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Exercises

12. Is the collection of all values of logi 1/2  the same as the collection of all values of
1
log i ? Explain.
2


13. Is the collection of all values of logi 2  the same as the collection of all values of
2 log i ? Explain.


3.8
14. Find all values of logz 1/2 . (in rectangular form)

15. At what points is the function given by Log z 2 « 1 analytic? Explain.

16. Find the principal value of
a) i i . b) 1  i 4i

17. a)Find all values of |i i |.




3.9
Chapter Four

Integration

4.1. Introduction. If  : D ‚ C is simply a function on a real interval D  ,  , then the

integral ˜ tdt is, of course, simply an ordered pair of everyday 3 rd grade calculus

integrals:

  

˜ tdt  ˜ xtdt « i ˜ ytdt,
  



where t  xt « iyt. Thus, for example,

1

˜t 2 « 1 « it 3 dt  4 « i.
3 4
0



Nothing really new here. The excitement begins when we consider the idea of an integral
of an honest-to-goodness complex function f : D ‚ C, where D is a subset of the complex
plane. Let™s define the integral of such things; it is pretty much a straight-forward extension
to two dimensions of what we did in one dimension back in Mrs. Turner™s class.

Suppose f is a complex-valued function on a subset of the complex plane and suppose a
and b are complex numbers in the domain of f. In one dimension, there is just one way to
get from one number to the other; here we must also specify a path from a to b. Let C be a
path from a to b, and we must also require that C be a subset of the domain of f.




4.1
(Note we do not even require that a ‚® b; but in case a  b, we must specify an orientation
for the closed path C.) Next, let P be a partition of the curve; that is, P  z 0 , z 1 , z 2 , µ , z n 
is a finite subset of C, such that a  z 0 , b  z n , and such that z j comes immediately after
z j1 as we travel along C from a to b.




A Riemann sum associated with the partition P is just what it is in the real case:

n

 fz „ z j ,
SP  j
j1



where z „ is a point on the arc between z j1 and z j , and z j  z j  z j1 . (Note that for a
j
given partition P, there are many SP”depending on how the points z „ are chosen.) If
j
there is a number L so that given any   0, there is a partition P  of C such that


|SP  L|  

whenever P ‘ P  , then f is said to be integrable on C and the number L is called the
integral of f on C. This number L is usually written ˜ fzdz.
C


Some properties of integrals are more or less evident from looking at Riemann sums:


˜ cfzdz  c ˜ fzdz
C C



for any complex constant c.




4.2
˜fz « gzdz  ˜ fzdz « ˜ gzdz
C C C




4.2 Evaluating integrals. Now, how on Earth do we ever find such an integral? Let
 : ,  ‚ C be a complex description of the curve C. We partition C by partitioning the
,    t 0  t 1  t 2 µ  t n  .
interval in the usual way: Then
a  , t 1 , t 2 , µ ,   b is partition of C. (Recall we assume that   t ‚® 0
for a complex description of a curve C.) A corresponding Riemann sum looks like

n

 ft „ t j   t j1 .
SP  j
j1



We have chosen the points z „  t „ , where t j1 ‚ t „ ‚ t j . Next, multiply each term in the
j j j
sum by 1 in disguise:

n

 ft „  t jtj  tt j1  t j  t j1 .

SP  j
j1
j1



I hope it is now reasonably convincing that ”in the limit”, we have



˜ fzdz  ˜ ft  tdt.
C 



(We are, of course, assuming that the derivative   exists.)


Example

We shall find the integral of fz  x 2 « y « ixy from a  0 to b  1 « i along three
different paths, or contours, as some call them.

First, let C 1 be the part of the parabola y  x 2 connecting the two points. A complex
description of C 1 is  1 t  t « it 2 , 0 ‚ t ‚ 1:




4.3
1


0.8


0.6


0.4


0.2


0 0.2 0.4 0.6 0.8 1
x




Now,  1 t  1 « 2ti, and f  1 t  t 2 « t 2  « itt 2  2t 2 « it 3 . Hence,

1

˜ fzdz  ˜ f  1 t 1 tdt
C1 0
1

 ˜2t 2 « it 3 1 « 2tidt
0
1

 ˜2t 2  2t 4 « 5t 3 idt
0

 4 « 5i
15 4

Next, let™s integrate along the straight line segment C 2 joining 0 and 1 « i.

