy can be written as z (r cos u) i(r sin u) or
z r (cos u i sin u).
17.2 Powers and Roots
r sin θ
θ
x
r cos θ
FIGURE 17.2.1 Polar coordinates
real axis is positive when measured counterclockwise and negative when measured clockw
The angle u is called an argument of z and is written u arg z. From Figure 17.2.1 we see
an argument of a complex number must satisfy the equation tan u y/x. The solutions of
equation are not unique, since if u0 is an argument of z, then necessarily the angles u0
u0 4p, . . ., are also arguments. The argument of a complex number in the interval p u is called the principal argument of z and is denoted by Arg z. For example, Arg(i) p/2.
A Complex Number in Polar Form
EXAMPLE 1
Express 1 "3i in polar form.
y
5π /3
– π /3
1 – √3i
x
SOLUTION With x 1 and y "3, we obtain r ZzZ #(1)2 ("3)2 2. Now si
the point (1, "3) lies in the fourth quadrant, we can take the solution of tan u "3/1 "
to be u arg z 5p/3. It follows from (1) that a polar form of the number is
z 2 acos
As we see in FIGURE 17.2.2, the argument of 1 "3i that lies in the interval (p, p], the p
cipal argument of z, is Arg z p/3. Thus, an alternative polar form of the complex numbe
p
p
z 2 c cos a b i sin a b d .
3
3
FIGURE 17.2.2 Two arguments of
z 1 "3i in Example 1
5p
5p
i sin
b.
3
3
Multiplication and Division The polar form of a complex number is especially c
venient to use when multiplying or dividing two complex numbers. Suppose
z1 r1(cos u1 i sin u1)
z2 r2(cos u2 i sin u2),
and
where u1 and u2 are any arguments of z1 and z2, respectively. Then
z1z2 r1r2[(cos u1 cos u2 sin u1 sin u2) i(sin u1 cos u2 cos u1 sin u2)]
and for z2 0,
z1
r1
[(cos u1 cos u2 sin u1 sin u2) i(sin u1 cos u2 cos u1 sin u2)].
z2
r2
From the addition formulas from trigonometry, (2) and (3) can be rewritten, in turn, as
z1z2 r1r2 [cos(u1 u2) i sin(u1 u2)]
z1
r1
[cos(u1 u2) i sin(u1 u2)].
z2
r2
and
Inspection of (4) and (5) shows that
Zz1z2 Z Zz1 Z Zz2 Z ,
arg(z1z2) arg z1 arg z2,
and
EXAMPLE 2
2
Zz1 Z
z1
2 ,
z2
Zz2 Z
z1
arga b arg z1 arg z2.
z2
Argument of a Product and of a Quotient
We have seen that Arg z1 p/2 for z1 i. In Example 1 we saw that Arg z2 p/3
z2 1 "3i. Thus, for
z1z2 i(1 2 "3i ) "3 i
824
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CHAPTER 17 Functions of a Complex Variable
and
z1
"3
i
1
i
z2
4
4
1 2 "3i
In Example 2 we used the principal arguments of z1 and z2 and obtained arg(z1z2) arg(z1/z2) Arg(z1/z2). It should be observed, however, that this was a coincidence.
is true for any arguments of z1 and z2, it is not true, in general, that Arg(z1z2) Arg
and Arg(z1/z2) Arg z1 Arg z2. See Problem 39 in Exercises 17.2.
Integer Powers of z We can find integer powers of the complex numbe
results in (4) and (5). For example, if z r (cos u i sin u), then with z1 z and z2 z 2 r 2 [cos (u u) i sin (u u)] r 2 (cos 2u i sin 2u).
Since z 3 z2z, it follows that
z3 r3 (cos 3u i sin 3u).
Moreover, since arg(1) 0, it follows from (5) that
1
z2 r2 [cos(2u) i sin(2u)].
z2
Continuing in this manner, we obtain a formula for the nth power of z for any integ
z n r n (cos nu i sin nu).
EXAMPLE 3
Power of a Complex Number
3
Compute z for z 1 "3i.
SOLUTION In Example 1 we saw that
p
p
z 2 c cos a b i sin a b d .
3
3
Hence from (8) with r 2, u p/3, and n 3, we get
p
p
(1 2 "3i)3 23 c cos a3a b b i sin a3a b b d
3
3
8[cos(p) i sin(p)] 8.
DeMoivre’s Formula When z cos u i sin u, we have ZzZ r 1 and s
(cos u i sin u)n cos nu i sin nu.
This last result is known as DeMoivre’s formula and is useful in deriving certain tr
identities. See Problems 37 and 38 in Exercises 17.2.
Roots A number w is said to be an nth root of a nonzero complex number z
we let w r(cos f i sin f) and z r (cos u i sin u) be the polar forms of w a
view of (8) , w n z becomes
rn (cos nf i sin nf) r (cos u i sin u).
From this we conclude that rn r or r r1/n and
cos nf i sin nf cos u i sin u.
By equating the real and imaginary parts, we get from this equation
cos nf cos u
and
sin nf sin u.
17.2 Powers and Roots
As k takes on the successive integer values k 0, 1, 2, . . ., n 1, we obtain n distinct roots w
the same modulus but different arguments. But for k n we obtain the same roots because
sine and cosine are 2p-periodic. To see this, suppose k n m, where m 0, 1, 2, . . . . The
f
u 2(n m)p
u 2mp
2p
n
n
sin f sin a
and so
u 2mp
b,
n
cos f cos a
u 2mp
b.
n
We summarize this result. The n nth roots of a nonzero complex number z r (cos u i si
are given by
wk r 1>n c cos a
where k 0, 1, 2, . . ., n 1.
EXAMPLE 4
u 2kp
u 2kp
b i sin a
bd,
n
n
Roots of a Complex Number
Find the three cube roots of z i.
SOLUTION With r 1, u arg z p/2, the polar form of the given numbe
z cos(p/2) i sin(p/2). From (10) with n 3 we obtain
wk (1)1>3 c cos a
p>2 2kp
3
Hence, the three roots are
p>2 2kp
3
b d , k 0, 1, 2.
k 0, w0 cos
p
p
"3
1
i sin i
6
6
2
2
k 1, w1 cos
5p
5p
"3
1
i sin
i
6
6
2
2
k 2, w2 cos
3p
3p
i sin
i.
2
2
The root w of a complex number z obtained by using the principal argument of z w
k 0 is sometimes called the principal nth root of z. In Example 4, since Arg(i) p
y
w1
b i sin a
w0
x
w2
FIGURE 17.2.3 Three cube roots of i
w0 "3/2 (1/2)i is the principal third root of i.
Since the roots given by (8) have the same modulus, the n roots of a nonzero comp
number z lie on a circle of radius r1/n centered at the origin in the complex plane. Moreo
since the difference between the arguments of any two successive roots is 2p/n, the nth ro
of z are equally spaced on this circle. FIGURE 17.2.3 shows the three cube roots of i equally spa
on a unit circle; the angle between roots (vectors) wk and wk 1 is 2p/3.
As the next example will show, the roots of a complex number do not have to be “ni
numbers as in Example 3.
EXAMPLE 5
Roots of a Complex Number
Find the four fourth roots of z 1 i.
SOLUTION In this case, r "2 and u arg z p/4. From (10) with n 4, we obtai
wk ( "2)1>4 c cos a
826
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CHAPTER 17 Functions of a Complex Variable
p>4 2kp
4
b i sin a
p>4 2kp
4
b d , k 0, 1, 2, 3.
k 1, w1 ( "2)1>4 c cos
k 2, w2 ( "2)1>4 c cos
k 3, w3 ( "2)1>4 c cos
17.2
Exercises
1. 2
2. 10
3. 3i
4. 6i
5. 1 i
6. 5 5i
7. "3 i
8. 2 2 "3i
10.
12
"3 i
In Problems 11–14, write the number given in polar form in the
form a ib.
11. z 5 acos
7p
7p
i sin
b
6
6
11p
11p
12. z 8 "2 acos
i sin
b
4
4
13. z 6 acos
p
p
i sin b
8
8
14. z 10 acos
p
p
i sin b
5
5
In Problems 15 and 16, find z1z2 and z1/z2. Write the number in
the form a ib.
15. z1 2 acos
p
p
3p
3p
i sin b, z2 4 acos
i sin
b
8
8
8
8
16. z1 "2 acos
z2 "3 acos
p
p
i sin b,
4
4
p
p
i sin b
12
12
In Problems 17–20, write each complex number in polar form.
Then use either (4) or (5) to obtain a polar form of the given
number. Write the polar form in the form a ib.
17. (3 3i)(5 5 "3i)
17p
17p
i sin
d 1.0696 2 0.2127i
16
16
25p
25p
i sin
d 0.2127 2 1.0696i.
16
16
Answers to selected odd-numbered problems begin on page ANS-39.
In Problems 1–10, write the given complex number in polar form.
3
9.
1 i
9p
9p
i sin
d 0.2127 1.0696i
16
16
18. (4 4i)(1 i)
19.
i
2 2 2i
20.
"2 "6i
1 "3i
In Problems 21–26, use (8) to compute the indicated
21. (1 "3i)9
22. (2 2i)5
23. ( 12 12 i)10
25. acos
p
p
i sin b
8
8
26. c "3 acos
12
24. ("2 "6i)
2p
2p 6
i sin
bd
9
9
In Problems 27–32, use (10) to compute all roots. Sk
roots on an appropriate circle centered at the origin.
27. (8)1/3
28. (1)1/8
29. (i)1/2
30. (1 i)1/3
31. (1 "3i)1/2
32. (1 "3i)1/4
33. z4 1 0
34. z8 2z4 1 0
In Problems 33 and 34, find all solutions of the given
In Problems 35 and 36, express the given complex n
in polar form and then in the form a ib.
35. acos
p
p
p 12
p 5
i sin b c 2 acos i sin b d
9
9
6
6
3p
3p 3
i sin
bd
8
8
36.
p
p 10
c 2 acos
i sin b d
16
16
c 8 acos
37. Use the result (cos u i sin u)2 cos 2u i s
trigonometric identities for cos 2u and sin 2u.
38. Use the result (cos u i sin u)3 cos 3u i s
trigonometric identities for cos 3u and sin 3u.
17.2 Powers and Roots
(b) If z1 1 and z2 5i, verify that
Arg(z1/z2) Arg(z1) Arg(z2).
17.3
z1
arg a b arg(z1) arg(z2).
z2
and
Sets in the Complex Plane
INTRODUCTION In the preceding sections we examined some rudiments of the algebra
geometry of complex numbers. But we have barely scratched the surface of the subject kno
as complex analysis; the main thrust of our study lies ahead. Our goal in the sections and chap
that follow is to examine functions of a single complex variable z x iy and the calculu
these functions.
Before introducing the notion of a function of a complex variable, we need to state so
essential definitions and terminology about sets in the complex plane.
Terminology Before discussing the concept of functions of a complex variable, we n
to introduce some essential terminology about sets in the complex plane.
z0
ρ
Suppose z0 x0 iy0. Since Zz z0 Z "(x 2 x0)2 ( y 2 y0)2 is the distance between
points z x iy and z0 x0 iy0, the points z x iy that satisfy the equation
Zz z0 Z r,
|z – z0| = ρ
FIGURE 17.3.1 Circle of radius r
r
0, lie on a circle of radius r centered at the point z0. See FIGURE 17.3.1.
EXAMPLE 1
Circles
(a) Zz Z 1 is the equation of a unit circle centered at the origin.
(b) Zz 1 2i Z 5 is the equation of a circle of radius 5 centered at 1 2i.
z0
FIGURE 17.3.2 Open set
The points z satisfying the inequality Zz z0 Z r, r 0, lie within, but not on, a circl
radius r centered at the point z0. This set is called a neighborhood of z0 or an open disk
point z0 is said to be an interior point of a set S of the complex plane if there exists some nei
borhood of z0 that lies entirely within S. If every point z of a set S is an interior point, then
said to be an open set. See FIGURE 17.3.2. For example, the inequality Re(z) 1 defines a r
half-plane, which is an open set. All complex numbers z x iy for which x 1 are in
set. If we choose, for example, z0 1.1 2i, then a neighborhood of z0 lying entirely in the
is defined by Zz (1.1 2i)Z 0.05. See FIGURE 17.3.3. On the other hand, the set S of point
the complex plane defined by Re(z) 1 is not open, since every neighborhood of a point on
line x 1 must contain points in S and points not in S. See FIGURE 17.3.4.
|z – (1.1 + 2i)| < 0.05
y
y
in S
z = 1.1 + 2i
not in S
x
x
x=1
FIGURE 17.3.3 Open set magnified view
of a point near x 1
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CHAPTER 17 Functions of a Complex Variable
x=1
FIGURE 17.3.4 Set S is not open
y
x
Im(z) < 0
lower half-plane
(a)
–1 < Re(z) < 1
infinite strip
(b)
y
y
x
|z| > 1
exterior of unit circle
(c)
1 < |z| < 2
circular ring
(d)
FIGURE 17.3.5 Four examples of open sets
The set of numbers satisfying the inequality
r1
z2
z1
FIGURE 17.3.6 Connected set
Zz z0 Z
r2,
such as illustrated in Figure 17.3.5(d), is called an open annulus.
If every neighborhood of a point z0 contains at least one point that is in a set S
one point that is not in S, then z0 is said to be a boundary point of S. The bounda
is the set of all boundary points of S. For the set of points defined by Re(z) 1, t
the line x 1 are boundary points. The points on the circle Z z i Z 2 are bounda
the disk Z z i Z 2.
If any pair of points z1 and z2 in an open set S can be connected by a polygonal
entirely in the set, then the open set S is said to be connected. See FIGURE 17.3.6. An op
set is called a domain. All the open sets in Figure 17.3.5 are connected and so are d
set of numbers satisfying Re(z) 4 is an open set but is not connected, since it is
to join points on either side of the vertical line x 4 by a polygonal line without le
(bear in mind that the points on x 4 are not in the set).
A region is a domain in the complex plane with all, some, or none of its boun
Since an open connected set does not contain any boundary points, it is automatica
A region containing all its boundary points is said to be closed. The disk defined b
is an example of a closed region and is referred to as a closed disk. A region may be
nor closed; the annular region defined by 1 Zz 5Z 3 contains only some of
points and so is neither open nor closed.
REMARKS
Often in mathematics the same word is used in entirely different contexts. Do not
concept of “domain” defined in this section with the concept of the “domain of a
17.3 Sets in the Complex Plane
1. Re(z) 5
17. 0 arg (z) 2p/3
2. Im(z) 2
p/4
19. Zz i Z
3. Im(z 3i) 6
4. Im(z i) Re(z 4 3i)
21.
5. Zz 3i Z 2
6. Z2z 1 Z 4
23.
7. Zz 4 3i Z 5
8. Zz 2 2i Z 2
24.
In Problems 9–22, sketch the set of points in the complex plane
satisfying the given inequality. Determine whether the set is a
domain.
1
11. Im(z) 3
13. 2 Re(z 1)
18. Zarg (z) Z
10. ZRe(z) Z
9. Re(z)
2
12. Im(z i)
26.
5
14. 1 Im(z)
4
25.
17.4
4
1
20. Zz iZ
0
2 Zz i Z 3
22. 1 Zz 1 iZ
2
Describe the set of points in the complex plane that satis
Zz 1 Z Zz iZ.
Describe the set of points in the complex plane that satis
ZRe(z)Z Zz Z.
Describe the set of points in the complex plane that satis
z2 z 2 2.
Describe the set of points in the complex plane that satis
Zz i Z Zz i Z 1.
Functions of a Complex Variable
INTRODUCTION One of the most important concepts in mathematics is that of a function. Y
may recall from previous courses that a function is a certain kind of correspondence between
sets; more specifically: A function f from a set A to a set B is a rule of correspondence that assi
to each element in A one and only one element in B. If b is the element in the set B assigned to
element a in the set A by f, we say that b is the image of a and write b f (a). The set A is ca
the domain of the function f (but is not necessarily a domain in the sense defined in Section 17
The set of all images in B is called the range of the function. For example, suppose the set A is a
of real numbers defined by 3 x q and the function is given by f (x) !x 2 3; then f (3) f (4) 1, f (8) !5, and so on. In other words, the range of f is the set given by 0 y q. Si
A is a set of real numbers, we say f is a function of a real variable x.
