2 The Functions e^z, sin z, cos z

Exercises



41



Proof

Set α = r cis θ = r eiθ , r > 0. Since e z = e x eiy , we will have e z = α if x = log r

and eiy = eiθ . Hence e z = α for all points z = x + i y with x = log r , y = Arg α =

θ ± 2kπ, k = 0, 1, 2 . . ..

v. (e z ) = e z .

Recall that (e z ) = (e z )x = e z .

To define sin z and cos z, note that for real y

eiy = cos y + i sin y

e−iy = cos y − i sin y

so that

sin y =



1 iy

(e − e−iy )

2i



and



1 iy

(e + e−iy ).

2

Thus we can define entire extensions of sin x and cos x by setting

cos y =



1 iz

(e − e−iz )

2i

1

cos z = (eiz + e−iz ).

2

sin z =



Many of the familiar properties of the sin and cos functions remain valid in the larger

setting of the complex plane. For example,

sin 2z = 2 sin z cos z

sin2 z + cos 2 z = 1

(sin z) = cos z.

These identities are easily verified and are left as an exercise. Moreover, in Section 6.3,

we will see that, in general, functional equations of the above form, known to be true

on the real axis, remain valid throughout the complex plane.

On the other hand, unlike sin x, sin z is not bounded in modulus by 1. For example,

| sin 10i | = 12 (e10 − e−10 ) > 10, 000.



Exercises

1. Show that

f x = lim



h→0

h real



provided the limits exist.



f (z + h) − f (z)

f (z + ih) − f (z)

; f y = lim

,

h

h

h→0

h real



42



3 Analytic Functions



2. a. Show that f (z) = x 2 + iy 2 is differentiable at all points on the line y = x.

b. Show that it is nowhere analytic.

3. Prove that the composition of differentiable functions is differentiable. That is, if f is differentiable

at z, and if g is differentiable at f (z), then g ◦ f is differentiable at z. [Hint: Begin by noting

g( f (z + h)) − g( f (z)) = [g ( f (z)) + ][ f (z + h) − f (z)]

where



→ 0 as h → 0.]



√

4. Suppose that g is a continuous “ z” (i.e., g 2 (z) = z) in some neighborhood of z. Verify that

√

g (z) = 1/2 z. [Hint: Use

g 2 (z) − g 2 (z 0 )

1=

z − z0

to evaluate

lim



z→z 0



g(z) − g(z 0 )

.]

z − z0



5. Suppose f is analytic in a region and f ≡ 0 there. Show that f is constant.

6. Assume that f is analytic in a region and that at every point, either f = 0 or f = 0. Show that f is

constant. [Hint: Consider f 2 .]

7. Show that a nonconstant analytic function cannot map a region into a straight line or into a circular

arc.

8. Find all analytic functions f = u + iv with u(x, y) = x 2 − y 2 .

9. Show that there are no analytic functions f = u + iv with u(x, y) = x 2 + y 2 .

10. Suppose f is an entire function of the form

f (x, y) = u(x) + iv(y).

Show that f is a linear polynomial.

11. a. Show that ez is entire by verifying the Cauchy-Riemann equations for its real and imaginary

parts.

b. Prove:

ez 1 +z 2 = ez 1 ez 2 .

12. Show: |ez | = e x .

13. Discuss the behavior of ez as z → ∞ along the various rays from the origin.

14. Find all solutions of

a. ez = 1,

c. ez = −3,



b. ez = i,

d. ez = 1 + i.



15. Verify the identities

a. sin 2z = 2 sin z cos z,

b. sin2 z + cos2 z = 1,

c. (sin z) = cos z.

16.* Show that

a. sin( π2 + iy) = 12 (e y + e−y ) = cosh y

b. | sin z| ≥ 1 at all points on the square with vertices ±(N + 12 )π ± (N + 12 )π i, for any positive

integer N.

c. | sin z| → ∞, as Imz = y → ±∞.



Exercises



43



17. Find (cos z) .

18. Find sin−1 (2)– that is, find the solutions of sin z = 2. [Hint: First set w = eiz and solve for ω.]

19.* Find all solutions of the equation:



z



ee = 1.

20. Show that sin(x + iy) = sin x cosh y + i cos x sinh y.

21. Show that the power series

f (z) = 1 + z +



z2

+ ... =

2!



∞

n=0



zn

n!



is equal to ez . [Hint: First show that f (z) f (w) = f (z + w), then show

f (x) = e x

f (iy) = cos y + i sin y

using the power series representations for the real functions

e x , cos x, sin x.]

22. Show:

g(z) = z −



z5

z3

+

− +...

3!

5!



is equal to sin z. [Hint: Use the power series representation for ez given in (21) to show that

g(z) =

23. Find a power series representation for cos z.



1 iz

(e − e−iz ).]