1


0.8


0.6


0.4


0.2


0 0.2 0.4 0.6 0.8 1
x



Here we have  2 t  t « it, 0 ‚ t ‚ 1. Thus,  2 t  1 « i, and our integral looks like




4.4
1

˜ fzdz  ˜ f  2 t 2 tdt
C2 0
1

 ˜t 2 « t « it 2 1 « idt
0
1

˜ t « it « 2t 2 dt

0

 1 « 7i
2 6

Finally, let™s integrate along C 3 , the path consisting of the line segment from 0 to 1
together with the segment from 1 to 1 « i.
1


0.8


0.6


0.4


0.2


0 0.2 0.4 0.6 0.8 1




We shall do this in two parts: C 31 , the line from 0 to 1 ; and C 32 , the line from 1 to 1 « i.
Then we have


˜ fzdz  ˜ fzdz « ˜ fzdz.
C3 C 31 C 32



For C 31 we have t  t, 0 ‚ t ‚ 1. Hence,

1

˜ fzdz  ˜ dt  1.
3
C 31 0



For C 32 we have t  1 « it, 0 ‚ t ‚ 1. Hence,

1

˜ fzdz  ˜1 « t « ititdt   1 « 5 i.
3 6
C 32 0




4.5
Thus,


˜ fzdz  ˜ fzdz « ˜ fzdz
C3 C 31 C 32

 5 i.
6

Suppose there is a number M so that |fz| ‚ M for all zC. Then



˜ fzdz ˜ ft  tdt

C 


‚ ˜|ft  t|dt



‚ M ˜|  t|dt  ML,


where L  ˜|  t|dt is the length of C.



Exercises

1. Evaluate the integral ˜ z dz, where C is the parabola y  x 2 from 0 to 1 « i.
C


2. Evaluate ˜ 1
dz, where C is the circle of radius 2 centered at 0 oriented
z
C
counterclockwise.

4. Evaluate ˜ fzdz, where C is the curve y  x 3 from 1  i to 1 « i , and
C



for y  0
1
fz  .
4y for y ‚ 0


5. Let C be the part of the circle t  e it in the first quadrant from a  1 to b  i. Find as
small an upper bound as you can for ˜ z 2  z 4 « 5dz .
C




4.6
6. Evaluate ˜ fzdz where fz  z « 2 z and C is the path from z  0 to z  1 « 2i
C
consisting of the line segment from 0 to 1 together with the segment from 1 to 1 « 2i.


4.3 Antiderivatives. Suppose D is a subset of the reals and  : D ‚ C is differentiable at t.
Suppose further that g is differentiable at t. Then let™s see about the derivative of the
composition gt. It is, in fact, exactly what one would guess. First,


gt  uxt, yt « ivxt, yt,

where gz  ux, y « ivx, y and t  xt « iyt. Then,


d gt  u dx « u dy « i v dx « v dy .
x dt y dt x dt y dt
dt

The places at which the functions on the right-hand side of the equation are evaluated are
obvious. Now, apply the Cauchy-Riemann equations:

d gt  u dx  v dy « i v dx « u dy
x dt x dt x dt x dt
dt
dx « i dy
 u « i v
x x dt dt
 g  t  t.

The nicest result in the world!

Now, back to integrals. Let F : D ‚ C and suppose F  z  fz in D. Suppose moreover
that a and b are in D and that C  D is a contour from a to b. Then



˜ fzdz  ˜ ft  tdt,
C 



where  : ,  ‚ C describes C. From our introductory discussion, we know that
d
Ft  F  t  t  ft  t. Hence,
dt




4.7


˜ fzdz  ˜ ft  tdt
C 


˜ d Ftdt  F  F

dt


 Fb  Fa.

This is very pleasing. Note that integral depends only on the points a and b and not at all
on the path C. We say the integral is path independent. Observe that this is equivalent to
saying that the integral of f around any closed path is 0. We have thus shown that if in D
the integrand f is the derivative of a function F, then any integral ˜ fzdz for C  D is path
C
independent.

Example

1 i
Let C be the curve y  from the point z  1 « i to the point z  3 « . Let™s find
9
x2



˜ z 2 dz.
C


1
This is easy”we know that F  z  z 2 , where Fz  z 3 . Thus,
3


3
˜ z 2 dz  1 1 « i  3  3 « 9
i
3
C

  260  728 i
27 2187

Now, instead of assuming f has an antiderivative, let us suppose that the integral of f
between any two points in the domain is independent of path and that f is continuous.
Assume also that every point in the domain D is an interior point of D and that D is
connected. We shall see that in this case, f has an antiderivative. To do so, let z 0 be any
point in D, and define the function F by

˜ fzdz,
Fz 
Cz

where C z is any path in D from z 0 to z. Here is important that the integral is path
independent, otherwise Fz would not be well-defined. Note also we need the assumption
that D is connected in order to be sure there always is at least one such path.