Functions of a Complex Variable When the domain A in the foregoing definit
of a function is a set of complex numbers z, we naturally say that f is a function of a comp
variable z or a complex function for short. The image w of a complex number z will be so
complex number u iv; that is,
w f (z) u(x, y) iv(x, y),
where u and v are the real and imaginary parts of w and are real-valued functions. Inherent in
mathematical statement (1) is the fact that we cannot draw a graph of a complex function w since a graph would require four axes in a four-dimensional coordinate system.
Some examples of functions of a complex variable are
f (z) z2 4z,
y
w = f (z)
z
domain of f
v
f (z) range of f
w
x
u
z
,
z 1
2
f (z) z Re(z),
z any complex number
z 2 i and z 2 i
z any complex number.
Each of these functions can be expressed in form (1). For example,
f (z) z2 4z (x iy)2 4(x iy) (x2 y2 4x) i(2xy 4y).
(a) z-plane
(b) w-plane
FIGURE 17.4.1 Mapping from z-plane to
w-plane
830
|
Thus, u(x, y) x2 y2 4x, and v(x, y) 2xy 4y.
Although we cannot draw a graph, a complex function w f (z) can be interpreted as a mapp
or transformation from the z-plane to the w-plane. See FIGURE 17.4.1.
CHAPTER 17 Functions of a Complex Variable
Re(z) ⫽ x and so by substituting x ⫽ 1 into the functions u and v, we obtain u ⫽
v ⫽ 2y. These are parametric equations of a curve in the w-plane. Substituting
the first equation eliminates the parameter y to give u ⫽ 1 ⫺ v 2 /4. In other word
of the line in FIGURE 17.4.2(a) is the parabola shown in Figure 17.4.2(b).
x=1
(a) z-plane
(b) w-plane
FIGURE 17.4.2 Image of x ⫽ 1 is a
parabola
We shall pursue the idea of f (z) as a mapping in greater detail in Chapter 20.
It should be noted that a complex function is completely determined by the
functions u and v. This means a complex function w ⫽ f (z) can be defined by arbitrari
u(x, y) and v(x, y), even though u ⫹ iv may not be obtainable through the familiar oper
symbol z alone. For example, if u(x, y) ⫽ xy2 and v(x, y) ⫽ x2 ⫺ 4y3, then f (z) ⫽ xy2 ⫹
is a function of a complex variable. To compute, say, f (3 ⫹ 2i), we substitute x ⫽
into u and v to obtain f (3 ⫹ 2i) ⫽ 12 ⫺ 23i.
y
i
x
–i
FIGURE 17.4.3 f1(z) ⫽ z (normalized)
Complex Functions as Flows We also may interpret a complex funct
as a two-dimensional fluid flow by considering the complex number f (z) as a v
at the point z. The vector f (z) specifies the speed and direction of the flow at a g
FIGURES 17.4.3 and 17.4.4 show the flows corresponding to the complex functions f
f2(z) ⫽ z 2, respectively.
If x(t) ⫹ iy(t) is a parametric representation for the path of a particle in the flow
vector T ⫽ x⬘(t) ⫹ iy⬘(t) must coincide with f (x(t) ⫹ iy(t)). When f (z) ⫽ u(x, y)
follows that the path of the particle must satisfy the system of differential equation
dx
⫽ u(x, y)
dt
y
dy
⫽ v(x, y).
dt
We call the family of solutions of this system the streamlines of the flow
with f (z).
x
EXAMPLE 2
Streamlines
Find the streamlines of the flows associated with the complex functions (a) f
(b) f2(z) ⫽ z2.
SOLUTION
FIGURE 17.4.4 f2(z) ⫽ z 2 (normalized)
(a) The streamlines corresponding to f1(z) ⫽ x ⫺ iy satisfy the sy
dx
⫽x
dt
dy
⫽ ⫺y
dt
and so x(t) ⫽ c1et and y(t) ⫽ c2e⫺t. By multiplying these two parametric equat
that the point x(t) ⫹ iy(t) lies on the hyperbola xy ⫽ c1c2.
(b) To find the streamlines corresponding to f2(z) ⫽ (x2 ⫺ y2) ⫹ i 2xy, note that dx/
dy/dt ⫽ 2xy, and so
dy
2xy
⫽ 2
.
dx
x 2 y2
This homogeneous differential equation has the solution x2 ⫹ y2 ⫽ c2 y, which
family of circles that have centers on the y-axis and pass through the origin.
Limits and Continuity The definition of a limit of a complex function f
has the same outward appearance as the limit in real variables.
17.4 Functions of a Complex Variable
lim f (z) L
zSz0
if, for each e
y
v
z0
f (z)
z
δ
ε
L
D
x
(a) δ -neighborhood
u
(b) ε -neighborhood
FIGURE 17.4.5 Geometric meaning of a
complex limit
0, there exists a d
0 such that Z f (z) LZ
e whenever 0
Zz z0 Z
d
In words, lim zSz0 f (z) L means that the points f (z) can be made arbitrarily close to
point L if we choose the point z sufficiently close to, but not equal to, the point z0. As show
FIGURE 17.4.5, for each e-neighborhood of L (defined by Z f (z) L Z e) there is a d-neighborh
of z0 (defined by Zz z0 Z d) so that the images of all points z z0 in this neighborhood li
the e-neighborhood of L.
The fundamental difference between this definition and the limit concept in real variab
lies in the understanding of z S z0. For a function f of a single real variable x, lim xSx0 f (x) means f (x) approaches L as x approaches x0 either from the right of x0 or from the lef
x0 on the real number line. But since z and z0 are points in the complex plane, when we
that lim zSz0 f (z) exists, we mean that f (z) approaches L as the point z approaches z0 fr
any direction.
The following theorem summarizes some properties of limits:
Limit of Sum, Product, Quotient
Theorem 17.4.1
Suppose lim zSz0 f (z) L1 and lim zSz0 g(z) L2. Then
(i) lim [ f (z) g(z)] L1 L2
zSz0
(ii) lim f (z)g(z) L1L2
zSz0
(iii) lim
zSz0
f (z)
L1
,
g(z)
L2
L2 0.
Continuity at a Point
Definition 17.4.2
A function f is continuous at a point z0 if
lim f (z) f (z0).
zSz0
As a consequence of Theorem 17.4.1, it follows that if two functions f and g are continu
at a point z0, then their sum and product are continuous at z0. The quotient of the two functi
is continuous at z0 provided g(z0) 0.
A function f defined by
f (z) anzn an1zn1 p a2z2 a1z a0,
an 0,
where n is a nonnegative integer and the coefficients ai, i 0, 1, . . ., n, are complex consta
is called a polynomial of degree n. Although we shall not prove it, the limit result lim z zSz0
indicates that the simple polynomial function f (z) z is continuous everywhere—that is
the entire z-plane. With this result in mind and with repeated applications of Theorem 17.4.1
and (ii), it follows that a polynomial function (2) is continuous everywhere. A ratio
function
f (z) g(z)
,
h(z)
where g and h are polynomial functions, is continuous except at those points at which h(z) is z
832
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CHAPTER 17 Functions of a Complex Variable
Definition 17.4.3
Derivative
Suppose the complex function f is defined in a neighborhood of a point z0. The de
f at z0 is
f 9(z0) lim
DzS0
f (z0 Dz) 2 f (z0)
Dz
provided this limit exists.
If the limit in (3) exists, the function f is said to be differentiable at z0. The de
function w f (z) is also written dw/dz.
As in real variables, differentiability implies continuity:
If f is differentiable at z0, then f is continuous at z0.
Moreover, the rules of differentiation are the same as in the calculus of real variab
are differentiable at a point z, and c is a complex constant, then
Constant Rules:
d
d
c 0,
cf (z) c f 9(z)
dz
dz
Sum Rule:
d
f f (z) g(z)g f 9(z) g9(z)
dz
Product Rule:
d
f f (z)g(z)g f (z)g9(z) g(z) f 9(z)
dz
Quotient Rule:
Chain Rule:
g(z) f 9(z) 2 f (z)g9(z)
d f (z)
c
d dz g(z)
fg(z)g 2
d
f (g(z)) f 9(g(z))g9(z).
dz
The usual Power Rule for differentiation of powers of z is also valid:
d n
z nz n 2 1 , n an integer.
dz
EXAMPLE 3
Using the Rules of Differentiation
Differentiate (a) f (z) 3z4 5z3 2z and (b) f (z) SOLUTION
z2
.
4z 1
(a) Using the Power Rule (9) along with the Sum Rule (5), we ob
f (z) 3 4z3 5 3z2 2 12z3 15z2 2.
(b) From the Quotient Rule (7),
f 9(z) (4z 1) 2z 2 z 2 4
4z 2 2z
.
2
(4z 1)
(4z 1)2
In order for a complex function f to be differentiable at a point z0, lim
f (z0 DzS0
must approach the same complex number from any direction. Thus in the study
variables, to require the differentiability of a function is a greater demand than in re
If a complex function is made up, such as f (z) x 4iy, there is a good chance
differentiable.
17.4 Functions of a Complex Variable
f (z z) f (z) (x x) 4i( y y) x 4iy x 4i y
and so
lim
DzS0
f (z Dz) 2 f (z)
Dx 4iDy
lim
.
Dz
DzS0 Dx iDy
Now, if we let z S 0 along a line parallel to the x-axis, then y 0 and the value of (10) i
On the other hand, if we let z S 0 along a line parallel to the y-axis, then x 0 and the va
of (10) is seen to be 4. Therefore, f (z) x 4iy is not differentiable at any point z.
Analytic Functions While the requirement of differentiability is a stringent dema
there is a class of functions that is of great importance whose members satisfy even more sev
requirements. These functions are called analytic functions.
Definition 17.4.4
Analyticity at a Point
A complex function w f (z) is said to be analytic at a point z0 if f is differentiable at z0 an
at every point in some neighborhood of z0.
A function f is analytic in a domain D if it is analytic at every point in D.
The student should read Definition 17.4.4 carefully. Analyticity at a point is a neighborh
property. Analyticity at a point is, therefore, not the same as differentiability at a point. It is
as an exercise to show that the function f (z) Zz Z 2 is differentiable at z 0 but is differentia
nowhere else. Hence, f (z) Zz Z 2 is nowhere analytic. In contrast, the simple polynomial f (z) is differentiable at every point z in the complex plane. Hence, f (z) z 2 is analytic everywh
A function that is analytic at every point z is said to be an entire function. Polynomial functi
are differentiable at every point z and so are entire functions.
REMARKS
Recall from algebra that a number c is a zero of a polynomial function if and only if x is a factor of f (x). The same result holds in complex analysis. For example, sin
f (z) z4 5z2 4 (z2 1)(z2 4), the zeros of f are i, i, 2i, and 2i. Henc
f (z) (z i)(z i)(z 2i)(z 2i). Moreover, the quadratic formula is also valid. F
example, using this formula, we can write
f (z) z2 2z 2 (z (1 i))(z (1 i))
(z 1 i)(z 1 i).
See Problems 21 and 22 in Exercises 17.4.
Exercises
17.4
Answers to selected odd-numbered problems begin on page ANS-40.
In Problems 1–6, find the image of the given line under the mapping f (z) z2.
1. y 2
3. x 0
5. y x
2. x 3
4. y 0
6. y x
In Problems 7–14, express the given function in the form
f (z) u iv.
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CHAPTER 17 Functions of a Complex Variable
7. f (z) 6z 5 9i
8. f (z) 7z 9i z 3 2i
9. f (z) z 2 3z 4i
10. f (z) 3z 2 2z
11. f (z) z3 4z
12. f (z) z4
13. f (z) z 1/z
14. f (z) z
z1
(a) 2i
(b) 2 i
(c) 5 3i
(b) 2 i
(c) 1 4i
3z 2 4 8i
2z i
33. f (z) 17. f (z) 4z i z Re(z)
(c) 2 7i
In Problems 35–38, give the points at which the give
will not be analytic.
(c) 3 pi/3
35. f (z) (a) 1 i
(a) 4 6i
(b) 5 12i
18. f (z) ex cos y iex sin y
(a) pi/4
(b) 1 pi
In Problems 19–22, the given limit exists. Find its value.
zSi
5z 2 2 2z 2
zS1 2 i
z1
z4 2 1
21. lim
zSi z 2 i
lim
22.
z 2 2 2z 2
zS1 i
z 2 2 2i
lim
In Problems 23 and 24, show that the given limit does not exist.
23. lim
zS0
z
z
24. lim
zS1
xy21
z21
In Problems 25 and 26, use (3) to obtain the indicated derivative
of the given function.
25. f (z) z2, f (z) 2z
26. f (z) 1/z, f (z) 1/z2
In Problems 27–34, use (4)–(8) to find the derivative f (z) for
the given function.
27. f (z) 4z3 (3 i)z2 5z 4
36. f (z) 2i
z 2 2z
2
z3 z
z24
38. f (z) 2
2
z 4
z 2 6z
39. Show that the function f (z) z is nowhere diffe
40. The function f (z) |z|2 is continuous througho
complex plane. Show, however, that f is differen
the point z 0. [Hint: Use (3) and consider two
and z 0. In the second case let z approach
line parallel to the x-axis and then let z approac
a line parallel to the y-axis.]
37. f (z) 19. lim (4z3 5z2 4z 1 5i)
20.
z
z 2 3i
34. f (z) 5z 2 2 z
z3 1
16. f (z) (x 1 1/x) i(4x2 2y2 4)
In Problems 41–44, find the streamlines of the flow
with the given complex function.
41. f (z) 2z
43. f (z) 1/ z
42. f (z) iz
44. f (z) x2 iy2
In Problems 45 and 46, use a graphics calculator or
to obtain the image of the given parabola under the m
f (z) z2.
45. y 12 x2
46. y (x 1)2
28. f (z) 5z4 iz3 (8 i)z2 6i
17.5
Cauchy–Riemann Equations
INTRODUCTION In the preceding section we saw that a function f of a complex
analytic at a point z when f is differentiable at z and differentiable at every point in
borhood of z. This requirement is more stringent than simply differentiability at a p
a complex function can be differentiable at a point z but yet be differentiable now
function f is analytic in a domain D if f is differentiable at all points in D. We shall
a test for analyticity of a complex function f (z) u(x, y) iv(x, y).
A Necessary Condition for Analyticity In the next theorem we see that
f (z) u(x, y) iv(x, y) is differentiable at a point z, then the functions u and v must
of equations that relate their first-order partial derivatives. This result is a necessa
for analyticity.
Theorem 17.5.1
Cauchy–Riemann Equations
Suppose f (z) u(x, y) iv(x, y) is differentiable at a point z x iy. Then at z the
partial derivatives of u and v exist and satisfy the Cauchy–Riemann equations
0u
0v
0x
0y
and
0u
0v
.
0y
0x
17.5 Cauchy–Riemann Equations
Dz
DzS0
By writing f (z) u(x, y) iv(x, y) and z x i y, we get from (2)
f 9(z) lim
DzS0
u(x Dx, y Dy) iv(x Dx, y Dy) 2 u(x, y) 2 iv(x, y)
.
Dx iDy
Since this limit exists, z can approach zero from any convenient direction. In particula
z S 0 horizontally, then z x and so (3) becomes
f 9(z) lim
DxS0
u(x Dx, y) 2 u(x, y)
v(x Dx, y) 2 v(x, y)
i lim
.
Dx
DxS0
Dx
Since f (z) exists, the two limits in (4) exist. But by definition the limits in (4) are the first par
derivatives of u and v with respect to x. Thus, we have shown that
f 9(z) 0u
0v
i .
