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. 3
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a |a n |
Now choose R large enough to insure that for each j  1, 2, µ , n, we have znj  2n
j

whenever |z|  R. (We are assuming that a n ‚® 0. ) Hence, for |z|  R, we know that

a0
a a
|pz| ‚ |a n |  n1 « n2 «µ « n |z| n
z z
2
z
‚ |a n |  a n1  a n2 µ  an |z| n
0
z z
2
z
|a | |a | |a |
 |a n |  n  n µ  n |z| n
2n 2n 2n
|a |
 n |z| n .
2

Hence, for |z|  R,

1 2 2.
n‚
|a n |R n
|a n ||z|
pz

1 1
Now suppose pz ‚® 0 for all z. Then pz is also bounded on the disk |z| ‚ R. Thus, pz
is a bounded entire function, and hence, by Liouville™s Theorem, constant! Hence the
polynomial is constant if it has no zeros. In other words, if pz is of degree at least one,
there must be at least one z 0 for which pz 0   0. This is, of course, the celebrated


6.9
Fundamental Theorem of Algebra.

Exercises

9. Suppose f is an entire function, and suppose there is an M such that Re fz ‚ M for all
z. Prove that f is a constant function.

10. Suppose w is a solution of 5z 4 « z 3 « z 2  7z « 14  0. Prove that |w| ‚ 3.

11. Prove that if p is a polynomial of degree n, and if pa  0, then pz  z  aqz,
where q is a polynomial of degree n  1.

12. Prove that if p is a polynomial of degree n ‚ 1, then

pz  cz  z 1  k 1 z  z 2  k 2 µ z  z j  k j ,

where k 1 , k 2 , µ , k j are positive integers such that n  k 1 « k 2 «µ «k j .

13. Suppose p is a polynomial with real coefficients. Prove that p can be expressed as a
product of linear and quadratic factors, each with real coefficients.


6.4. Maximum moduli. Suppose f is analytic on a closed domain D. Then, being
continuous, |fz| must attain its maximum value somewhere in this domain. Suppose this
happens at an interior point. That is, suppose |fz| ‚ M for all z µ D and suppose that
|fz 0 |  M for some z 0 in the interior of D. Now z 0 is an interior point of D, so there is a
number R such that the disk  centered at z 0 having radius R is included in D. Let C be a
positively oriented circle of radius  ‚ R centered at z 0 . From Cauchy™s formula, we
know

fs
˜
1
fz 0   s  z 0 ds.
2i
C

Hence,

2

˜ fz 0 « e it dt,
fz 0   1
2
0

and so,




6.10
2

˜ |fz 0 « e it |dt ‚ M.
M  |fz 0 | ‚ 1
2
0

since |fz 0 « e it | ‚ M. This means

2

˜ |fz 0 « e it |dt.
M 1
2
0



Thus,

2 2

˜ |fz 0 « e it |dt  ˜ M  |fz 0 « e it | dt  0.
M 1 1
2 2
0 0



This integrand is continuous and non-negative, and so must be zero. In other words,
|fz|  M for all z µ C. There was nothing special about C except its radius  ‚ R, and so
we have shown that f must be constant on the disk .

I hope it is easy to see that if D is a region (connected and open), then the only way in
which the modulus |fz| of the analytic function f can attain a maximum on D is for f to be
constant.

Exercises

14. Suppose f is analytic and not constant on a region D and suppose fz ‚® 0 for all z µ D.
Explain why |fz| does not have a minimum in D.

15. Suppose fz  ux, y « ivx, y is analytic on a region D. Prove that if ux, y attains a
maximum value in D, then u must be constant.