4.8
Now, for the computation of the derivative of F:


˜ fsds,
Fz « z  Fz 
L z



where L z is the line segment from z to z « z.




Next, observe that ˜ ds  z. Thus, fz  ˜ fzds, and we have
1
z
L z L z


Fz « z  Fz
˜ fs  fz ds.
 fz  1
z z
L z



Now then,

˜ fs  fz ds
1 1 |z| max|fs  fz| : sL z 

z z
L z

‚ max|fs  fz| : sL z .

We know f is continuous at z, and so lim max|fs  fz| : sL z   0. Hence,
z‚0



Fz « z  Fz
˜ fs  fz ds
1
 fz  lim
lim
z z
z‚0 z‚0
L z

 0.




4.9
In other words, F  z  fz, and so, just as promised, f has an antiderivative! Let™s
summarize what we have shown in this section:

Suppose f : D ‚ C is continuous, where D is connected and every point of D is an interior
point. Then f has an antiderivative if and only if the integral between any two points of D is
path independent.


Exercises

7. Suppose C is any curve from 0 to  « 2i. Evaluate the integral


˜ cos z dz.
2
C



8. a)Let Fz  log z, 0  arg z  2. Show that the derivative F  z  1 .
z
 7
b)Let Gz  log z,  4  arg z  4 . Show that the derivative G z  1 .

z
c)Let C 1 be a curve in the right-half plane D 1  z : Re z ‚ 0 from i to i that does not
pass through the origin. Find the integral


˜ 1 dz.
z
C1



d)Let C 2 be a curve in the left-half plane D 2  z : Re z ‚ 0 from i to i that does not
pass through the origin. Find the integral.


˜ 1 dz.
z
C2



9. Let C be the circle of radius 1 centered at 0 with the clockwise orientation. Find


˜ 1 dz.
z
C



10. a)Let Hz  z c ,   arg z  . Find the derivative H  z.
b)Let Kz  z c ,    arg z  7 . What is the largest subset of the plane on which
4 4
Hz  Kz?
c)Let C be any path from 1 to 1 that lies completely in the upper half-plane. (Upper


4.10
half-plane  z : Im z ‚ 0.) Find


˜ Fzdz,
C



where Fz  z i ,   arg z ‚ .

11. Suppose P is a polynomial and C is a closed curve. Explain how you know that
˜ Pzdz  0.
C




4.11
Chapter Five

Cauchy™s Theorem

5.1. Homotopy. Suppose D is a connected subset of the plane such that every point of D is
an interior point”we call such a set a region”and let C 1 and C 2 be oriented closed curves
in D. We say C 1 is homotopic to C 2 in D if there is a continuous function H : S ‚ D,
where S is the square S  t, s : 0 ‚ s, t ‚ 1, such that Ht, 0 describes C 1 and Ht, 1
describes C 2 , and for each fixed s, the function Ht, s describes a closed curve C s in D.
The function H is called a homotopy between C 1 and C 2 . Note that if C 1 is homotopic to
C 2 in D, then C 2 is homotopic to C 1 in D. Just observe that the function
Kt, s  Ht, 1  s is a homotopy.

It is convenient to consider a point to be a closed curve. The point c is a described by a
constant function t  c. We thus speak of a closed curve C being homotopic to a
constant”we sometimes say C is contractible to a point.

Emotionally, the fact that two closed curves are homotopic in D means that one can be
continuously deformed into the other in D.




Example

Let D be the annular region D  z : 1  |z|  5. Suppose C 1 is the circle described by
 1 t  2e i2t , 0 ‚ t ‚ 1; and C 2 is the circle described by  2 t  4e i2t , 0 ‚ t ‚ 1. Then
Ht, s  2 « 2se i2t is a homotopy in D between C 1 and C 2 . Suppose C 3 is the same
circle as C 2 but with the opposite orientation; that is, a description is given by
 3 t  4e i2t , 0 ‚ t ‚ 1. A homotopy between C 1 and C 3 is not too easy to construct”in
fact, it is not possible! The moral: orientation counts. From now on, the term ”closed
curve” will mean an oriented closed curve.




5.1
Another Example

Let D be the set obtained by removing the point z  0 from the plane. Take a look at the
picture. Meditate on it and convince yourself that C and K are homotopic in D, but  and 
are homotopic in D, while K and  are not homotopic in D.