0x
0x
Now if we let z S 0 vertically, then z i y and (3) becomes
f 9(z) lim
DyS0
u(x, y Dy) 2 u(x, y)
v(x, y Dy) 2 v(x, y)
i lim
,
iDy
DyS0
iDy
which is the same as
f 9(z) i
0u
0v
.
0y
0y
Equating the real and imaginary parts of (5) and (7) yields the pair of equations in (1).
If a complex function f (z) u(x, y) iv(x, y) is analytic throughout a domain D, then the
functions u and v must satisfy the Cauchy–Riemann equations (1) at every point in D.
EXAMPLE 1
Using the Cauchy–Riemann Equations
The polynomial f (z) z2 z is analytic for all z and f (z) x2 y2 x i(2xy y). Th
u(x, y) x2 y2 x and v(x, y) 2xy y. For any point (x, y), we see that the Cauc
Riemann equations are satisfied:
0u
0v
2x 1 0x
0y
EXAMPLE 2
and
0u
0v
2y .
0y
0x
Using the Cauchy–Riemann Equations
Show that the function f (z) (2x2 y) i( y2 x) is not analytic at any point.
SOLUTION We identify u(x, y) 2x2 y and v(x, y) y2 x. Now from
0u
4x
0x
and
0v
2y
0y
0u
1x
0y
and
0v
1
0x
we see that 0u/0y 0v/0x but that the equality 0u/0x 0v/0y is satisfied only on the
y 2x. However, for any point z on the line, there is no neighborhood or open disk about
which f is differentiable. We conclude that f is nowhere analytic.
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CHAPTER 17 Functions of a Complex Variable
Theorem 17.5.2
Criterion for Analyticity
Suppose the real-valued functions u(x, y) and v(x, y) are continuous and have
first-order partial derivatives in a domain D. If u and v satisfy the Cauchy–Rieman
at all points of D, then the complex function f (z) u(x, y) iv(x, y) is analytic in
EXAMPLE 3
Using Theorem 17.5.2
For the function f (z) y
x
2i 2
we have
x 2 y2
x y2
y2 2 x 2
0u
0v
2
0x
0y
(x y 2)2
and
2xy
0u
0v
2
.
0y
0x
(x y 2)2
In other words, the Cauchy–Riemann equations are satisfied except at the
x2 y2 0; that is, at z 0. We conclude from Theorem 17.5.2 that f is an
domain not containing the point z 0.
The results in (5) and (7) were obtained under the basic assumption that f was d
at the point z. In other words, (5) and (7) give us a formula for computing f (z):
f 9(z) 0u
0v
0v
0u
i
2i .
0x
0x
0y
0y
For example, we know that f (z) z2 is differentiable for all z. With u(x, y) x2 y
v(x, y) 2xy, and 0v/0x 2y, we see that
f (z) 2x i2y 2(x iy) 2z.
Recall that analyticity implies differentiability but not vice versa. Theorem 17.5.2 has
that gives sufficient conditions for differentiability:
If the real-valued functions u(x, y) and v(x, y) are continuous and have continu
order partial derivatives in a neighborhood of z, and if u and v satisfy the Cauchy–
equations at the point z, then the complex function f (z) u(x, y) iv(x, y) is diffe
at z and f (z) is given by (8).
The function f (z) x2 y2 i is nowhere analytic. With the identifications u(x,
v(x, y) y2, we see from
0u
0v
0u
0v
2x,
2y
and
0,
0
0x
0y
0y
0x
that the Cauchy–Riemann equations are satisfied only when y x. But since the
0u/0x, 0u/0y, v, 0v/0x, and 0v/0y are continuous at every point, it follows that f is diffe
the line y x and on that line (8) gives the derivative f (z) 2x 2y.
Harmonic Functions We saw in Chapter 13 that Laplace’s equation 02u/0x2 occurs in certain problems involving steady-state temperatures. This partial differen
also plays an important role in many areas of applied mathematics. Indeed, as we
real and imaginary parts of an analytic function cannot be chosen arbitrarily, since
must satisfy Laplace’s equation. It is this link between analytic functions and Laplac
that makes complex variables so essential in the serious study of applied mathema
Definition 17.5.1
Harmonic Functions
A real-valued function f(x, y) that has continuous second-order partial derivatives in
and satisfies Laplace’s equation is said to be harmonic in D.
17.5 Cauchy–Riemann Equations
We will see in Chapter 18 that
an analytic function possesses
derivatives of all orders.
PROOF: In this proof we shall assume that u and v have continuous second-order par
derivatives. Since f is analytic, the Cauchy–Riemann equations are satisfied. Differentiating b
sides of 0u/0x 0v/0y with respect to x and differentiating both sides of 0u/0y 0v/0x with resp
to y then give
0 2u
0 2v
2
0x 0y
0x
and
0 2u
0 2v
.
2
0y 0x
0y
With the assumption of continuity, the mixed partials are equal. Hence, adding these two eq
tions gives
0 2u
0 2u
0.
0x 2
0y 2
This shows that u(x, y) is harmonic.
Now differentiating both sides of 0u/0x 0v/0y with respect to y and differentiating both si
of 0u/0y 0v/0x with respect to x and subtracting yield
0 2v
0 2v
0.
0x 2
0y 2
Harmonic Conjugate Functions If f (z) u(x, y) iv(x, y) is analytic in a domain
then u and v are harmonic in D. Now suppose u(x, y) is a given function that is harmonic in
It is then sometimes possible to find another function v(x, y) that is harmonic in D so
u(x, y) iv(x, y) is an analytic function in D. The function v is called a harmonic conjug
function of u.
Harmonic Function/Harmonic Conjugate Function
EXAMPLE 4
(a) Verify that the function u(x, y) x3 3xy2 5y is harmonic in the entire complex pla
(b) Find the harmonic conjugate function of u.
SOLUTION
(a) From the partial derivatives
0u
3x 2 2 3y 2,
0x
0 2u
6x,
0x 2
0u
6xy 2 5,
0y
0 2u
6x
0y 2
we see that u satisfies Laplace’s equation:
0 2u
0 2u
6x 2 6x 0.
0x 2
0y 2
(b) Since the harmonic conjugate function v must satisfy the Cauchy–Riemann equatio
we must have
0v
0u
3x 2 2 3y 2
0y
0x
and
0v
0u
6xy 5.
0x
0y
Partial integration of the first equation in (9) with respect to y gives v(x, y) 3x2y y3 h
From this we get
0v
6xy h9(x).
0x
Substituting this result into the second equation in (9) gives h (x) 5, and so h(x) 5x Therefore, the harmonic conjugate function of u is v(x, y) 3x2y y3 5x C. The anal
function is f (z) x3 3xy2 5y i(3x2y y3 5x C ).
838
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CHAPTER 17 Functions of a Complex Variable
analytic function f (z). The level curves u(x, y) c1 and v(x, y) c2 defined by thes
form two orthogonal families of curves. See Problem 32 in Exercises 17.5. For ex
level curves generated by the simple analytic function f (z) z x iy are x c1
The family of vertical lines defined by x c1 is clearly orthogonal to the family of
lines defined by y c2. In electrostatics, if u(x, y) c1 defines the equipotential c
the other, and orthogonal, family v (x, y) c2 defines the lines of force.
Exercises
17.5
Answers to selected odd-numbered problems begin on page ANS-40.
In Problems 1 and 2, the given function is analytic for all z.
Show that the Cauchy–Riemann equations are satisfied at every
point.
1. f (z) z
3
2
2. f (z) 3z 5z 6i
In Problems 3–8, show that the given function is not analytic at
any point.
3. f (z) Re(z)
4. f (z) y ix
5. f (z) 4z 6z 3
7. f (z) x 2 y 2
6. f (z) z 2
8. f (x) In Problems 9–14, use Theorem 17.5.2 to show that the given
function is analytic in an appropriate domain.
9. f (z) ex cos y iex sin y
10. f (z) x sin x cosh y i(y cos x sinh y)
2
2
f (z) x 2 y 2 2xyi; x-axis
f (z) 3x 2y 2 6x 2y 2i; coordinate axes
f (z) x 3 3xy 2 x i( y 3 3x 2y y); coord
f (z) x 2 x y i( y 2 5y x); y x 2
Use (8) to find the derivative of the function in P
Use (8) to find the derivative of the function in P
23. u(x, y) x
24. u(x, y) 2x 2
2
25. u(x, y) x y
2
26. u(x, y) 4xy3 4x3y x
27. u(x, y) loge(x2 y2)
28. u(x, y) ex (x cos y y sin y)
2
11. f (z) e x 2 y cos 2xy ie x 2 y sin 2xy
12. f (z) 4x2 5x 4y2 9 i(8xy 5y 1)
29. Sketch the level curves u(x, y) c1 and v(x, y)
y
x21
13. f (z) 2i
2
2
(x 2 1) y
(x 2 1)2 y 2
x 3 xy 2 x
x 2y y 3 2 y
14. f (x) i
x 2 y2
x 2 y2
In Problems 15 and 16, find real constants a, b, c, and d so that
the given function is analytic.
15. f (z) 3x y 5 i(ax by 3)
16. f (z) x2 axy by2 i(cx2 dxy y2)
17.6
17.
18.
19.
20.
21.
22.
In Problems 23–28, verify that the given function u i
Find v, the harmonic conjugate function of u. Form t
sponding analytic function f (z) u iv.
y
x
i 2
x 2 y2
x y2
2
In Problems 17–20, show that the given function is n
at any point but is differentiable along the indicated
analytic function f (z) z2.
30. Consider the function f (z) 1/z. Describe the le
31. Consider the function f (z) z 1/z. Describe th
v(x, y) 0.
32. Suppose u and v are the harmonic functions form
and imaginary parts of an analytic function. S
level curves u(x, y) c1 and v(x, y) c2 are ortho
Consider the gradient of u and the gradient of
case where a gradient vector is the zero vector.]
Exponential and Logarithmic Functions
INTRODUCTION In this and the next section, we shall examine the exponential,
trigonometric, and hyperbolic functions of a complex variable z. Although the d
these complex functions are motivated by their real variable analogues, the prope
complex functions will yield some surprises.
Exponential Function Recall that in real variables the exponential funct
has the properties
f (x) f (x)
and
f (x1 x2) f (x1)f (x2).
17.6 Exponential and Logarithmic Functions
eiy cos y i sin y,
y a real number,
played an important role in Section 3.3. We can formally establish the result in (2) by using
Maclaurin series for ex and replacing x by iy and rearranging terms:
q (iy)k
(iy)2
(iy)3
(iy)4
e iy a
1 iy p
2!
3!
4!
k 0 k!
Maclaurin series for
cos y and sin y.
a1 2
y2
y4
y6
y3
y5
y7
2
p b i ay 2
2
pb
2!
4!
6!
3!
5!
7!
cos y i sin y.
For z x iy, it is natural to expect that
e xiy e xe iy
e xiy ex (cos y i sin y).
and so by (2),
Inspired by this formal result, we make the following definition.
Exponential Function
Definition 17.6.1
ez e xiy ex (cos y i sin y).
(
The exponential function ez is also denoted by the symbol exp z. Note that (3) reduces to ex w
y 0.
Complex Value of the Exponential Function
EXAMPLE 1
Evaluate e
1.74.2i
.
SOLUTION With the identifications x 1.7 and y 4.2 and the aid of a calculator, we ha
to four rounded decimal places,
e1.7 cos 4.2 2.6837
e1.7 sin 4.2 4.7710.
and
It follows from (3) that e1.74.2i 2.6837 4.7710i.
The real and imaginary parts of e z, u(x, y) e x cos y and v(x, y) e x sin y, are continu
and have continuous first partial derivatives at every point z of the complex plane. Moreover,
Cauchy–Riemann equations are satisfied at all points of the complex plane:
0u
0v
e x cos y 0x
0y
and
0u
0v
e x sin y .
0y
0x
It follows from Theorem 17.5.2 that f (z) e z is analytic for all z; in other words, f is an en
function.
Properties We shall now demonstrate that e z possesses the two desired properties gi
in (1). First, the derivative of f is given by (5) of Section 17.5:
f (z) e x cos y i(e x sin y) e x (cos y i sin y) f (z).
As desired, we have established that
d z
e e z.
dz
840
|
CHAPTER 17 Functions of a Complex Variable
z + 2π i
⫽ e x1 ⫹ x2 f( cos y1 cos y2 2 sin y1 sin y2) ⫹ i ( sin y1 cos y2 ⫹ cos y1 s
πi
z
x
z – 2πi
⫽ e x1 ⫹ x2 f( cos( y1 ⫹ y2) ⫹ i sin( y1 ⫹ y2)g ⫽ f (z1 ⫹ z2).
e z1e z2 ⫽ e z1 ⫹ z2.
In other words,
–π i
It is left as an exercise to prove that
e z1
⫽ e z1 2 z2.
e z2
–3π i
FIGURE 17.6.1 Values of f (z) ⫽ e z at the
four points are the same
Periodicity Unlike the real function ex, the complex function f (z) ⫽ ez is p
y
πi
x
–πi
the complex period 2pi. Since e2pi ⫽ cos 2p ⫹ i sin 2p ⫽ 1 and, in view of (4), ez⫹2pi
for all z, it follows that f (z ⫹ 2pi) ⫽ f (z). Because of this complex periodicity,
functional values of f (z) ⫽ ez are assumed in any infinite horizontal strip of wid
if we divide the complex plane into horizontal strips defined by (2n ⫺ 1)p ⬍ y ⱕ
n ⫽ 0, ⫾1, ⫾2, . . ., then, as shown in FIGURE 17.6.1, for any point z in the strip ⫺p ⬍
values f (z), f (z ⫹ 2pi), f (z ⫺ 2pi), f (z ⫹ 4pi), and so on, are the same. The strip ⫺
is called the fundamental region for the exponential function f (z) ⫽ ez. The corresp
over the fundamental region is shown in FIGURE 17.6.2.
Polar Form of a Complex Number In Section 17.2, we saw that the comp
could be written in polar form as z ⫽ r (cos u ⫹ i sin u). Since eiu ⫽ cos u ⫹ i sin u,
write the polar form of a complex number as
z ⫽ reiu.
FIGURE 17.6.2 Flow over the fundamental
region
For example, in polar form z ⫽ 1 ⫹ i is z ⫽ "2e pi>4 .
Circuits In applying mathematics, mathematicians and engineers often appro
problem in completely different ways. Consider, for example, the solution of Ex
Section 3.8. In this example we used strictly real analysis to find the steady-state cu
an LRC-series circuit described by the differential equation
L
d 2q
dq
1
⫹R
⫹ q ⫽ E0 sin gt.
2
dt
C
dt
Electrical engineers often solve circuit problems such as this using complex
illustrate, let us first denote the imaginary unit !⫺1 by the symbol j to avoid co
the current i. Since current i is related to charge q by i ⫽ dq/dt, the differential eq
same as
L
1
di
⫹ Ri ⫹ q ⫽ E0 sin gt.
dt
C
Moreover, the impressed voltage E0 sin gt can be replaced by Im(E0e jgt ), where I
“imaginary part of.” Because of this last form, the method of undetermined coefficie
that we assume a solution in the form of a constant multiple of complex exponen
ip(t) ⫽ Im(Ae jgt). We substitute this expression into the last differential equation, use
q is an antiderivative of i, and equate coefficients of e jgt:
a jLg ⫹ R ⫹
1
b A ⫽ E0 gives A ⫽
jCg
E0
1
R ⫹ j aLg 2
b
Cg
.
17.6 Exponential and Logarithmic Functions
Now, in polar form the complex impedance is
Lg 2
Z ZZ Ze ju
tan u where
1
Cg
R
.
Hence, A E0 /Z E0 /(ZZ Ze ju), and so the steady-state current can be written as
ip(t) Im
E0 ju jgt
e e .
ZZZ
The reader is encouraged to verify that this last expression is the same as (35) in Section 3.8
Logarithmic Function The logarithm of a complex number z x iy, z 0, is defi
as the inverse of the exponential function—that is,
w ln z
z ew.
if
In (5) we note that ln z is not defined for z 0, since there is no value of w for which ew 0
find the real and imaginary parts of ln z , we write w u iv and use (3) and (5):
x iy euiv eu (cos v i sin v) eu cos v ieu sin v.