6.11
Chapter Seven

Harmonic Functions

7.1. The Laplace equation. The Fourier law of heat conduction says that the rate at which
heat passes across a surface S is proportional to the flux, or surface integral, of the
temperature gradient on the surface:


k ˜˜ T  dA.
S



Here k is the constant of proportionality, generally called the thermal conductivity of the
substance (We assume uniform stuff. ). We further assume no heat sources or sinks, and we
assume steady-state conditions”the temperature does not depend on time. Now if we take
S to be an arbitrary closed surface, then this rate of flow must be 0:


k ˜˜ T  dA  0.
S



Otherwise there would be more heat entering the region B bounded by S than is coming
out, or vice-versa. Now, apply the celebrated Divergence Theorem to conclude that


˜˜˜  TdV  0,
B



where B is the region bounded by the closed surface S. But since the region B is completely
arbitrary, this means that

2 2 2
  T   T «  T «  T  0.
x 2 y 2 z 2

This is the world-famous Laplace Equation.

Now consider a slab of heat conducting material,




7.1
in which we assume there is no heat flow in the z-direction. Equivalently, we could assume
we are looking at the cross-section of a long rod in which there is no longitudinal heat
flow. In other words, we are looking at a two-dimensional problem”the temperature
depends only on x and y, and satisfies the two-dimensional version of the Laplace equation:

 2 T «  2 T  0.
x 2 y 2

Suppose now, for instance, the temperature is specified on the boundary of our region D,
and we wish to find the temperature Tx, y in region. We are simply looking for a solution
of the Laplace equation that satisfies the specified boundary condition.

Let™s look at another physical problem which leads to Laplace™s equation. Gauss™s Law of
electrostatics tells us that the integral over a closed surface S of the electric field E is
proportional to the charge included in the region B enclosed by S. Thus in the absence of
any charge, we have


˜˜ E dA  0.
S



But in this case, we know the field E is conservative; let ¤ be the potential function”that
is,

E  .

Thus,


˜˜ E dA  ˜˜  dA.
S S



Again, we call on the Divergence Theorem to conclude that  must satisfy the Laplace
equation. Mathematically, we cannot tell the problem of finding the electric potential in a


7.2
region D, given the potential on the boundary of D, from the previous problem of finding
the temperature in the region, given the temperature on the boundary. These are but two of
the many physical problems that lead to the Laplace equation”You probably already know
of some others. Let D be a domain and let  be a given function continuous on the
boundary of D. The problem of finding a function  harmonic on the interior of D and
which agrees with  on the boundary of D is called the Dirichlet problem.

7.2. Harmonic functions. If D is a region in the plane, a real-valued function ux, y
having continuous second partial derivatives is said to be harmonic on D if it satisfies
Laplace™s equation on D :

 2 u «  2 u  0.
x 2 y 2

There is an intimate relationship between harmonic functions and analytic functions.
Suppose f is analytic on D, and let fz  ux, y « ivx, y. Now, from the Cauchy-Riemann
equations, we know

u  v , and
x y
u   v .
y x

If we differentiate the first of these with respect to x and the second with respect to y, and
then add the two results, we have

 2 u «  2 u   2 v   2 v  0.
xy yx
x 2 y 2

Thus the real part of any analytic function is harmonic! Next, if we differentiate the first of
the Cauchy-Riemann equations with respect to y and the second with respect to x, and then
subtract the second from the first, we have

 2 v «  2 v  0,
x 2 y 2

and we see that the imaginary part of an analytic function is also harmonic.

There is even more excitement. Suppose we are given a function  harmonic in a simply
connected region D. Then there is a function f analytic on D which is such that Re f  .
Let™s see why this is so. First, define g by


7.3
 
gz  i .
x y

We™ll show that g is analytic by verifying that the real and imaginary parts satisfy the
Cauchy-Riemann equations:

2 2
 
  2   
 ,
x x y y
x 2 y

since  is harmonic. Next,

2 2
 
   
  .
y x yx xy x y

Since g is analytic on the simply connected region D, we know that the integral of g around
any closed curve is zero, and so it has an antiderivative Gz  u « iv. This antiderivative
is, of course, analytic on D, and we know that

 
G  z  u  i u  i .
x y x y

Thus, ux, y  x, y « hy. From this,

u   « h  y,
y y

and so h  y  0, or h  constant, from which it follows that ux, y  x, y « c. In other
words, Re G  u, as we promised to show.