Exercises

1. Suppose C 1 is homotopic to C 2 in D, and C 2 is homotopic to C 3 in D. Prove that C 1 is
homotopic to C 3 in D.

2. Explain how you know that any two closed curves in the plane C are homotopic in C.

3. A region D is said to be simply connected if every closed curve in D is contractible to a
point in D. Prove that any two closed curves in a simply connected region are homotopic in
D.


5.2 Cauchy™s Theorem. Suppose C 1 and C 2 are closed curves in a region D that are
homotopic in D, and suppose f is a function analytic on D. Let Ht, s be a homotopy
between C 1 and C 2 . For each s, the function  s t describes a closed curve C s in D. Let
Is be given by


˜ fzdz.
Is 
Cs



Then,




5.2
1

˜ fHt, s Ht, s dt.
Is 
t
0



Now let™s look at the derivative of Is. We assume everything is nice enough to allow us
to differentiate under the integral:

1

˜ fHt, s Ht, s dt
I  s  d
t
ds
0
1
 2 Ht, s
Ht, s Ht, s
˜ 
 f Ht, s « fHt, s dt
s t st
0
1
 2 Ht, s
Ht, s Ht, s
˜ 
 f Ht, s « fHt, s dt
s t ts
0
1
 fHt, s Ht, s dt
˜

t s
0
H1, s H0, s
 fH1, s  fH0, s .
s s
But we know each Ht, s describes a closed curve, and so H0, s  H1, s for all s. Thus,

H1, s H0, s
I  s  fH1, s  fH0, s  0,
s s

which means Is is constant! In particular, I0  I1, or


˜ fzdz  ˜ fzdz.
C1 C2



This is a big deal. We have shown that if C 1 and C 2 are closed curves in a region D that are
homotopic in D, and f is analytic on D, then ˜ fzdz  ˜ fzdz.
C1 C2


An easy corollary of this result is the celebrated Cauchy™s Theorem, which says that if f is
analytic on a simply connected region D, then for any closed curve C in D,




5.3
˜ fzdz  0.
C



In court testimony, one is admonished to tell the truth, the whole truth, and nothing but the
truth. Well, so far in this chapter, we have told the truth and nothing but the truth, but we
have not quite told the whole truth. We assumed all sorts of continuous derivatives in the
preceding discussion. These are not always necessary”specifically, the results can be
proved true without all our smoothness assumptions”think about approximation.


Example

Look at the picture below and convince your self that the path C is homotopic to the closed
path consisting of the two curves C 1 and C 2 together with the line L. We traverse the line
twice, once from C 1 to C 2 and once from C 2 to C 1 .




Observe then that an integral over this closed path is simply the sum of the integrals over
C 1 and C 2 , since the two integrals along L , being in opposite directions, would sum to
zero. Thus, if f is analytic in the region bounded by these curves (the region with two holes
in it), then we know that


˜ fzdz  ˜ fzdz « ˜ fzdz.
C C1 C2



Exercises

4. Prove Cauchy™s Theorem.

5. Let S be the square with sides x  ‚±100, and y  ‚±100 with the counterclockwise
orientation. Find


5.4
˜ 1 dz.
z
S



6. a)Find ˜ 1
dz, where C is any circle centered at z  1 with the usual counterclockwise
z1
C
orientation: t  1 « Ae 2it , 0 ‚ t ‚ 1.
b)Find ˜ z«1 dz, where C is any circle centered at z  1 with the usual counterclockwise
1

C
orientation.

c)Find ˜ 1
dz, where C is the ellipse 4x 2 « y 2  100 with the counterclockwise
z 2 1
C
orientation. [Hint: partial fractions]

d)Find ˜ 1
dz, where C is the circle x 2  10x « y 2  0 with the counterclockwise
z 2 1
C
orientation.

8. Evaluate ˜Log z « 3dz, where C is the circle |z|  2 oriented counterclockwise.
C


9. Evaluate ˜ 1
dz where C is the circle described by t  e 2it , 0 ‚ t ‚ 1, and n is an
zn
C
integer ‚® 1.

10. a)Does the function fz  1 have an antiderivative on the set of all z ‚® 0? Explain.
z
1
b)How about fz  z n , n an integer ‚® 1 ?

2
11. Find as large a set D as you can so that the function fz  e z have an antiderivative
on D.

12. Explain how you know that every function analytic in a simply connected (cf. Exercise
3) region D is the derivative of a function analytic in D.