The last equality implies x eu cos v and y eu sin v. We can solve these two equations fo
and v. First, by squaring and adding the equations, we find
e2u x2 y2 r 2 Zz Z 2
and so
u loge Zz Z,
where loge Zz Z denotes the real natural logarithm of the modulus of z. Second, to solve for v,
divide the two equations to obtain
y
tan v .
x
This last equation means that v is an argument of z; that is, v u arg z. But since ther
no unique argument of a given complex number z x iy, if u is an argument of z, then s
u 2np, n 0, 1, 2, . . . .
Definition 17.6.2
Logarithm of a Complex Number
For z 0, and u arg z,
ln z loge|z| i(u 2np),
n 0,
1,
2, . . . .
(
As is clearly indicated in (6), there are infinitely many values of the logarithm o
complex number z. This should not be any great surprise since the exponential funct
is periodic.
In real calculus, logarithms of negative numbers are not defined. As the next example w
show, this is not the case in complex calculus.
Complex Values of the Logarithmic Function
EXAMPLE 2
Find the values of (a) ln(2), (b) ln i, and (c) ln(1 i).
SOLUTION
(a) With u arg(2) p and loge|2| 0.6932, we have from (6)
ln(2) 0.6932 i(p 2np).
842
|
CHAPTER 17 Functions of a Complex Variable
In other words, ln i pi/2, 3pi/2, 5pi/2, 7pi/2, and so on.
(c) With u arg(1 i) 5p/4 and loge|1 i| loge "2 0.3466, we ha
ln(1 i) 0.3466 i a
EXAMPLE 3
5p
2npb .
4
Solving an Exponential Equation
Find all values of z such that ez !3 i.
SOLUTION From (5), with the symbol w replaced by z, we have z ln( !
| !3 i | 2 and tan u 1/ !3 imply that arg( !3 i) p/6, and so (6) give
z log e2 ia
p
2npb
6
or z 0.6931 ia
p
2npb.
6
Principal Value It is interesting to note that as a consequence of (6), th
of a positive real number has many values. For example, in real calculus, loge5 h
value: loge 5 1.6094, whereas in complex calculus, ln 5 1.6094 2npi. The
corresponding to n 0 is the same as the real logarithm loge 5 and is called the pri
of ln 5. Recall that in Section 17.2 we stipulated that the principal argument of a com
written Arg z, lies in the interval (p, p]. In general, we define the principal val
that complex logarithm corresponding to n 0 and u Arg z. To emphasize the pr
of the logarithm, we shall adopt the notation Ln z. In other words,
Ln z loge|z| i Arg z.
Since Arg z is unique, there is only one value of Ln z for each z 0.
EXAMPLE 4
Principal Values
The principal values of the logarithms in Example 2 are as follows:
(a) Since Arg(2) p, we need only set n 0 in the result given in part (a) of
Ln(2) 0.6932 pi.
(b) Similarly, since Arg(i) p/2, we set n 0 in the result in part (b) of E
obtain
Ln i p
i.
2
(c) In part (c) of Example 2, arg(1 i) 5p/4 is not the principal argument o
The argument of z that lies in the interval (p, p] is Arg(1 i) 3p/4. Hen
from (7) that
Ln(1 i) 0.3466 3p
i.
4
Up to this point we have avoided the use of the word function for the obviou
ln z defined in (6) is not a function in the strictest interpretation of that word. Non
customary to write f (z) ln z and to refer to f (z) ln z by the seemingly contradi
multiple-valued function. Although we shall not pursue the details, (6) can be inte
infinite collection of logarithmic functions (standard meaning of the word). Each fu
collection is called a branch of ln z. The function f (z) Ln z is then called the princ
of ln z, or the principal logarithmic function. To minimize the confusion, we sh
simply use the words logarithmic function when referring to either f (z) ln z or f (
17.6 Exponential and Logarithmic Functions
Equations (8) and (9) are to be interpreted in the sense that if values are assigned to two of
terms, then a correct value is assigned to the third term.
Properties of Logarithms
EXAMPLE 5
Suppose z1 ⫽ 1 and z2 ⫽ ⫺1. Then if we take ln z1 ⫽ 2pi and ln z2 ⫽ pi, we get
ln(z1z2) ⫽ ln(⫺1) ⫽ ln z1 ⫹ ln z2 ⫽ 2pi ⫹ pi ⫽ 3pi
z1
ln a b ⫽ ln(⫺1) ⫽ ln z1 ⫺ ln z2 ⫽ 2pi ⫺ pi ⫽ pi.
z2
Just as (7) of Section 17.2 was not valid when arg z was replaced with Arg z , so too (8
not true, in general, when ln z is replaced by Ln z. See Problems 45 and 46 in Exercises 17.6
y
Analyticity The logarithmic function f (z) ⫽ Ln z is not continuous at z ⫽ 0 since f (0
not defined. Moreover, f (z) ⫽ Ln z is discontinuous at all points of the negative real axis. T
is because the imaginary part of the function, v ⫽ Arg z, is discontinuous only at these poi
To see this, suppose x0 is a point on the negative real axis. As z S x0 from the upper half-pla
Arg z S p, whereas if z S x0 from the lower half-plane, then Arg z S ⫺p. This means
f (z) ⫽ Ln z is not analytic on the nonpositive real axis. However, f (z) ⫽ Ln z is analytic through
the domain D consisting of all the points in the complex plane except those on the nonposi
real axis. It is convenient to think of D as the complex plane from which the nonpositive
axis has been cut out. Since f (z) ⫽ Ln z is the principal branch of ln z, the nonpositive real a
is referred to as a branch cut for the function. See FIGURE 17.6.3. It is left as exercises to sh
that the Cauchy–Riemann equations are satisfied throughout this cut plane and that the der
tive of Ln z is given by
branch
cut
x
FIGURE 17.6.3 Branch cut for Ln z
y
d
1
Ln z ⫽
z
dz
for all z in D.
i
FIGURE 17.6.4 shows w ⫽ Ln z as a flow. Note that the vector field is not continuous along
x
–i
branch cut.
Complex Powers Inspired by the identity xa ⫽ ea ln x in real variables, we can define co
plex powers of a complex number. If a is a complex number and z ⫽ x ⫹ iy, then za is defined
za ⫽ ea ln z,
FIGURE 17.6.4 w ⫽ Ln z as a flow
z ⫽ 0.
In general, za is multiple-valued since ln z is multiple-valued. However, in the special case w
a ⫽ n, n ⫽ 0, ⫾1, ⫾2, . . ., (10) is single-valued since there is only one value for z2, z3, z⫺1,
so on. To see that this is so, suppose a ⫽ 2 and z ⫽ reiu, where u is any argument of z. Then
e2 ln z ⫽ e 2 (loger ⫹ iu) ⫽ e 2 loger ⫹ 2iu ⫽ e 2 loger e2iu ⫽ r 2 eiueiu ⫽ (reiu )(reiu ) ⫽ z2.
If we use Ln z in place of ln z, then (10) gives the principal value of z a.
Complex Power
EXAMPLE 6
2i
Find the value of i .
SOLUTION With z ⫽ i, arg z ⫽ p/2, and a ⫽ 2i , it follows from (10) that
i2i ⫽ e2i[loge1⫹i(p/2⫹2np)] ⫽ e⫺(1⫹4n)p
where n ⫽ 0, ⫾1, ⫾2, . . . . Inspection of the equation shows that i2i is real for every value o
Since p/2 is the principal argument of z ⫽ i, we obtain the principal value of i2i for n ⫽ 0
four rounded decimal places, this principal value is i 2i ⫽ e⫺p ⫽ 0.0432.
844
|
CHAPTER 17 Functions of a Complex Variable
1. z p
i
6
3. z 1 p
i
4
p
i
3
p
z22 i
2
3p
z p i
2
z 0.3 0.5i
z 0.23 i
35. ez 4i
4.
In Problems 39–42, find all values of the given quan
5. z p pi
6.
7. z 1.5 2i
9. z 5i
8.
10.
37. e
13. f (z) e
15. f (z) e
z2
e
e3 pi>2
14. f (z) e
16. f (z) e
38. e2z ez 1 0
40. 3i/p
42. (1 "3i)3i
41. (1 i)
In Problems 43 and 44, find the principal value of th
quantity. Express answers in the form a ib.
43. (1)(2i/p)
44. (1 i)2i
45. If z1 i and z2 1 i, verify that
Ln(z1z2) Ln z1 Ln z2.
In Problems 13–16, use Definition 17.6.1 to express the given
function in the form f (z) u iv.
iz
ie
2
(1 i)
2 3pi
12.
z1
39. (i)4i
In Problems 11 and 12, express the given number in the form a ib.
11. e15pi/4 e1pi/3
36. e1/z 1
2. z 46. Find two complex numbers z1 and z2 such that
2z
Ln(z1/z2) Ln z1 Ln z2.
1/z
47. Determine whether the given statement is true.
In Problems 17–20, verify the given result.
e z1
17. |ez| ex
18. z2 e z1 2 z2
e
19. ezpi ezpi
20. (ez)n enz, n an integer
21. Show that f (z) e z is nowhere analytic.
2
22. (a) Use the result in2 Problem 15 to show that f (z) e z is an
entire function.
2
(b) Verify that u(x, y) Re(e z ) is a harmonic function.
In Problems 23–28, express ln z in the form a ib.
23. z 5
24. z ei
25. z 2 2i
26. z 1 i
27. z "2 "6i
28. z "3 i
In Problems 29–34, express Ln z in the form a ib.
29. z 6 6i
30. z e3
31. z 12 5i
32. z 3 4i
5
33. z (1 "3i)
34. z (1 i)4
17.7
(a) Ln(1 i)2 2 Ln(1 i)
(b) Ln i 3 3 Ln i
(c) ln i 3 3 ln i
48. The laws of exponents hold for complex numbe
zazb zab,
za
zab,
zb
(za)n zna,
na
However, the last law is not valid if n is a comp
Verify that (i i)2 i 2i, but (i 2)i i 2i.
49. For complex numbers z satisfying Re(z)
0, s
can be written as
Ln z y
1
loge(x2 y2) i tan1 .
x
2
50. The function given in Problem 49 is analytic.
(a) Verify that u(x, y) loge(x2 y2) is a harmo
(b) Verify that v(x, y) tan1( y/x) is a harmon
Trigonometric and Hyperbolic Functions
INTRODUCTION In this section we define the complex trigonometric and hyp
tions. Analogous to the complex functions ez and Ln z defined in the previous s
functions will agree with their real counterparts for real values of z. In addition, w
that the complex trigonometric and hyperbolic functions have the same derivative
many of the same identities as the real trigonometric and hyperbolic functions.
Trigonometric Functions If x is a real variable, then Euler’s formula giv
eix cos x i sin x
and
eix cos x i sin x.
By subtracting and then adding these equations, we see that the real functions sin x a
be expressed as a combination of exponential functions:
sin x e ix 2 eix
,
2i
cos x e ix eix
.
2
17.7 Trigonometric and Hyperbolic Functions
For any complex number z x iy,
sin z e iz 2 eiz
2i
and
cos z e iz eiz
.
2
(
As in trigonometry, we define four additional trigonometric functions in terms of s
and cos z:
sin z
1
1
1
, cot z , csc z , sec z .
cos z
cos
z
tan z
sin z
tan z When y 0, each function in (2) and (3) reduces to its real counterpart.
Analyticity Since the exponential functions eiz and eiz are entire functions, it follo
that sin z and cos z are entire functions. Now, as we shall see shortly, sin z 0 only for the
numbers z np, n an integer, and cos z 0 only for the real numbers z (2n 1)p/2, n
integer. Thus, tan z and sec z are analytic except at the points z (2n 1)p/2, and cot z
csc z are analytic except at the points z np.
Derivatives Since (d/dz)ez ez, it follows from the Chain Rule that (d/dz)eiz ieiz
(d /dz)eiz ieiz. Hence,
d
d e iz 2 eiz
e iz eiz
sin z cos z.
dz
dz
2i
2
In fact, it is readily shown that the forms of the derivatives of the complex trigonometric fu
tions are the same as the real functions. We summarize the results:
d
sin z cos z
dz
d
cos z sin z
dz
d
tan z sec 2z
dz
d
cot z csc 2z
dz
d
sec z sec z tan z
dz
d
csc z csc z cot z.
dz
Identities The familiar trigonometric identities are also the same in the complex cas
sin(z) sin z
2
cos(z) cos z
2
cos z sin z 1
sin(z1
z2) sin z1 cos z2
cos(z1
z2) cos z1 cos z2 sin z1 sin z2
sin 2z 2 sin z cos z
cos z1 sin z2
cos 2z cos2z sin2z.
Zeros To find the zeros of sin z and cos z we need to express both functions in the fo
u iv. Before proceeding, recall from calculus that if y is real, then the hyperbolic sine
hyperbolic cosine are defined in terms of the real exponential functions ey and ey:
sinh y e y 2 ey
2
and
cosh y e y ey
.
2
Now from Definition 17.7.1 and Euler’s formula we find, after simplifying,
sin z e i(x iy) 2 ei(x iy)
e y ey
e y 2 ey
sin x a
b i cos x a
b.
2i
2
2
Thus from (5) we have
sin z sin x cosh y i cos x sinh y.
846
|
CHAPTER 17 Functions of a Complex Variable
From (6), (7), and cosh y 1 sinh y, we find
|sin z|2 sin2 x sinh2y
|cos z|2 cos2 x sinh2y.
and
Now a complex number z is zero if and only if |z|2 0. Thus, if sin z 0, then from
have sin2 x sinh2y 0. This implies that sin x 0 and sinh y 0, and so x np
Thus the only zeros of sin z are the real numbers z np 0i np, n 0, 1, 2, .
it follows from (9) that cos z 0 only when z (2n 1)p/2, n 0, 1, 2, . . . .
EXAMPLE 1
Complex Value of the Sine Function
From (6) we have, with the aid of a calculator,
sin(2 i) sin 2 cosh 1 i cos 2 sinh 1 1.4031 0.4891i.
In ordinary trigonometry we are accustomed to the fact that |sin x| 1 and |cos x| of (8) and (9) shows that these inequalities do not hold for the complex sine and
sinh y can range from q to q. In other words, it is perfectly feasible to have s
equations such as cos z 10.
EXAMPLE 2
Solving a Trigonometric Equation
Solve the equation cos z 10.
SOLUTION From (2), cos z 10 is equivalent to (eiz eiz)/2 10. Multipl
equation by eiz then gives the quadratic equation in eiz:
e2iz 20e iz 1 0.
From the quadratic formula we find eiz 10 3 !11. Thus, for n 0, 1,
have iz loge(10
3 !11) 2npi. Dividing by i and utilizing loge(10 loge(10 3 "11), we can express the solutions of the given equation as
i loge(10 3 "11).
Hyperbolic Functions We define the complex hyperbolic sine and cosine
analogous to the real definitions given in (5).
Definition 17.7.2
Hyperbolic Sine and Cosine
For any complex number z x iy,
sinh z e z 2 ez
2
and
cosh z e z ez
.
2
The hyperbolic tangent, cotangent, secant, and cosecant functions are defined
sinh z and cosh z:
tanh z sinh z
1
1
1
, coth z , sech z , csch z cosh z
tanh z
cosh z
sinh z
The hyperbolic sine and cosine are entire functions, and the functions defined in
lytic except at points where the denominators are zero. It is also easy to see from (1
d
sinh z cosh z
dz
and
d
cosh z sinh z .
dz
It is interesting to observe that, in contrast to real calculus, the trigonometric and hyp
tions are related in complex calculus. If we replace z by iz everywhere in (10) and compa
with (2), we see that sinh(iz) i sin z and cosh(iz) cos z. These equations enable
17.7 Trigonometric and Hyperbolic Functions
sinh z i sin(iz),
cosh z cos(iz).