Example

The function x, y  x 3  3xy 2 is harmonic everywhere. We shall find an analytic
function G so that Re G  . We know that Gz  x 3  3xy 2  « iv, and so from the
Cauchy-Riemann equations:

v   u  6xy
x y




7.4
Hence,


vx, y  3x 2 y « ky.

To find ky differentiate with respect to y :

v  3x 2 « k  y  u  3x 2  3y 2 ,
y x

and so,

k  y  3y 2 , or
ky  y 3 « any constant.

If we choose the constant to be zero, this gives us

v  3x 2 y « ky  3x 2 y  y 3 ,

and finally,


Gz  u « iv  x 3  3xy 2  « i3x 2 y  y 3 .



Exercises

1. Suppose  is harmonic on a simply connected region D. Prove that if  assumes its
maximum or its minimum value at some point in D, then  is constant in D.

2. Suppose  and  are harmonic in a simply connected region D bounded by the curve C.
Suppose moreover that x, y  x, y for all x, y µ C. Explain how you know that
   everywhere in D.

3. Find an entire function f such that Re f  x 2  3x  y 2 , or explain why there is no such
function f.

4. Find an entire function f such that Re f  x 2 « 3x  y 2 , or explain why there is no such
function f.



7.5
7.3. Poisson™s integral formula. Let  be the disk bounded by the circle
C   z : |z|  . Suppose  is harmonic on  and let f be a function analytic on  and
such that Re f  . Now then, for fixed z with |z|  , the function

fs z
gs 
2  s z

is analyic on . Thus from Cauchy™s Theorem

fs z
˜ gsds  ˜ ds  0.
2  s z
C C

We know also that

fs
˜
1
fz  s  z ds.
2i
C



Adding these two equations gives us


˜
1 1 z
fz  s  z « 2  s z fsds
2i
C

 2  |z| 2
˜
1 fsds.
2i s  z 2  s z 
C



Next, let t  e it , and our integral becomes

2
 2  |z| 2
˜
fz  1 fe it ie it dt
2i it 2 it
e  z  e z 
0
2
2
2
fe it 
  |z|
˜
 dt
2 e it  ze it  z 
0
2
2
2
fe it 
  |z|
˜
 dt
2 2
it
|e  z|
0



Now,


7.6
2
2
2
e it 
  |z|
˜
x, y  Re f  dt.
2 2
it
|e  z|
0



Next, use polar coordinates: z  re i“ :

2
2  r2 e it 
˜
r, “  dt.
2 e it  re i“ | 2
|
0



Now,
2
|e it  re i“ |  e it  re i“ e it  re i“    2 « r 2  re it“ « e it“ 
  2 « r 2  2r cost  “.

Substituting this in the integral, we have Poisson™s integral formula:

2
2 2
e it 
 r
˜
r, “  dt
2  2 « r 2  2r cost  “
0



This famous formula essentially solves the Dirichlet problem for a disk.

Exercises

2
5. Evaluate ˜ 1
dt. [Hint: This is easy.]
 2 «r 2 2r cost“
0


6. Suppose  is harmonic in a region D. If x 0 , y 0  µ D and if C  D is a circle centered at
x 0 , y 0 , the inside of which is also in D, then x 0 , y 0  is the average value of  on the
circle C.

7. Suppose  is harmonic on the disk   z : |z| ‚ . Prove that


˜˜ dA.
1
0, 0 
 2





7.7
Chapter Eight

Series

8.1. Sequences. The basic definitions for complex sequences and series are essentially the
same as for the real case. A sequence of complex numbers is a function g : Z « ‚ C from
the positive integers into the complex numbers. It is traditional to use subscripts to indicate
the values of the function. Thus we write gn ‚ z n and an explicit name for the sequence
is seldom used; we write simply z n  to stand for the sequence g which is such that
i i
gn  z n . For example,  n  is the sequence g for which gn  n .