5.5
Chapter Six

More Integration

6.1. Cauchy™s Integral Formula. Suppose f is analytic in a region containing a simple
closed contour C with the usual positive orientation and its inside , and suppose z 0 is inside
C. Then it turns out that

fz
˜
1
fz 0   z  z 0 dz.
2i
C



This is the famous Cauchy Integral Formula. Let™s see why it™s true.

Let   0 be any positive number. We know that f is continuous at z 0 and so there is a
number  such that |fz  fz 0 |   whenever |z  z 0 |  . Now let   0 be a number
such that    and the circle C 0  z : |z  z 0 |   is also inside C. Now, the function
fz
zz 0 is analytic in the region between C and C 0 ; thus



fz fz
˜ ˜
z  z 0 dz  z  z 0 dz.
C C0



We know that ˜ 1
dz  2i, so we can write
zz 0
C0



fz fz
˜ ˜ ˜ 1
z  z 0 dz  2ifz 0   z  z 0 dz  fz 0  z  z 0 dz
C0 C0 C0

fz  fz 0 
˜
 z  z 0 dz.
C0



For zC 0 we have
|fz  fz 0 |
fz  fz 0 

z  z0 |z  z 0 |

‚ .


Thus,




6.1
fz  fz 0 
fz
˜ ˜
z  z 0 dz  2ifz 0   z  z 0 dz
C0 C0

‚  2  2.


But  is any positive number, and so


fz
˜ z  z 0 dz  2ifz 0   0,
C0

or,
fz fz
˜ ˜
1 dz  1
fz 0   z  z 0 dz,
z  z0
2i 2i
C0 C



which is exactly what we set out to show.

Meditate on this result. It says that if f is analytic on and inside a simple closed curve and
we know the values fz for every z on the simple closed curve, then we know the value for
the function at every point inside the curve”quite remarkable indeed.

Example

Let C be the circle |z|  4 traversed once in the counterclockwise direction. Let™s evaluate
the integral


˜ cos z dz.
2
z  6z « 5
C

We simply write the integrand as

fz
cos z cos z
  ,
z1
z  5z  1
z 2  6z « 5
where
fz  cos z .
z5
Observe that f is analytic on and inside C, and so,




6.2
fz
˜ ˜
cos z dz  dz  2if1
z1
2
z  6z « 5
C C

 2i cos 1   i cos 1
15 2

Exercises

1. Suppose f and g are analytic on and inside the simple closed curve C, and suppose
moreover that fz  gz for all z on C. Prove that fz  gz for all z inside C.

2. Let C be the ellipse 9x 2 « 4y 2  36 traversed once in the counterclockwise direction.
Define the function g by


˜ s 2 « s « 1 ds.
gz  sz
C

Find a) gi b) g4i

3. Find


˜ e 2z dz,
z2  4
C

where C is the closed curve in the picture:




4. Find ˜ e 2z
dz, where  is the contour in the picture:
z 2 4





6.3
6.2. Functions defined by integrals. Suppose C is a curve (not necessarily a simple closed
curve, just a curve) and suppose the function g is continuous on C (not necessarily analytic,
just continuous). Let the function G be defined by

gs
˜
Gz  s  z ds
C

for all z  C. We shall show that G is analytic. Here we go.

Consider,
Gz « z  Gz
˜
1 1 1
 s  z gsds
z z s  z  z
C
gs
˜
 ds.
s  z  zs  z
C



Next,

Gz « z  Gz gs
˜ ˜ 1 1
ds   gsds
z s  z  zs  z
s  z 2 s  z 2
C C

s  z  s  z  z
˜
 gsds
s  z  zs  z 2
C
gs
 z ˜ ds.
s  z  zs  z 2
C



Now we want to show that




6.4
gs
z ˜  0.
lim ds
s  z  zs  z 2
z‚0
C



To that end, let M  max|gs| : s µ C, and let d be the shortest distance from z to C.
Thus, for s µ C, we have |s  z| ‚ d  0 and also
|s  z  z| ‚ |s  z|  |z| ‚ d  |z|.

Putting this all together, we can estimate the integrand above:

gs M

s  z  zs  z 2 d  |z|d 2
for all s µ C. Finally,


gs
z ˜ M
ds ‚ |z| lengthC,
2
d  |z|d 2
s  z  zs  z
C



and it is clear that


gs
z ˜  0,
lim ds
s  z  zs  z 2
z‚0
C



just as we set out to show. Hence G has a derivative at z, and

gs
˜
G  z  ds.
s  z 2
C



Truly a miracle!