Zeros The relationships given in (14) enable us to derive identities for the hyperbolic fu
tions utilizing results for the trigonometric functions. For example, to express sinh z in the fo
u iv we write sinh z i sin(iz) in the form sinh z i sin(y ix) and use (6):
sinh z i [sin(y) cosh x i cos(y) sinh x].
Since sin(y) sin y and cos(y) cos y, the foregoing expression simplifies to
sinh z sinh x cos y i cosh x sin y.
cosh z cosh x cos y i sinh x sin y.
Similarly,
It also follows directly from (14) that the zeros of sinh z and cosh z are pure imaginary and
respectively,
pi
z npi and z (2n 1) , n 0, 1, 2, . . . .
2
Periodicity Since sin x and cos x are 2p-periodic, we can easily demonstrate that s
and cos z are also periodic with the same real period 2p. For example, from (6), note that
sin(z 2p) sin(x 2p iy)
sin(x 2p) cosh y i cos(x 2p) sinh y
sin x cosh y i cos x sinh y;
that is, sin(z 2p) sin z. In exactly the same manner, it follows from (7) that cos(z 2p) co
In addition, the hyperbolic functions sinh z and cosh z have the imaginary period 2pi. This
result follows from either Definition 17.7.2 and the fact that ez is periodic with period 2pi
from (15) and (16) and replacing z by z 2pi.
Exercises
17.7
Answers to selected odd-numbered problems begin on page ANS-41.
In Problems 1–12, express the given quantity in the form a ib.
1. cos(3i)
2. sin(2i)
p
3. sina ib
4
4. cos(2 4i)
p
3ib
2
8. csc(1 i)
3p
ib
10. sinha
2
5. tan(i)
6. cota
7. sec(p i)
9. cosh(pi)
11. sinha1 p
ib
3
12. cosh(2 3i)
In Problems 13 and 14, verify the given result.
17. sinh z i
19. cos z sin z
18. sinh z 1
20. cos z i sin z
In Problems 21 and 22, use the definition of equality of comp
numbers to find all values of z satisfying the given equation.
21. cos z cosh 2
22. sin z i sinh 2
23. Prove that cos z cos x cosh y i sin x sinh y.
24. Prove that sinh z sinh x cos y i cosh x sin y.
25. Prove that cosh z cosh x cos y i sinh x sin y.
26. Prove that |sinh z|2 sin2y sinh2 x.
27. Prove that |cosh z|2 cos2y sinh2 x.
28. Prove that cos2z sin2z 1.
p
5
i ln 2b 2
4
p
3
14. cosa i ln 2b i
2
4
29. Prove that cosh2z sinh2z 1.
In Problems 15–20, find all values of z satisfying the given
equation.
31. Prove that tanh z is periodic with period pi.
13. sina
15. sin z 2
848
|
30. Show that tan z u iv, where
u
16. cos z 3i
CHAPTER 17 Functions of a Complex Variable
sinh 2y
sin 2x
and v cos 2x cosh 2y
cos 2x cosh 2y
32. Prove that (a) sin z sin z and (b) cos z cos z.
1. sin1(i)
3. sin1 0
5. cos1 2
17
2. sin1 "2
4. sin1 13
5
6. cos1 2i
Chapter in Review
10. tan1 3i
12. cosh1 i
13. tanh1(1 2i)
14. tanh1( "3i)
Answers to selected odd-numbered problems begin on page A
Answer Problems 1–16 without referring back to the text. Fill in
the blank or answer true/false.
10
9. tan1 1
11. sinh1 43
26. Let z and w be complex numbers such that |z| 1
Prove that
10
1. Re(1 i ) _____ and Im(1 i ) _____.
2. If z is a point in the third quadrant, then iz is in the _____
quadrant.
z
z
i127 5i 9 2i1 _____
4i
If z , then |z| _____.
3 2 4i
Describe the region defined by 1 |z 2| 3. _____
Arg(z z) 0 _____
5
, then Arg z _____.
If z "3 i
If ez 2i, then z _____.
If |ez| 1, then z is a pure imaginary number. _____
The principal value of (1 i)(2i) is _____.
If f (z) x 2 3xy 5y 3 i(4x 2y 4x 7y), then
f (1 2i) _____.
If the Cauchy–Riemann equations are satisfied at a point, then
the function is necessarily analytic there. _____
f (z) e z is periodic with period _____.
Ln(ie3) _____
f (z) sin(x iy) is nowhere analytic. _____
3. If z 3 4i, then Re a b _____.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
In Problems 17–20, write the given number in the form a ib.
17. i (2 3i)2 (4 2i)
19.
(1 2 i)10
(1 i)3
32i
2 2 2i
18.
2 3i
1 5i
2
z2w
2 1.
1 2 zw
In Problems 27 and 28, find all solutions of the given
1
22i
1i
29. If f (z) z24 3z20 4z12 5z6, find f a
"2
30. Write f (z) Im(z 3z) z Re(z 2 ) 5z
f (z) u(x, y) iv(x, y).
27. z4 1 i
28. z 3>2 In Problems 31 and 32, find the image of the line x w-plane under the given mapping.
31. f (z) x2 y i ( y2 x)
32. f (z) 1
z
In Problems 33–36, find all complex numbers for wh
given statement is true.
33. z z1
35. z z
1
z
2
36. z ( z )2
34. z 37. Show that the function f (z) (2xy 5x) i(x
is analytic for all z. Find f (z).
38. Determine whether the function
f (z) x3 xy2 4x i(4y y3 x
20. 4epi/3 epi/4
is differentiable. Is it analytic?
In Problems 21–24, sketch the set of points in the complex plane
satisfying the given inequality.
In Problems 39 and 40, verify the given equality.
21. Im(z 2) 2
39. Ln[(1 i)(1 i)] Ln(1 i) Ln(1 i)
22. Im(z 5i)
3
1
23.
1
24. Im(z) Re(z)
ZzZ
25. Look up the definitions of conic sections in a calculus text.
Now describe the set of points in the complex plane that satisfy the equation |z 2i| |z 2i| 5.
40. Lna
1i
b Ln(1 i) Ln(1 i)
12i
CHAPTER 17 in Review
CHAPTER
18
To define an integral of a
complex function f, we start with
f defined along some curve C or
contour in the complex plane. We
will see in this chapter that the
definition of a complex integral,
its properties, and method of
evaluation are quite similar to
those of a real line integral in the
plane (Section 9.8).
Integration in the
Complex Plane
CHAPTER CONTENTS
18.1
18.2
18.3
18.4
Contour Integrals
Cauchy–Goursat Theorem
Independence of the Path
Cauchy’s Integral Formulas
Chapter 18 in Review
INTRODUCTION In Section 9.8 we saw that the definition of the definite integral eab f (x
starts with a real function y f (x) that is defined on an interval [a, b] on the x-axis. Becau
planar curve is the two-dimensional analogue of an interval, we then generalized the defini
of the definite integral to integrals of real functions of two variables defined on a curve C in
Cartesian plane. We shall see in this section that a complex integral is defined in a manner
is quite similar to that of a line integral in the Cartesian plane. In case you have not stud
Sections 9.8 and 9.9, a review of those sections is recommended.
z*n
zn –1
zn
C
z0
z*1
z1
z*2
z2
FIGURE 18.1.1 Sample points are the
red dots
A Definition Integration in the complex plane is defined in a manner similar to that
line integral in the plane. In other words, we shall be dealing with an integral of a complex fu
tion f (z) that is defined along a curve C in the complex plane. These curves are defined in te
of parametric equations x x(t), y y(t), a t b, where t is a real parameter. By using x(t)
y(t) as real and imaginary parts, we can also describe a curve C in the complex plane by mean
a complex-valued function of a real variable t: z(t) x(t) iy(t), a t b. For example, x co
y sin t, 0 t 2p, describes a unit circle centered at the origin. This circle can also be descri
by z(t) cos t i sin t, or even more compactly by z(t) eit, 0 t 2p. The same definiti
of smooth curve, piecewise-smooth curve, closed curve, and simple closed curve given in Sec
9.8 carry over to this discussion. As before, we shall assume that the positive direction on C c
responds to increasing values of t. In complex variables, a piecewise-smooth curve C is also ca
a contour or path. An integral of f (z) on C is denoted by C f (z) dz or C f (z) dz if the contou
is closed; it is referred to as a contour integral or simply as a complex integral.
1. Let f (z) u(x, y) iv(x, y) be defined at all points on a smooth curve C defined by x x
y y(t), a t b.
2. Divide C into n subarcs according to the partition a t0 t1 . . . tn b of [a,
The corresponding points on the curve C are z0 x0 iy0 x(t0) iy(t0), z1 x1 iy
x(t1) iy(t1), . . ., zn xn iyn x(tn) iy(tn). Let zk zk zk1, k 1, 2, . . ., n.
3. Let P be the norm of the partition, that is, the maximum value of |zk|.
4. Choose a sample point z*k 5 x*k 1 iy*k on each subarc. See FIGURE 18.1.1.
5. Form the sum a f (z *k ) zk.
n
k1
Definition 18.1.1
Contour Integral
Let f be defined at points of a smooth curve C defined by x x(t), y y(t), a t b. T
contour integral of f along C is
#
C
f (z) dz lim a f (z *k ) Dzk.
iPiS0
n
(
k1
The limit in (1) exists if f is continuous at all points on C and C is either smooth or piecew
smooth. Consequently we shall, hereafter, assume these conditions as a matter of course.
Method of Evaluation We shall turn now to the question of evaluating a cont
integral. To facilitate the discussion, let us suppress the subscripts and write (1) in
abbreviated form
# f (z) dz lim (u iv)(x i y)
C
lim [(ux vy) i (vx uy)].
This means
# f (z) dz # u dx v dy i # v dx u dy.
C
854
|
CHAPTER 18 Integration in the Complex Plane
C
C
#
b
a
b
# [v(x(t), y(t))x (t) u(x(t), y(t))y
[u(x(t), y(t))x (t) v(x(t), y(t))y (t)] dt i
a
But if we use z(t) x(t) iy(t) to describe C, the last result is the same as ba f
when separated into two integrals. Thus we arrive at a practical means of evaluati
integral:
Theorem 18.1.1
Evaluation of a Contour Integral
If f is continuous on a smooth curve C given by z(t) x(t) iy(t), a t b, the
#
C
f (z) dz #
b
f (z(t)) z9(t) dt.
a
If f is expressed in terms of the symbol z, then to evaluate f (z(t)) we simply replac
z by z(t). If f is not expressed in terms of z, then to evaluate f (z(t)), we replace x an
they appear by x(t) and y(t), respectively.
Evaluating a Contour Integral
EXAMPLE 1
Evaluate
# z dz, where C is given by x 3t, y t , 1 t 4.
2
C
SOLUTION We write z(t) 3t it 2 so that z (t) 3 2it and f (z(t)) 3t 1 it
Thus,
#
C
z dz #
4
#
4
(3t it 2)(3 2it) dt
21
(2t 3 9t) dt i
21
BC
䉲
SOLUTION
3t 2 dt 195 65
21
Evaluating a Contour Integral
EXAMPLE 2
Evaluate
#
4
1
dz, where C is the circle x cos t, y sin t, 0 t 2p.
z
In this case z(t) cos t i sin t eit, z (t) ieit, and f (z) 1/z BC
䉲
1
dz z
#
2p
0
(e it )ie itdt i
#
2p
0
dt 2pi.
For some curves, the real variable x itself can be used as the parameter. For exam
ate C (8x2 iy) dz on y 5x, 0 x 2, we write C (8x2 iy) dz 20 (8x2 5i
and integrate in the usual manner.
Properties The following properties of contour integrals are analogous to th
of line integrals.
18.1 Contour Integrals
# kf (z) dz k # f (z) dz, k a constant
(ii) # [ f (z) g(z)] dz # f (z) dz # g(z) dz
(iii) # f (z) dz # f (z) dz # f (z) dz, where C is the union of the smooth curves C and C
(iv) # f (z) dz # f (z) dz, where C denotes the curve having the opposite orientation of C
(i)
C
C
C
C
C
C
C1
1
C2
C
2C
The four parts of Theorem 18.1.2 also hold when C is a piecewise-smooth curve in D.
y
Evaluating a Contour Integral
EXAMPLE 3
2
2
1 + 2i
Evaluate C (x iy ) dz, where C is the contour shown in FIGURE 18.1.2.
C2
SOLUTION In view of Theorem 18.1.2(iii) we write
1+i
# (x iy ) dz #
2
C1
2
C
x
C1
(x2 iy2) dz #
C2
(x2 iy2) dz.
Since the curve C1 is defined by y x, it makes sense to use x as a parameter. Theref
z(x) x ix, z (x) 1 i, f (z(x)) x 2 ix 2, and
#
FIGURE 18.1.2 Piecewise-smooth contour
in Example 3
C1
1
(x2 iy2) dz # (x ix )(1 i) dx
2
2
0
#
1
(1 i)2 x 2dx 0
(1 i)2
2
i.
3
3
The curve C2 is defined by x 1, 1 y 2. Using y as a parameter, we have z( y) 1 z ( y) i, and f (z( y)) 1 iy2. Thus,
#
C2
(x2 iy2) dz #
2
1
(1 iy2) i dy #
2
1
y2 dy i
#
2
1
7
dy i.
3
Finally, we have eC (x 2 iy 2) dz 23– i (73– i) 73– 53– i.
There are times in the application of complex integration that it is useful to find an up
bound for the absolute value of a contour integral. In the next theorem we shall use the fact
b
the length of a plane curve is s ea "fx9(t)g 2 fy9(t)g 2 dt. But if z (t) x (t) iy (t), t
|z (t)| "fx9(t)g 2 fy9(t)g 2 and consequently s ab |z (t)| dt.
Theorem 18.1.3
A Bounding Theorem
If f is continuous on a smooth curve C and if | f (z)| M for all z on C, then Z eC f (z) dz Z # M
where L is the length of C.
PROOF: From the triangle inequality (6) of Section 17.1 we can write
2 a f (z *k )Dzk 2 # a Z f (z *k )Z ZDzk Z # M a ZDzk Z.
856
|
CHAPTER 18 Integration in the Complex Plane
n
n
n
k1
k1
k1
Theorem 18.1.3 is used often in the theory of complex integration and is someti
to as the ML-inequality.
A Bound for a Contour Integral
EXAMPLE 4
Find an upper bound for the absolute value of
ez
dz, where C is the circl
BC z 1
䉲
SOLUTION First, the length s of the circle of radius 4 is 8p. Next, from the ineq
Section 17.1, it follows that |z 1| |z| 1 4 1 3, and so
2
Zez Z
Zez Z
ez
2#
5
.
z11
ZzZ 2 1
3
In addition, |ez| |ex (cos y i sin y)| ex. For points on the circle |z| 4, the m
x can be is 4, and so (5) becomes
2
ez
e4
2# .
z11
3
Hence from Theorem 18.1.3 we have
2
ez
8p e 4
dz 2 #
.
3
BC z 1
䉲
Circulation and Net Flux Let T and N denote the unit tangent vector
normal vector to a positively oriented simple closed contour C. When we interpret
function f (z) u(x, y) iv(x, y) as a vector, the line integrals
BC
䉲
BC
and
䉲
f T ds f N ds BC
䉲
u dx v dy
BC
u dy 2 v dx
䉲
have special interpretations. The line integral in (6) is called the circulation around
sures the tendency of the flow to rotate the curve C. See Section 9.8 for the derivat
flux across C is the difference between the rate at which fluid enters and the rate a
leaves the region bounded by C. The net flux across C is given by the line integral
nonzero value for C f N ds indicates the presence of sources or sinks for the flu
curve C. Note that
䉲
and so
a
BC
䉲
f T dsb i a
BC
䉲
f N dsb BC
䉲
(u 2 iv)(dx i dy) circulation Re a
net flux Ima
BC
䉲
f (z) dzb
BC
䉲
f (z) dzb.
BC
䉲
f (z) dz
Thus, both of these key quantities may be found by computing a single complex in
18.1 Contour Integrals
SOLUTION
BC
x
C
䉲
Since f (z) (1 i)z and z(t) eit, 0 t 2p, we have
f (z) dz #
2p
0
(1 2 i) eitie it dt (1 i)
#
2p
0
dt 2p(1 i) 2p 2pi.