2i



Chapter 4



Line Integrals and Entire Functions



Introduction

Recall that, according to Theorem 2.9, an everywhere convergent power series

represents an entire function. Our main goal in the next two chapters is the somewhat

surprising converse of that result: namely, that every entire function can be expanded

as an everywhere convergent power series. As an immediate corollary, we will be

able to prove that every entire function is infinitely differentiable. To arrive at these

results, however, we must begin by discussing integrals rather than derivatives.



4.1 Properties of the Line Integral

4.1 Definition

Let f (t) = u(t) + i v(t) be any continuous complex-valued function of the real

variable t, a ≤ t ≤ b.

b

a



b



f (t)dt =

a



b



u(t)dt + i



v(t)dt.



a



4.2 Definition

a. Let z(t) = x(t) + i y(t), a ≤ t ≤ b. The curve determined by z(t) is called

piecewise differentiable and we set

z˙ (t) = x (t) + i y (t)

if x and y are continuous on [a, b] and continuously differentiable on each subinterval [a, x 1 ], [x 1 , x 2 ], . . . , [x n−1 , b] of some partition of [a, b].

b. The curve is said to be smooth if, in addition, z˙ (t) = 0 (i.e., x (t) and y (t) do not

both vanish) except at a finite number of points.



45



46



4 Line Integrals and Entire Functions



Throughout the remainder of the text, all curves will be assumed to be smooth

unless otherwise stated.

Finally, we define the important concept of a line integral.

4.3 Definition

Let C be a smooth curve given by z(t), a ≤ t ≤ b, and suppose f is continuous at

all the points z(t). Then, the integral of f along C is

b



f (z)dz =



f (z(t))˙z (t)dt.



a



C



Note that the integral along the curve C depends not only on the points of C but on

the direction as well. However, we will show that it is independent of the particular

parametrization. Intuitively, if z(t), a ≤ t ≤ b, and ω(t), c ≤ t ≤ d, trace the same

curve in the same direction, then λ = z −1 ◦ ω will be a 1-1 mapping of [c, d] onto

[a, b] such that

ω(t) = z(λ(t)).

(1)

However, if z is not 1-1, it is difficult to define z −1 . Instead, we take the existence of

some λ that satisfies (1) as the definition for equivalent curves.

C



z

a



ω

b



c



d



4.4 Definition

The two curves

C1 : z(t), a ≤ t ≤ b

and

C2 : ω(t), c ≤ t ≤ d

are smoothly equivalent if there exists a 1-1 C 1 mapping λ(t) : [c, d] → [a, b] such

that λ(c) = a, λ(d) = b, λ (t) > 0 for all t, and

ω(t) = z(λ(t)).

(It is easy to verify that the above is an equivalence relation. See Exercise 1.)

4.5 Proposition

If C1 and C2 are smoothly equivalent, then

f =

C1



f.

C2



4.1 Properties of the Line Integral



47



Proof

Suppose f (z) = u(z) + i v(z), C1 and C2 as above. Then, by definition

b



f =

a



C1



b



u(z(t))x (t)dt −



v(z(t))y (t)dt



a

b



+i



b



u(z(t))y (t)dt + i



a



v(z(t))x (t)dt



(1)



a



while

d



f =

C2



[u(z(λ(t))) + i v(z(λ(t)))][x (λ(t)) + i y (λ(t))]λ (t)dt.



(2)



c



Expanding the integrand in (2) and analyzing the four terms separately, we find that

they are exactly equal to the four corresponding terms in (1).

For example

d



b



u(z(λ(t)))x (λ(t))λ (t)dt =



c



u(z(t))x (t)dt,



a



by the change-of-variable theorem for ordinary real integrals, and the proof is

complete.

The following proposition points out the dependence of the line integral on the

direction of the curve.

4.6 Definition

Suppose C is given by z(t), a ≤ t ≤ b. Then −C is defined by z(b + a − t),

a ≤ t ≤ b. (Intuitively, −C is the point set of C traced in the opposite direction.)

4.7 Proposition



−C



f =



f.

C



Proof



−C



b



f =−



f (z(b + a − t))˙z (b + a − t)dt.



a



Again, expanding the integral into real and imaginary parts and applying the changeof-variable theorem to each (real) integral, we find



−C



a



f =

b



f (z(t))˙z (t)dt = −



f.

C



48



4 Line Integrals and Entire Functions



E XAMPLE 1

Suppose f (z) = x 2 + i y 2 (where x and y denote the real and imaginary parts of z,

respectively), and consider

C : z(t) = t + i t, 0 ≤ t ≤ 1.

Then

z˙ (t) = 1 + i

and

1



f (z)dz =

0



C



1



(t 2 + i t 2 )(1 + i )dt = (1 + i )2



t 2 dt = 2i /3.



0



♦

E XAMPLE 2

Let

f (z) =



x

1

y

= 2

−i 2

,

z

x + y2

x + y2



and set

C : z(t) = R cos t + i R sin t, 0 ≤ t ≤ 2π,



R = 0.



Then

2π



f (z)dz =

0



C



2π



=



sin t

cos t

−i

R

R



(−R sin t + i R cos t)dt



i dt = 2πi (See Exercise 8.)