Next we see that G  has a derivative and it is just what you think it should be. Consider




6.5
G  z « z  G  z
˜
1 1 1
 gsds
z z s  z  z 2 s  z 2
C

s  z 2  s  z  z 2
˜
1 gsds
z s  z  z 2 s  z 2
C

2s  zz  z 2
˜
1 gsds
z s  z  z 2 s  z 2
C

2s  z  z
˜
 gsds
s  z  z 2 s  z 2
C

Next,

G  z « z  G  z gs
 2˜ ds
z s  z 3
C

2s  z  z
˜ 2
  gsds
2 2
s  z 3
s  z  z s  z
C

2s  z 2  zs  z  2s  z  z 2
˜
 gsds
s  z  z 2 s  z 3
C

2s  z 2  zs  z  2s  z 2 « 4zs  z  2z 2
˜
 gsds
s  z  z 2 s  z 3
C

3zs  z  2z 2
˜
 gsds
s  z  z 2 s  z 3
C

Hence,


G  z « z  G  z 3zs  z  2z 2
gs
 2˜ ˜
ds  gsds
z s  z 3 s  z  z 2 s  z 3
C C
|3m| « 2|z|M
‚ |z| ,
d  z 2 d 3

where m  max|s  z| : s µ C. It should be clear then that


G  z « z  G  z gs
 2˜ ds  0,
lim
z s  z 3
z‚0
C



or in other words,



6.6
gs
G  z  2 ˜ ds.
s  z 3
C



Suppose f is analytic in a region D and suppose C is a positively oriented simple closed
curve in D. Suppose also the inside of C is in D. Then from the Cauchy Integral formula,
we know that

fs
˜
2ifz  s  z ds
C



and so with g  f in the formulas just derived, we have

fs fs
˜ ˜
1 ds, and f  z  2
f  z  ds
2i 2i
2
s  z 3
s  z
C C



for all z inside the closed curve C. Meditate on these results. They say that the derivative
of an analytic function is also analytic. Now suppose f is continuous on a domain D in
which every point of D is an interior point and suppose that ˜ fzdz  0 for every closed
C
curve in D. Then we know that f has an antiderivative in D”in other words f is the
derivative of an analytic function. We now know this means that f is itself analytic. We
thus have the celebrated Morera™s Theorem:

If f:D ‚ C is continuous and such that ˜ fzdz  0 for every closed curve in D, then f is
C
analytic in D.

Example

Let™s evaluate the integral

˜ e z dz,
z3
C

where C is any positively oriented closed curve around the origin. We simply use the
equation

fs
˜
2
f  z  ds
2i s  z 3
C




6.7
with z  0 and fs  e s . Thus,


˜ e z dz.
ie 0  i 
z3
C



Exercises

5. Evaluate

˜ sin z dz
z2
C

where C is a positively oriented closed curve around the origin.

6. Let C be the circle |z  i|  2 with the positive orientation. Evaluate

a) ˜ b) ˜
1 1
dz dz
z 2 «4 z 2 «4 2
C C


7. Suppose f is analytic inside and on the simple closed curve C. Show that

f  z fz
˜ ˜
z  w dz  dz
z  w 2
C C



for every w  C.

8. a) Let  be a real constant, and let C be the circle t  e it ,  ‚ t ‚ . Evaluate

e z dz.
˜ z
C

b) Use your answer in part a) to show that



˜ e  cos t cos sin tdt  .
0



6.3. Liouville™s Theorem. Suppose f is entire and bounded; that is, f is analytic in the
entire plane and there is a constant M such that |fz| ‚ M for all z. Then it must be true
that f  z  0 identically. To see this, suppose that f  w ‚® 0 for some w. Choose R large
enough to insure that M  |f  w|. Now let C be a circle centered at 0 and with radius
R



6.8
  maxR, |w|. Then we have :


fs
˜
M  |f  w| ‚ 1 dz
 2i s  w 2
C

‚ 1 M 2  M ,

2  2


a contradiction. It must therefore be true that there is no w for which f  w ‚® 0; or, in other
words, f  z  0 for all z. This, of course, means that f is a constant function. What we
have shown has a name, Liouville™s Theorem:

The only bounded entire functions are the constant functions.

Let™s put this theorem to some good use. Let pz  a n z n « a n1 z n1 «µ «a 1 z « a 0 be a
polynomial. Then


pz  a n « a n1 « a n2 «µ « an0
zn.
z z
2
z

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