Using (8) and (9), the circulation around C is 2p and the net flux across C is 2p.
FIGURE 18.1.3.
FIGURE 18.1.3 Flow f (z) (1 i)z
Exercises
18.1
Answers to selected odd-numbered problems begin on page ANS-41.
In Problems 1–16, evaluate the given integral along the
indicated contour.
19.
1. C (z 3) dz, where C is x 2t, y 4t 1, 1 t 3
BC
䉲
z 2 dz
20.
BC
䉲
z 2 dz
y
2. C (2z z) dz, where C is x t, y t 2 2, 0 t 2
1+i
3. C z 2 dz, where C is z(t) 3t 2it, 2 t 2
4. C (3z 2 2z) dz, where C is z(t) t it 2, 0 t 1
1 1 z dz, where C is the right half of the circle |z| 1
z
from z i to z i
5. C
x
1
6. C |z|2 dz, where C is x t 2, y 1/t, 1 t 2
7. C Re(z) dz, where C is the circle |z| 1
䉲
8.
9.
10.
11.
12.
13.
14.
15.
16.
1
5
2
8b dz, where C is the circle | z i | 1,
C a
3
zi
(z i)
0 t 2p
C (x2 iy3) dz, where C is the straight line from z 1 to z i
C (x3 iy3) dz, where C is the lower half of the circle |z| 1
from z 1 to z 1
C ez dz, where C is the polygonal path consisting of the line
segments from z 0 to z 2 and from z 2 to z 1 pi
C sin z dz, where C is the polygonal path consisting of the line
segments from z 0 to z 1 and from z 1 to z 1 i
C Im(z i) dz, where C is the polygonal path consisting of
the circular arc along |z| 1 from z 1 to z i and the line
segment from z i to z 1
C dz, where C is the left half of the ellipse x 2 /36 y 2 /4 1
from z 2i to z 2i
C zez dz, where C is the square with vertices z 0, z 1,
z 1 i, and z i
2,
x,0
C f (z) dz, where f (z) e
and C is the parabola
6x, x . 0
2
y x from z 1 i to z 1 i
䉲
In Problems 21–24, evaluate C (z2 z 2) dz from i to 1 alo
the indicated contours.
21.
y
22.
y
i
i
x
x
1
1
FIGURE 18.1.6 Contour in
Problem 22
FIGURE 18.1.5 Contour in
Problem 21
23.
1+i
24.
y
y
䉲
In Problems 17–20, evaluate the given integral along the
contour C given in FIGURE 18.1.4.
17.
FIGURE 18.1.4 Contour in Problems 17–20
BC
䉲
x dz
858
|
18.
BC
䉲
(2z 2 1) dz
CHAPTER 18 Integration in the Complex Plane
i
i
y = 1 – x2
x2 + y2 = 1
x
1
FIGURE 18.1.7 Contour in
Problem 23
x
1
FIGURE 18.1.8 Contour in
Problem 24
BC z 2 1
26.
#z
31. Use the results of Problems 29 and 30 to evaluate 䉲
where C is
(a) the straight line from 1 i to 2 3i, and
(b) the closed contour x4 y4 4.
1
dz, where C is the right half of the circle |z| 6
2 2i
C
from z 6i to z 6i
2
In Problems 32–35, compute the circulation and net
given flow and the indicated closed contour.
27. C (z2 4) dz, where C is the line segment from z 0 to
z1i
1
28.
dz, where C is one quarter of the circle |z| 4 from
3
C z
z 4i to z 4
32. f (z) 1>z, where C is the circle |z| 2
33. f (z) 2z, where C is the circle |z| 1
#
34. f (z) 1>(z 2 1), where C is the circle |z 1| 35. f (z) z, where C is the square with vertices z
29. (a) Use Definition 18.1.1 to show for any smooth curve C
z 1 i, z i
between z0 and zn that C dz zn z0.
(b) Use the result in part (a) to verify the answer to Problem 14.
18.2 Cauchy–Goursat Theorem
INTRODUCTION In this section we shall concentrate on contour integrals wher
C is a simple closed curve with a positive (counterclockwise) orientation. Specific
see that when f is analytic in a special kind of domain D, the value of the contour integ
is the same for any simple closed curve C that lies entirely within D. This theorem
Cauchy–Goursat theorem, is one of the fundamental results in complex analysis.
to discussing the Cauchy–Goursat theorem and some of its ramifications, we first n
guish two kinds of domains in the complex plane: simply connected and multiply c
D
(a) Simply connected domain
D
(b) Multiply connected domain
FIGURE 18.2.1 Two kinds of domains
Simply and Multiply Connected Domains In the discussion that follo
concentrate on contour integrals where the contour C is a simple closed curve wi
(counterclockwise) orientation. Before doing this, we need to distinguish two kinds
A domain D is said to be simply connected if every simple closed contour C lying
can be shrunk to a point without leaving D. In other words, in a simply connected d
simple closed contour C lying entirely within it encloses only points of the domain D
yet another way, a simply connected domain has no “holes” in it. The entire comple
example of a simply connected domain. A domain that is not simply connected is calle
connected domain; that is, a multiply connected domain has “holes” in it. See FIGU
in Section 9.9, we call a domain with one “hole” doubly connected, a domain with
triply connected, and so on.
Cauchy’s Theorem In 1825, the French mathematician Louis-Augustin Ca
one of the most important theorems in complex analysis. Cauchy’s theorem says:
Suppose that a function f is analytic in a simply connected domain D and
continuous in D. Then for every simple closed contour C in D, C f (z) dz 0.
䉲
The proof of this theorem is an immediate consequence of Green’s theorem and the Cauc
equations. Since f is continuous throughout D, the real and imaginary parts of f (z) their first partial derivatives are continuous throughout D. By (2) of Section 18.1 we wr
in terms of real-line integrals and use Green’s theorem on each line integral:
BC
䉲
f (z) dz BC
䉲
u(x, y) dx v(x, y) dy i
BC
䉲
v(x, y) dx u(x, y) dy
0v
0u
0u
0v
6 a 2 b dA i 6 a
2 b dA.
0x
0y
0x
0y
D
D
18.2 Cauchy–Goursat Theorem
is known as the Cauchy–Goursat theorem.
Cauchy–Goursat Theorem
Theorem 18.2.1
Suppose a function f is analytic in a simply connected domain D. Then for every simple close
contour C in D, C f (z) dz 0.
䉲
Since the interior of a simple closed contour is a simply connected domain, the Cauchy–Gou
theorem can be stated in the slightly more practical manner:
y
If f is analytic at all points within and on a simple closed contour C, then
C f (z) dz 0.
䉲
C
Applying the Cauchy–Goursat Theorem
EXAMPLE 1
Evaluate
x
BC
䉲
ez dz, where C is the curve shown in FIGURE 18.2.2.
SOLUTION The function f (z) ez is entire and C is a simple closed contour. It follows fr
the form of the Cauchy–Goursat theorem given in (2) that C ez dz 0.
FIGURE 18.2.2 Contour in Example 1
䉲
y
EXAMPLE 2
i
Evaluate
x
1
–1
–i
Applying the Cauchy–Goursat Theorem
( y 2 5)2
dz
2
,
where
C
is
the
ellipse
(x
2)
1.
4
BC z 2
䉲
SOLUTION The rational function f (z) 1/z2 is analytic everywhere except at z 0. But z
is not a point interior to or on the contour C. Thus, from (2) we have C dz/z2 0.
䉲
EXAMPLE 3
FIGURE 18.2.3 Flow f (z) cos z
(2
Applying the Cauchy–Goursat Theorem
Given the flow f (z) cos z, compute the circulation around and net flux across C, wher
is the square with vertices z 1, z i, z 1, and z i.
SOLUTION We must compute C f (z) dz C cos z dz and then take the real and imagin
parts of the integral to find the circulation and net flux, respectively. The function cos
analytic everywhere, and so C f (z) dz 0 from (2). The circulation and net flux are theref
both 0. FIGURE 18.2.3 shows the flow f (z) cos z and the contour C.
䉲
䉲
䉲
C1
D
Cauchy–Goursat Theorem for Multiply Connected Domains If f is analyti
a multiply connected domain D, then we cannot conclude that C f (z) dz 0 for every sim
closed contour C in D. To begin, suppose D is a doubly connected domain and C and
are simple closed contours such that C1 surrounds the “hole” in the domain and is interior to
See FIGURE 18.2.4(a). Suppose, also, that f is analytic on each contour and at each point inte
to C but exterior to C1. When we introduce the cut AB shown in Figure 18.2.4(b), the reg
bounded by the curves is simply connected. Now the integral from A to B has the opposite va
of the integral from B to A, and so from (2) we have C f (z) dz C1 f (z) dz 0 or
䉲
C
(a)
A
䉲
B
BC
䉲
D
C
(b)
FIGURE 18.2.4 Doubly connected
domain D
860
|
f (z) dz BC
䉲
䉱
f (z) dz.
1
The last result is sometimes called the principle of deformation of contours, since we
think of the contour C1 as a continuous deformation of the contour C. Under this deformat
of contours, the value of the integral does not change. Thus, on a practical level, (3) allow
to evaluate an integral over a complicated simple closed contour by replacing that contour w
one that is more convenient.
CHAPTER 18 Integration in the Complex Plane
C
SOLUTION In view of (3), we choose the more convenient circular contour C1 in t
taking the radius of the circle to be r ⫽ 1, we are guaranteed that C1 lies within C. In
C1 is the circle |z ⫺ i| ⫽ 1, which can be parameterized by x ⫽ cos t, y ⫽ 1 ⫹ sin t
or equivalently by z ⫽ i ⫹ eit, 0 ⱕ t ⱕ 2p. From z ⫺ i ⫽ eit and dz ⫽ ieit dt we ob
C1
i
dz
dz
⫽
⫽
BC z 2 i BC1 z 2 i
x
–2
䉲
2 – 2i
–2i
FIGURE 18.2.5 We use the simpler
contour C1 in Example 4
䉲
#
2p
0
ie it
dt ⫽ i
e it
#
2p
dt ⫽ 2pi.
0
The result in Example 4 can be generalized. Using the principle of deformation of
and proceeding as in the example, we can show that if z0 is any constant complex num
to any simple closed contour C, then
dz
2pi,
n ⫽ e
0,
BC (z 2 z0)
n⫽1
n an integer 2 1.
䉲
The fact that the integral in (4) is zero when n is an integer ⫽ 1 follows only p
the Cauchy–Goursat theorem. When n is zero or a negative integer, 1/(z ⫺ z0)n is a
(for example, n ⫽ ⫺3, 1/(z ⫺ z0)⫺3 ⫽ (z ⫺ z0)3) and therefore entire. Theorem 18.2.1
养C dz/(z ⫺ z0)n ⫽ 0. It is left as an exercise to show that the integral is still zero whe
tive integer different from one. See Problem 22 in Exercises 18.2.
䉲
Applying Formula (4)
EXAMPLE 5
Evaluate
5z ⫹ 7
dz, where C is the circle |z ⫺ 2| ⫽ 2.
BC z ⫹ 2z 2 3
䉲
2
SOLUTION Since the denominator factors as z2 ⫹ 2z ⫺ 3 ⫽ (z ⫺ 1)(z ⫹ 3), t
fails to be analytic at z ⫽ 1 and z ⫽ ⫺3. Of these two points, only z ⫽ 1 lies withi
C, which is a circle centered at z ⫽ 2 of radius r ⫽ 2. Now by partial fractions,
3
5z 1 7
2
5
1
z21
z13
z2 1 2z 2 3
and so
5z ⫹ 7
dz
dz
dz ⫽ 3
⫹2
.
BC z ⫹ 2z 2 3
BC z 2 1
BC z ⫹ 3
䉲
䉲
2
䉲
In view of the result given in (4), the first integral in (5) has the value 2pi. By
Goursat theorem, the value of the second integral is zero. Hence, (5) becomes
BC z 2 ⫹ 2z 2 3
D
5z ⫹ 7
䉲
C1
C2
C
If C, C1, and C2 are the simple closed contours shown in FIGURE 18.2.6 and if f is ana
of the three contours as well as at each point interior to C but exterior to both C1 and C
troducing cuts, we get from Theorem 18.2.1 that 养C f (z) dz ⫹ 养C1 f (z) dz ⫹ 养C2 f (z) dz ⫽
BC
䉲
FIGURE 18.2.6 Triply connected domain D
dz ⫽ 3(2pi) ⫹ 2(0) ⫽ 6pi.
f (z) dz ⫽
BC
䉲
䉲
BC
f (z) dz ⫹
䉲
1
䉱
䉱
f (z) dz.
2
The next theorem summarizes the general result for a multiply connected domain with n
Theorem 18.2.2
Cauchy–Goursat Theorem for Multiply Connected Domains
Suppose C, C1, . . ., Cn are simple closed curves with a positive orientation such that C
are interior to C but the regions interior to each Ck, k ⫽ 1, 2, . . ., n, have no poin
mon. If f is analytic on each contour and at each point interior to C but exterio
Ck, k ⫽ 1, 2, . . ., n, then
f (z) dz ⫽ a
B
n
䉲
C
k⫽1
BC
䉲
f (z) dz.
k
18.2 Cauchy–Goursat Theorem
y
C
i
SOLUTION In this case the denominator of the integrand factors as z2 1 (z i)(z Consequently, the integrand 1/(z2 1) is not analytic at z i and z i. Both of these po
lie within the contour C. Using partial fraction decomposition once more, we have
C1
1>2i
1>2i
1
5
2
z2i
z1i
z 11
2
x
–i
1
dz
1
1
c
2
d dz.
BC z 1 2i BC z 2 i z i
C2
and
FIGURE 18.2.7 Contour in Example 6
䉲
䉲
2
We now surround the points z i and z i by circular contours C1 and C2, respectiv
that lie entirely within C. Specifically, the choice |z i| 12 for C1 and |z i| 12 for C2
suffice. See FIGURE 18.2.7. From Theorem 18.2.2 we can then write
1
dz
1
1
1
1
1
c
2
d dz c
2
d dz
2
2i
z
2
i
z
i
2i
z
2
i
z
i
z
1
BC
BC1
BC2
䉲
䉲
䉲
1
dz
1
dz
1
dz
1
dz
2
2
.
2i B
2i B
2i B
2i B
C1 z 2 i
C1 z i
C2 z 2 i
C2 z i
䉲
䉲
䉲
䉲
Because 1/(z i) is analytic on C1 and at each point in its interior and because 1/(z analytic on C2 and at each point in its interior, it follows from (4) that the second and th
integrals in (7) are zero. Moreover, it follows from (4), with n 1, that
dz
2pi
z
BC1 2 i
and
䉲
Thus (7) becomes
C
FIGURE 18.2.8 Contour C is closed but
not simple
Exercises
p 2 p 0.
䉲
Answers to selected odd-numbered problems begin on page ANS-41.
䉲
|
dz
Throughout the foregoing discussion we assumed that C was a simple closed contour; in oth
words, C did not intersect itself. Although we shall not give the proof, it can be shown that t
Cauchy–Goursat theorem is valid for any closed contour C in a simply connected domain D
As shown in FIGURE 18.2.8, the contour C is closed but not simple. Nevertheless, if f is analyt
in D, then C f (z) dz 0.
In Problems 1–8, prove that C f (z) dz 0, where f is the given
function and C is the unit circle |z| 1.
1
1. f (z) z3 1 3i
2. f (z) z2 z24
z
z23
3. f (z) 4. f (z) 2
2z 1 3
z 1 2z 1 2
862
䉲
REMARKS
D
18.2
BC z 2 1
dz
2pi.
z
BC2 i
䉲
CHAPTER 18 Integration in the Complex Plane
5. f (z) sin z
(z 2 2 25) (z 2 9)
7. f (z) tan z
ez
2z2 1 11z 1 15
z2 2 9
8. f (z) cosh z
6. f (z) BC z 2 z
y
16.