0



That is, the integral of 1/z around any circle centered at the origin (traversed counterclockwise) is 2πi .

♦

E XAMPLE 3

Suppose f (z) ≡ 1, and let C be any smooth curve. Then

b



f (z)dz =

C



z˙ (t)dt = z(b) − z(a).



a



♦

The integrals defined above are natural generalizations of the definite integral and,

not too surprisingly, they share many of the same properties.

4.8 Proposition

Let C be a smooth curve; let f and g be continuous functions on C; and let α be any

complex number. Then



4.1 Properties of the Line Integral



I.



C



[ f (z) + g(z)]dz =



II.



C



α f (z)dz = α



C



49



f (z)dz +



C



C



g(z)dz



f (z)dz.



Proof

Exercise 4.

Notation: If α and β are complex numbers, the symbol α

the inequality |α| ≤ |β|.



β will be used to denote



4.9 Lemma

Suppose G(t) is a continuous complex-valued function of t. Then

b



b



G(t)dt

a



|G(t)|dt.



a



Proof

Suppose

b



G(t)dt = Reiθ , R ≥ 0.



(1)



a



By Proposition 4.8, then

b



e−iθ G(t)dt = R.



(2)



a



Suppose further that e−iθ G(t) = A(t) + i B(t), with A and B real-valued. Then,

according to (2),

b



R=



b



A(t)dt =



a



Re (e−iθ G(t))dt.



a



But Re z ≤ |Re z| ≤ |z|, hence

b



R≤



|G(t)|dt.



(3)



a



A comparison of (1) and (3) then gives the desired result.

4.10 M-L Formula

Suppose that C is a (smooth) curve of length L, that f is continuous on C, and that

f

M throughout C. Then

f (z)dz

C



M L.



50



4 Line Integrals and Entire Functions



Proof

Suppose C is given by z(t) = x(t)+i y(t), a ≤ t ≤ b. Then, by the previous lemma,

b



f (z)dz =



b



f (z(t))˙z dt



a



C



| f (z(t))˙z (t)|dt.



a



According to the Mean-Value Theorem for Integrals applied to the positive functions

| f (z(t))| and |˙z (t)|

f (z)dz)



b



max | f (z)|

z∈C



C



|˙z (t)|dt.



(4)



a



Finally, recall that for any curve given parametrically by (x(t), y(t)), a ≤ t ≤ b,

the arc length L is given by

b



L=

a



so that according to (4)



C



b



(x (t))2 + (y (t))2 dt =



|˙z (t)|dt,



a



f (z)dz



M L.



E XAMPLE

Let C be the unit circle and suppose f



1 on C. Then M = 1, L = 2π, and



f (z)dz



2π.



C



To see that the upper bound of 2π can actually be achieved, consider Example 2

above.

♦

4.11 Proposition

Suppose { f n } is a sequence of continuous functions and f n → f uniformly on the

smooth curve C. Then

f (z)dz = lim



n→∞ C



C



f n (z)dz.



Proof

f (z)dz −

C



f n (z)dz =

C



[ f (z) − f n (z)]dz

C



by Proposition 4.8. Taking n large enough so that | f (z) − f n (z)| <

and applying Proposition 4.10, shows that

f (z) −

C



fn (z)dz

C



· (length of C)



for all z ∈ C,



4.1 Properties of the Line Integral



51



for any pre-assigned > 0, and hence that

lim



n→∞ C



f n (z)dz =



f (z)dz.

C



The following generalization of the Fundamental Theorem of Calculus will be

crucial in the development of this chapter.

4.12 Proposition

Suppose f is the derivative of an analytic function F–that is, f (z) = F (z), where

F is analytic on the smooth curve C. Then

f (z)dz = F(z(b)) − F(z(a)).

C



Proof

The proof depends on a complex analogue of the chain-rule for differentiation. Letting

γ (t) = F(z(t)), a ≤ t ≤ b,

we wish to show that

γ˙ (t) = f (z(t))˙z (t)

at the all-but-finite number of points where z˙ (t) exists and is nonzero.

Note first that for any smooth curve λ(t), by considering the real and imaginary

parts of λ separately, it is easily seen that

λ(t + h) − λ(t)

˙

λ(t)

= lim

.

h

h→0

h real



Hence

F(z(t + h)) − F(z(t))

h→0

h

F(z(t + h)) − F(z(t)) z(t + h) − z(t)

·

.

= lim

h→0

z(t + h) − z(t)

h



γ˙ (t) = lim



[Since z˙ (t) = 0, we can find δ > 0 so that |h| < δ implies z(t + h) − z(t) = 0.]

Thus

γ˙ (t) = f (z(t))˙z (t).

Proposition 4.12 follows then by noting that

b



f (z)dz =

C



a



b



f (z(t))˙z (t)dt =



γ˙ (t)dt



a



= γ (b) − γ (a) = F(z(b)) − F(z(a)).