C
17.
x
2
18.
19.
FIGURE 18.2.9 Contour in Problem 9
5
dz, where C is the contour shown in
z
1i
BC
FIGURE 18.2.10.
10. Evaluate
20.
䉲
21.
y
x 4 + y 4 = 16
2
2z
dz; (a) |z| 1, (b) |z 2i| 1, (c)
BC z 3
3z 2
dz; (a) |z 5| 2, (b) |z| 2
BC z 2 8z 12
3
1
a
2
b dz; (a) |z| 5, (b) |z BC z 2 z 2 2i
z21
dz; Zz 2 iZ 12
BC z(z 2 i)(z 2 3i)
1
dz; ZzZ 1
3
BC z 2iz 2
8z 2 3
dz, where C is the closed co
Evaluate
BC z 2 2 z
in FIGURE 18.2.11. [Hint: Express C as the union o
curves C1 and C2.]
䉲
2
䉲
䉲
䉲
䉲
䉲
y
C
C
x
1
x
FIGURE 18.2.11 Contour in Problem 21
22. Suppose z0 is any constant complex number in
FIGURE 18.2.10 Contour in Problem 10
simple closed contour C. Show that
dz
2pi,
n e
(z
2
z
)
0,
BC
0
In Problems 11–20, use any of the results in this section to
evaluate the given integral along the indicated closed contour(s).
1
az b dz; ZzZ 2
z
BC
1
12.
az 2 b dz; ZzZ 2
z
BC
z
13.
dz; ZzZ 3
BC z 2 2 p2
11.
䉲
n1
n a positive integ
In Problems 23 and 24, evaluate the given integral by
䉲
ez
2 3zb dz, C is the unit circle |z| BC z 3
24. C (z 3 z 2 Re(z)) dz, C is the triangle w
z 0, z 1 2i, z 1
23.
䉲
䉲
䉲
䉲
18.3
a
Independence of the Path
INTRODUCTION In real calculus when a function f possesses an elementary an
that is, a function F for which F (x) f (x), a definite integral can be evaluated by the F
Theorem of Calculus:
b
# f (x) dx F(b) F(a).
a
b
ea f
Note that
(x) dx depends only on the numbers a and b at the initial and terminal
interval of integration. In contrast, the value of a real-line integral C P dx Q
depends on the curve C. However, we saw in Section 9.9 that there exist line inte
value depends only on the initial point A and terminal point B of the curve C, a
itself. In this case we say that the line integral is independent of the path. These
18.3 Independence of the Path
Can a contour integral eC f (z) dz be independent of the path?
In this section we will see that the answer to both of these questions is “yes.”
A Definition As the next definition shows, the definition of path independence fo
contour integral C f (z) dz is essentially the same as for a real-line integral C P dx Q dy.
Independence of the Path
Definition 18.3.1
Let z0 and z1 be points in a domain D. A contour integral C f (z) dz is said to be independent of th
path if its value is the same for all contours C in D with an initial point z0 and a terminal point z1.
At the end of the preceding section we noted that the Cauchy–Goursat theorem also holds
closed contours, not just simple closed contours, in a simply connected domain D. Now suppo
as shown in FIGURE 18.3.1, that C and C1 are two contours in a simply connected domain D, b
with initial point z0 and terminal point z1. Note that C and C1 form a closed contour. Thu
f is analytic in D, it follows from the Cauchy–Goursat theorem that
z1
C1
C
# f (z) dz #
D
C
z0
f (z) dz 0.
2C1
But (2) is equivalent to
# f (z) dz #
FIGURE 18.3.1 If f is analytic in D,
integrals on C and C1 are equal
C
f (z) dz.
C1
The result in (3) is also an example of the principle of deformation of contours introduced in
of Section 18.2. We summarize the last result as a theorem.
Analyticity Implies Path Independence
Theorem 18.3.1
If f is an analytic function in a simply connected domain D, then C f (z) dz is independent
of the path C.
y
Choosing a Different Path
EXAMPLE 1
Evaluate C 2z dz, where C is the contour with initial point z 1 and terminal po
z 1 i shown in FIGURE 18.3.2.
–1 + i
C1
x
–1
C
FIGURE 18.3.2 Contour in Example 1
SOLUTION Since the function f (z) 2z is entire, we can replace the path C by any conven
contour C1 joining z 1 and z 1 i. In particular, by choosing C1 to be the stra
line segment x 1, 0 y 1, shown in red in Figure 18.3.2, we have z 1 dz i dy. Therefore,
#
2z dz 5
C
#
2z dz 5 22
C1
#
1
0
y dy 2 2i
#
1
dy 5 21 2 2i.
0
A contour integral C f (z) dz that is independent of the path C is usually writ
z
ez01 f (z) dz, where z0 and z1 are the initial and terminal points of C. Hence in Example 1 we
1 i
write e1 2z dz.
There is an easier way to evaluate the contour integral in Example 1, but before proceed
we need another definition.
Definition 18.3.2
Antiderivative
Suppose f is continuous in a domain D. If there exists a function F such that F (z) f (z) f
each z in D, then F is called an antiderivative of f.
864
|
CHAPTER 18 Integration in the Complex Plane
necessarily analytic and hence continuous in D (recall that differentiability implies
We are now in a position to prove the complex analogue of (1).
Fundamental Theorem for Contour Integrals
Theorem 18.3.2
Suppose f is continuous in a domain D and F is an antiderivative of f in D. Then for an
in D with initial point z0 and terminal point z1,
# f (z) dz ⫽ F(z ) ⫺ F(z ).
1
C
0
PROOF: We will prove (4) in the case when C is a smooth curve defined by z ⫽ z(
Using (3) of Section 18.1 and the fact that F⬘(z) ⫽ f (z) for each z in D, we have
#
C
f (z) dz ⫽
⫽
#
b
#
b
a
a
f (z(t)) z9(t) dt ⫽
d
F(z(t)) dt
dt
⫽ F(z(t)) d
#
b
F9(z(t)) z9(t) dt
a
d Chain Rule
b
a
⫽ F(z(b)) 2 F(z(a)) ⫽ F(z1) 2 F(z0).
Using an Antiderivative
EXAMPLE 2
In Example 1 we saw that the integral 兰C 2z dz, where C is shown in Figure 18.3.2, is
of the path. Now since f (z) ⫽ 2z is an entire function, it is continuous. Moreov
is an antiderivative of f, since F⬘(z) ⫽ 2z. Hence by (4) we have
#
⫺1 ⫹ i
⫺1
2z dz ⫽ z 2 d
⫺1 ⫹ i
⫺1
⫽ (⫺1 ⫹ i)2 2 (⫺1)2 ⫽ ⫺1 2 2i.
Using an Antiderivative
EXAMPLE 3
Evaluate 兰C cos z dz, where C is any contour with initial point z ⫽ 0 and terminal po
SOLUTION F(z) ⫽ sin z is an antiderivative of f (z) ⫽ cos z, since F⬘(z) ⫽ cos
from (4) we have
#
C
cos z dz ⫽
#
2⫹i
0
cos z dz ⫽ sin z d
2⫹i
0
⫽ sin (2 ⫹ i) 2 sin 0 ⫽ sin (2 ⫹
If we desired a complex number of the form a ⫹ ib for an answer, we can use
1.4031 ⫺ 0.4891i (see Example 1 in Section 17.7). Hence,
# cos z dz ⫽ 1.4031 ⫺ 0.4891i.
C
We can draw several immediate conclusions from Theorem 18.3.2. First, obser
contour C is closed, then z0 ⫽ z1 and consequently
BC
䉲
f (z) dz ⫽ 0.
18.3 Independence of the Path
independent of the path.
In addition we have the following sufficient condition for the existence of an antiderivative:
If f is continuous and eC f (z) dz is independent of the path in a domain D,
then f has an antiderivative everywhere in D.
D
z
s
z + Δz
z0
FIGURE 18.3.3 Contour used in proof of (7)
The last statement is important and deserves a proof. Assume that f is continuous, 兰C f (z
z
is independent of the path in a domain D, and F is a function defined by F(z) ez0 f (s) ds whe
denotes a complex variable, z0 is a fixed point in D, and z represents any point in D. We wis
show that F(z) f (z); that is, F is an antiderivative of f in D. Now,
F(z Dz) 2 F(z) #
z Dz
#
f (s) ds 2
z0
z
z0
f (s) ds #
z Dz
f (s) ds.
z
Because D is a domain we can choose z so that z z is in D. Moreover, z and z z can
joined by a straight segment lying in D, as shown in FIGURE 18.3.3. This is the contour we us
the last integral in (8). With z fixed, we can write*
f (z) Dz f (z)
#
z Dz
z
ds #
z Dz
f (z) ds
f (z) and
z
1
Dz
#
z Dz
f (z) ds.
z
From (8) and (9) it follows that
F(z Dz) 2 F(z)
1
2 f (z) Dz
Dz
#
z Dz
f f (s) 2 f (z)g ds.
z
Now f is continuous at the point z. This means that for any e 0 there exists a d 0 so
| f (s) f (z)| e whenever |s z| d. Consequently, if we choose z so that | z| d, we ha
2
F(z Dz) 2 F(z)
1
2 f (z) 2 2
Dz
Dz
2
#
1
22
Dz
z Dz
f f (s) 2 f (z)g ds 2
z
#
z Dz
f f (s) 2 f (z)g ds 2 #
z
1
eZDzZ ZDzZ
Hence, we have shown that
lim
DzS0
F(z Dz) 2 F(z)
f (z)
Dz
or
F9(z) f (z).
If f is an analytic function in a simply connected domain D, it is necessarily continu
throughout D. This fact, when put together with the results in Theorem 18.3.1 and (7), lead
a theorem that states that an analytic function possesses an analytic antiderivative.
Theorem 18.3.3
Existence of an Antiderivative
If f is analytic in a simply connected domain D, then f has an antiderivative in D; that is, the
exists a function F such that F(z) f (z) for all z in D.
In (9) of Section 17.6 we saw that 1/z is the derivative of Ln z. This means that under some circu
stances Ln z is an antiderivative of 1/z. Care must be exercised in using this result. For example, supp
D is the entire complex plane without the origin. The function 1/z is analytic in this mult
*See Problem 29 in Exercises 18.1.
866
|
CHAPTER 18 Integration in the Complex Plane
BC z
In this case, Ln z is not an antiderivative of 1/z in D, since Ln z is not analytic in D. Re
fails to be analytic on the nonpositive real axis (the branch cut off the principal branch of th
Using the Logarithmic Function
EXAMPLE 4
y
Evaluate
2i
C
#
C
1
dz, where C is the contour shown in FIGURE 18.3.4.
z
SOLUTION Suppose that D is the simply connected domain defined by x y Im(z) 0. In this case, Ln z is an antiderivative of 1/z, since both these f
analytic in D. Hence by (4),
x
3
#
FIGURE 18.3.4 Contour in Example 4
2i
3
2i
1
dz Ln z d Ln 2i 2 Ln 3.
z
3
From (7) of Section 17.6, we have
Ln 2i loge2 #
and so
2i
3
p
i
2
and
Ln 3 loge3
1
2
p
dz 5 log e 1 i 5 20.4055 1 1.5708i.
z
3
2
REMARKS
Suppose f and g are analytic in a simply connected domain D that contains the contour
z1 are the initial and terminal points of C, then the integration by parts formula is va
#
z1
z0
z1
f (z) g9(z) dz f (z) g(z)d
2
z0
#
z1
f 9(z) g(z) dz.
z0
This can be proved in a straightforward manner using Theorem 18.3.2 on th
(d/dz)( fg). See Problems 21–24 in Exercises 18.3.
Exercises
18.3
Answers to selected odd-numbered problems begin on page ANS-41.
In Problems 1 and 2, evaluate the given integral, where C is the
contour given in the figure, by (a) finding an alternative path of
integration and (b) using Theorem 18.3.2.
1.
# (4z 1) dz
2.
C
3.
# e dz
z
y
4.
3 + 3i
3
i(t 4 4t 3 2),
# 6z dz, where C is z(t) 2 cos pt i sin
2
3
C
2
p
4
In Problems 5–24, use Theorem 18.3.2 to evaluate th
integral. Write each answer in the form a ib.
x
3+i
x
5.
FIGURE 18.3.6 Contour in
Problem 2
7.
|z| = 1
FIGURE 18.3.5 Contour in
Problem 1
# 2z dz, where C is z(t) 2t
C
C
y
i
–i
In Problems 3 and 4, evaluate the given integral alon
indicated contour C.
0
#
31i
#
11i
z2 dz
6.
#
2i
(3z 2 2 4z 2i
0
12i
#
1
z3 dz
8.
(z 3 2 z) dz
3i
18.3 Independence of the Path
11.
#e
dz
12.
13.
#
#
p 1 2i
sin
p
z
dz
2
14.
#
#
a
z
Re(z) ⬎ 0
12i
pi
cos z dz
⫹
z2
b dz, C is any contour in the right half-pl
i
1 ⫹ (p>2)i
16. #
sinh 3z dz
# cosh z dz
1
# z dz, C is the arc of the circle z ⫽ 4e , ⫺p/2 ⱕ t ⱕ p/2
1
# z dz, C is the straight line segment between z ⫽ 1 ⫹ i and
i
pi
17.
20.
1 2 2i
2pi
15.
ze dz
12i
i>2
21.
i
#e
z
cos z dz
22.
23.
#
11i
z
ze dz
24.
#
pi
z2ez dz
0
i
it
# z sin z dz
0
p
C
18.
C
z ⫽ 4 ⫹ 4i
18.4
Cauchy’s Integral Formulas
INTRODUCTION
In the last two sections we saw the importance of the Cauchy–Gou
theorem in the evaluation of contour integrals. In this section we are going to examine sev
more consequences of the Cauchy–Goursat theorem. Unquestionably, the most significan
these is the following result:
The value of an analytic function f at any point z0 in a simply connected domain can
be represented by a contour integral.
After establishing this proposition we shall use it to further show that
An analytic function f in a simply connected domain possesses derivatives of all orders.
The ramifications of these two results alone will keep us busy not only for the remainder of
section but in the next chapter as well.
First Formula We begin with the Cauchy integral formula. The idea in the next theo
is this: If f is analytic in a simply connected domain and z0 is any point D, then the quot
f (z)/(z ⫺ z0) is not analytic in D. As a consequence, the integral of f (z)/(z ⫺ z0) around a sim
closed contour C that contains z0 is not necessarily zero but has, as we shall now see, the va
2pi f (z0). This remarkable result indicates that the values of an analytic function f at points ins
a simple closed contour C are determined by the values of f on the contour C.
Theorem 18.4.1
Cauchy’s Integral Formula
Let f be analytic in a simply connected domain D, and let C be a simple closed contour lyin
entirely within D. If z0 is any point within C, then
f (z0) ⫽
f (z)
1
dz.
2pi B
C z 2 z0
䉲
(
PROOF: Let D be a simply connected domain, C a simple closed contour in D, and z0 an inte
point of C. In addition, let C1 be a circle centered at z0 with radius small enough that it is interior t
By the principle of deformation of contours, we can write
BC z 2 z0
䉲
868
|
CHAPTER 18 Integration in the Complex Plane
f (z)
dz ⫽
BC z 2 z0
f (z)
䉲
1
dz.
BC z 2 z0
䉲
dz ⫽
1
BC
䉲
dz
z 2 z0
1
⫽ f (z0)
f (z) 2 f (z0)
dz
⫹
dz.
z
2
z
BC1
BC1 z 2 z0
0
䉲
䉲
Now from (4) of Section 18.2 we know that
dz
⫽ 2pi.
BC1 z 2 z0
䉲
Thus, (3) becomes
f (z) 2 f (z0)
f(z)
dz ⫽ 2pi f (z0) ⫹
dz.
z
2
z
BC1
BC1 z 2 z0
0
䉲
䉲
Since f is continuous at z0 for any arbitrarily small e ⬎ 0, there exists a d ⬎
| f (z) ⫺ f (z0)| ⬍ e whenever |z ⫺ z0| ⬍ d. In particular, if we choose the cir
|z ⫺ z0| ⫽ d/2 ⬍ d, then by the ML-inequality (Theorem 18.1.3) the absolute valu
gral on the right side of (4) satisfies
2
BC
䉲
f (z) 2 f (z0)
e
d
dz 2 #
2p a b ⫽ 2pe.
z
2
z
d>2
2
0
1
In other words, the absolute value of the integral can be made arbitrarily small b
radius of the circle C1 to be sufficiently small. This can happen only if the integral
Cauchy integral formula (1) follows from (4) by dividing both sides by 2pi.
The Cauchy integral formula (1) can be used to evaluate contour integrals. Sin
work problems without a simply connected domain explicitly defined, a more prac
ment of Theorem 18.4.1 is
If f is analytic at all points within and on a simple closed contour C, and z 0 is
f (z)
1
any point interior to C, then f (z0) ⫽
dz.
z
2 z0
2pi B
C
䉲
Using Cauchy’s Integral Formula
EXAMPLE 1
z 2 4z ⫹ 4
dz, where C is the circle |z| ⫽ 2.
BC z ⫹ i
2
Evaluate
䉲
SOLUTION First, we identify f (z) ⫽ z2 ⫺ 4z ⫹ 4 and z0 ⫽ ⫺i as a point within
Next, we observe that f is analytic at all points within and on the contour C. Thus b
integral formula we obtain
y
z 2 2 4z ⫹ 4
dz ⫽ 2pi f (⫺i) ⫽ 2pi(3 ⫹ 4i) ⫽ 2p(⫺4 ⫹ 3i).
BC z ⫹ i
C
䉲
3i
x
Evaluate
–3i
FIGURE 18.4.1 Contour in Example 2
Using Cauchy’s Integral Formula
EXAMPLE 2
z
dz, where C is the circle |z ⫺ 2i| ⫽ 4.
BC z ⫹ 9
䉲
2
SOLUTION By factoring the denominator as z2 ⫹ 9 ⫽ (z ⫺ 3i)(z ⫹ 3i), we see
only point within the closed contour at which the integrand fails to be analytic. See
18.4 Cauchy’s Integral Formulas
z2 1 9
y
5
z 2 3i
,
we can identify f (z) ⫽ z/(z ⫹ 3i). This function is analytic at all points within and on
contour C. From the Cauchy integral formula we then have
z1
BC z 2 ⫹ 9
z
䉲
dz ⫽
z
z ⫹ 3i
3i
dz ⫽ 2pi f (3i) ⫽ 2pi ⫽ pi.
z 2 3i
6i
BC
䉲
x
Flux and Cauchy’s Integral Formula
EXAMPLE 3
(a) Source: k > 0
The complex function f (z) ⫽ k/(z 2 z1 ), where k ⫽ a ⫹ ib and z1 are complex numbers, gi
rise to a flow in the domain z ⫽ z1. If C is a simple closed contour containing z ⫽ z1 in
interior, then from the Cauchy integral formula we have
y
BC
z1
䉲
x
(b) Sink: k < 0
FIGURE 18.4.2 Vector fields in Example 3
f (z) dz ⫽
a 2 ib
dz ⫽ 2pi(a 2 ib).
BC z 2 z1
䉲
Thus the circulation around C is 2pb and the net flux across C is 2pa. If z1 were in the exte
of C, both the circulation and net flux would be zero by Cauchy’s theorem.
Note that when k is real, the circulation around C is zero but the net flux acros
is 2pk. The complex number z 1 is called a source for the flow when k ⬎ 0 and a s
when k ⬍ 0. Vector fields corresponding to these two cases are shown in FIGURE 18.4.
and 18.4.2(b).
Second Formula We can now use Theorem 18.4.1 to prove that an analytic funct
possesses derivatives of all orders; that is, if f is analytic at a point z0, then f ⬘, f ⬙, f ⵮, and so
are also analytic at z0. Moreover, the values of the derivatives f (n)(z0), n ⫽ 1, 2, 3, … , are gi
by a formula similar to (1).
Cauchy’s Integral Formula for Derivatives
Theorem 18.4.2
Let f be analytic in a simply connected domain D, and let C be a simple closed contour lyin
entirely within D. If z0 is any point interior to C, then
f (n)(z0) ⫽
f (z)
n!
dz.
n⫹1
2pi B
C (z 2 z0)
(
䉲
PARTIAL PROOF: We will prove (6) only for the case n ⫽ 1. The remainder of the proof
be completed using the principle of mathematical induction.
We begin with the definition of the derivative and (1):
f 9(z0) ⫽ lim
DzS0
⫽ lim
DzS0
⫽ lim
DzS0
870
|
CHAPTER 18 Integration in the Complex Plane
f (z0 ⫹ Dz) 2 f (z0)
Dz
f (z)
f (z)
1
c
dz 2
dz d
z
2pi Dz B
z
2
(z
⫹
Dz)
BC 2 z0
C
0
䉲
䉲
f (z)
1
dz.
2pi B
C (z 2 z0 2 Dz) (z 2 z0)
䉲
0
Zz 2 z0 Z $ d
or
1
1
# 2.
2
Zz 2 z0 Z
d
Furthermore, if we choose |z| d/2, then
Zz 2 z0 2 DzZ $ iz 2 z0Z 2 ZDzi $ d 2 Z DzZ $
d
1
and so
2
Zz 2 z0 2 Dz
Now,
2
f (z)
f (z)
Dz f (z)
dz 2
dz 2
dz 2 2
2
BC (z 2 z0)
BC (z 2 z0 2 Dz) (z 2 z0)
BC (z 2 z0)2(z 2 z0 2 Dz)
䉲
䉲
䉲
Because the last expression approaches zero as z S 0, we have shown that
f 9(z0) lim
f (z0 Dz) 2 f (z0)
f (z)
1
dz.
2
Dz
2pi B
C (z 2 z0)
䉲
DzS0
If f (z) u(x, y) iv(x, y) is analytic at a point, then its derivatives of all orders
point and are continuous. Consequently, from
f 9 (z) 0u
0v
0v
0u
i
2i
0x
0x
0y
0y
f 0 (z) 0 2u
0 2v
0 2v
0 2u
i
2
i
0y 0x
0y 0x
0x 2
0x 2
we can conclude that the real functions u and v have continuous partial derivatives
at a point of analyticity.
Like (1), (6) can sometimes be used to evaluate integrals.
Using Cauchy’s Integral Formula for Derivatives
EXAMPLE 4
Evaluate
BC z 4 4z 3
䉲
z1
dz, where C is the circle |z| 1.
SOLUTION Inspection of the integrand shows that it is not analytic at z 0 and
only z 0 lies within the closed contour. By writing the integrand as
z1
z1
z4
,
z 4 4z 3
z3
we can identify z0 0, n 2, and f (z) (z 1)/(z 4). By the Quotient R
6/(z 4)3 and so by (6) we have
2pi
3p
z1
dz f 0 (0) i.
3
2!
32
BC z 4z
䉲
4
18.4 Cauchy’s Integral Formulas
BC z(z 2 i)
i
x
0
SOLUTION Although C is not a simple closed contour, we can think of it as the union of
simple closed contours C1 and C2 as indicated in Figure 18.4.3. By writing
z3 3
z3 3
z3 3
dz dz dz
2
2
BC z(z 2 i)
BC1 z(z 2 i)
BC2 z(z 2 i)2
䉱
䉲
C1
FIGURE 18.4.3 Contour in Example 5
䉲
BC
䉲
1
z3 3
z3 3
z
(z 2 i)2
dz dz I1 I2,
z
BC2 (z 2 i)2
䉲
we are in a position to use both (1) and (6).
To evaluate I1, we identify z0 0 and f (z) (z3 3)/(z i)2. By (1) it follows that
I1 BC
䉲
1
z3 3
(z 2 i)2
dz 2pi f (0) 6pi.
z
To evaluate I2 we identify z0 i, n 1, f (z) (z3 3)/z, and f (z) (2z3 3)/z2. From
we obtain
I2 z3 3
z
BC2 (z 2 i)2
䉲
dz 2pi
f 9(i) 2pi(3 2i) 2p(2 3i).
1!
Finally we get
z3 3
dz I1 I2 6pi 2p(2 3i) 4p(1 3i).
BC z(z 2 i)2
䉲
Liouville’s Theorem If we take the contour C to be the circle |z z0| r, it follo
from (6) and the ML-inequality that
Z f (n)(z0)Z f (z)
n!
n!
1
n!M
2
dz 2 #
M n 1 2pr n ,
n1
2p B
2p r
r
C (z 2 z0)
䉲
where M is a real number such that | f (z)| M for all points z on C. The result in (7), ca
Cauchy’s inequality, is used to prove the next result.
Theorem 18.4.3
Liouville’s Theorem
The only bounded entire functions are constants.
PROOF: Suppose f is an entire function and is bounded; that is, | f (z)| M for all z. Then
any point z0, (7) gives | f (z0)| M/r. By making r arbitrarily large, we can make | f (z0)
small as we wish. This means f (z0) 0 for all points z0 in the complex plane. Hence f m
be a constant.
Fundamental Theorem of Algebra Liouville’s theorem enables us to prove, in tu
a result that is learned in elementary algebra:
If P(z) is a nonconstant polynomial, then the equation P(z) 0 has at least one root.
872
|
CHAPTER 18 Integration in the Complex Plane
to our underlying assumption that P was not a constant polynomial. We conclude th
exist at least one number z for which P(z) 0.
Exercises
18.4
Answers to selected odd-numbered problems begin on page ANS-41.
In Problems 1–24, use Theorems 18.4.1 and 18.4.2, when
appropriate, to evaluate the given integral along the indicated
closed contour(s).
1.
2.
4
dz; ZzZ 5
z
2
3i
BC
17.
18.
䉲
z2
dz; ZzZ 5
BC (z 2 3i)2
䉲
19.
e
dz; ZzZ 4
z
2
pi
BC
z
3.
䉲
20.
1 2e z
dz; ZzZ 1
4.
BC z
䉲
5.
6.
z 2 2 3z 4i
dz; ZzZ 3
BC z 2i
䉲
8.
9.
10.
22.
cos z
dz; ZzZ 1.1
BC 3z 2 p
䉲
z
dz;
BC z 4
2
7.
21.
䉲
2
(a) Zz 2 iZ 2,
z 2 3z 2i
dz;
BC z 2 3z 2 4
䉲
(a) ZzZ 2,
23.
(b) Zz 2iZ 1
(b) Zz 5Z BC
䉲
BC
䉲
3
a
a
(a) ZzZ 1,
(b) Zz 2
(b) Zz 2 2Z
e 2iz
z4
2
b dz; ZzZ 6
z4
(z 2 i)3
cosh z
sin 2z
2
b dz; ZzZ 3
(z 2 p)3
(2z 2 p)3
3
1
dz; Zz 2 iZ BC z (z 1)
䉲
2
2
3
2
3z 1
dz; C is given in FIGURE 18.4.4
BC z (z 2 2)2
䉲
y
C
0
sin z
dz; Zz 2 2iZ 2
BC z p2
䉲
(a) ZzZ 1,
1
dz; Zz 2 2Z 5
BC z (z 2 1)2
䉲
2
2
x
2
ez
dz; Zz 2 iZ 1
BC (z 2 i)3
FIGURE 18.4.4 Contour in Problem 23
䉲
24.
䉲
14.
1
dz;
BC z (z 2 4)
䉲
z2 4
dz; Zz 2 3iZ 1.3
BC z 2 5iz 2 4
䉲
z
12.
dz; ZzZ 2
BC (z i)4
13.
2
3
2
2
11.
z2
dz;
BC z (z 2 1 2 i)
䉲
BC
䉲
e iz
dz; C is given in FIGURE 18.4.5
BC (z 2 1)2
䉲
cos 2z
dz; ZzZ 1
z5
y
e z sin z
dz; Zz 2 1Z 3
BC z 3
i
䉲
2z 5
dz; (a) ZzZ 12, (b) Zz 1Z 2,
15.
BC z 2 2 2z
(c) Zz 2 3Z 2, (d) Zz 2iZ 1
x
䉲
16.
z
dz; (a) ZzZ 12,
BC (z 2 1)(z 2 2)
(c) Zz 2 1Z 12, (d) ZzZ 4
䉲
(b) Zz 1Z 1,
–i
C
FIGURE 18.4.5 Contour in Problem 24
18.4 Cauchy’s Integral Formulas
Answer Problems 1–12 without referring back to the text. Fill in
the blank or answer true/false.
14.
# (x iy) dz; C is the contour shown in Figure 18.R.1
C
1. The sector defined by p/6 arg z p/6 is a simply
connected domain. _____
2. If
BC
䉲
y
C
f (z) dz 0 for every simple closed contour C, then f is
analytic within and on C. _____
x
#
z22
dz is the same for any path C in the
z
C
right half-plane Re(z) 0 between z 1 i and
3. The value of
FIGURE 18.R.1 Contour in Problems 13 and 14
z 10 8i. _____
g(z)
g(z)
4. If g is entire, then
dz dz, where C is the
z
2
i
z
BC
BC1 2 i
circle |z| 3 and C1 is the ellipse x2 y2 /9 1. _____
䉲
15.
䉲
f (z) dz _____.
j2 6j 2 2
dj, where C is |z| 3, then
BC j 2 z
f (1 i) _____.
6. If f (z) 16.
f (z)
dz _____.
(z
pi)3
BC
17.
18.
20.
f 9 (z)
f (z)
dz dz _____ .
z
2
z
(z
2
z0)2
BC
BC
0
0,
if n _____
2pi, if n _____
BC
where n is an integer and C is |z| 1.
z ndz e
䉲
C
|
CHAPTER 18 Integration in the Complex Plane
4
i(1 t 3)2, 1 t 1
# (4z 3z 2z 1) dz; C is the line segment from 0 to
3
2
23.
e 2z
dz; C is the circle |z 1| 3
BC z 4
24.
cos z
dz; C is the circle |z| 12
BC z 2 z 2
25.
BC 2z 2 7z 3
26.
BC
f (z) dz 2 # _____ .
(x iy) dz; C is the contour shown in FIGURE 18.R.1
874
# sin z dz; C is z(t) t
3z 4
dz; C is the circle |z| 2
BC z 2 2 1
In Problems 13–28, evaluate the given integral using the
techniques considered in this chapter.
#
(4z 2 6) dz
22.
䉲
BC
12i
BC
z0 is a point within C, then
12. If |f (z)| 2 on |z| 3, then 2
#
epz dz; C is the ellipse x2 /100 y2 /64 1
21.
10. If f is analytic within and on the simple closed contour C and
13.
䉲
C
䉲
䉲
BC
C
1
9.
dz 0 for every simple closed contour C
BC (z 2 z0)(z 2 z1)
that encloses the points z0 and z1. _____
䉲
dz; C is the line segment from z i to z 1 i
pz
3i
19.
䉲
2
#e
䉲
8. If f is entire and | f (z)| 10 for all z, then f (z) _____.
11.
2
C
7. If f (z) z3 ez and C is the contour z 8eit, 0 t 2p,
then
# |z | dz; C is z(t) t it , 0 t 2
C
䉲
5. If f is a polynomial and C is a simple closed curve, then
BC
3
–4
䉲
(z2 z1 z z2) dz; C is the circle |z| 1
䉲
䉲
䉲
䉲
䉲
3
1
dz; C is the ellipse x2 /4 y2 1
z csc z dz; C is the rectangle with vertices 1 i, 1 2 i, 2 i
C
29. Let f (z) ⫽ zng(z), where n is a positive integer, g
x
–2
3
and g(z) ⫽ 0 for all z. Let C be a circle with
f 9(z)
origin. Evaluate
dz.
BC f (z)
䉲
30. Let C be the straight line segment from i to 2 ⫹
FIGURE 18.R.2 Contour in Problem 27
#
2 Ln (z ⫹ 1) dz 2 # log e10 ⫹
C
p
.
2
CHAPTER 18 in Review