The number L is a limit of the sequence z n  if given an   0, there is an integer N  such
that |z n  L|   for all n ‚ N  . If L is a limit of z n , we sometimes say that z n 
converges to L. We frequently write limz n   L. It is relatively easy to see that if the
complex sequence z n   u n « iv n  converges to L, then the two real sequences u n  and
v n  each have a limit: u n  converges to Re L and v n  converges to Im L. Conversely, if
the two real sequences u n  and v n  each have a limit, then so also does the complex
sequence u n « iv n . All the usual nice properties of limits of sequences are thus true:

limz n ‚± w n   limz n  ‚± limw n ;
limz n w n   limz n  limw n ; and
limz n 
z
lim wnn  .
limw n 

provided that limz n  and limw n  exist. (And in the last equation, we must, of course,
insist that limw n  ‚® 0.)

A necessary and sufficient condition for the convergence of a sequence a n  is the
celebrated Cauchy criterion: given   0, there is an integer N  so that |a n  a m |  
whenever n, m  N  .

A sequence f n  of functions on a domain D is the obvious thing: a function from the
positive integers into the set of complex functions on D. Thus, for each zD, we have an
ordinary sequence f n z. If each of the sequences f n z converges, then we say the
sequence of functions f n  converges to the function f defined by fz  limf n z. This
pretty obvious stuff. The sequence f n  is said to converge to f uniformly on a set S if
given an   0, there is an integer N  so that |f n z  fz|   for all n ‚ N  and all z µ S.

Note that it is possible for a sequence of continuous functions to have a limit function that
is not continuous. This cannot happen if the convergence is uniform. To see this, suppose
the sequence f n  of continuous functions converges uniformly to f on a domain D, let
z 0 D, and let   0. We need to show there is a  so that |fz 0   fz|   whenever


8.1

|z 0  z|  . Let™s do it. First, choose N so that |f N z  fz|  3 . We can do this because
of the uniform convergence of the sequence f n . Next, choose  so that

|f N z 0   f N z|  3 whenever |z 0  z|  . This is possible because f N is continuous.
Now then, when |z 0  z|  , we have

|fz 0   fz|  |fz 0   f N z 0  « f N z 0   f N z « f N z  fz|
‚ |fz 0   f N z 0 | « |f N z 0   f N z| « |f N z  fz|
  «  «   ,
3 3 3

and we have done it!

Now suppose we have a sequence f n  of continuous functions which converges uniformly
˜ f n zdz converges to ˜ fzdz. This
on a contour C to the function f. Then the sequence
C C

is easy to see. Let   0. Now let N be so that |f n z  fz|  for n  N, where A is the
A
length of C. Then,



˜ f n zdz  ˜ fzdz ˜f n z  fzdz

C C C

 A
A

whenever n  N.

Now suppose f n  is a sequence of functions each analytic on some region D, and suppose
the sequence converges uniformly on D to the function f. Then f is analytic. This result is in
marked contrast to what happens with real functions”examples of uniformly convergent
sequences of differentiable functions with a nondifferentiable limit abound in the real case.
To see that this uniform limit is analytic, let z 0 D, and let S  z : |z  z 0 |  r  D . Now
consider any simple closed curve C  S. Each f n is analytic, and so ˜ f n zdz  0 for every
C
n. From the uniform convergence of f n  , we know that ˜ fzdz is the limit of the sequence
C

˜ f n zdz , and so ˜ fzdz  0. Morera™s theorem now tells us that f is analytic on S, and
C C
hence at z 0 . Truly a miracle.


Exercises



8.2
1. Prove that a sequence cannot have more than one limit. (We thus speak of the limit of a
sequence.)

2. Give an example of a sequence that does not have a limit, or explain carefully why there
is no such sequence.

3. Give an example of a bounded sequence that does not have a limit, or explain carefully
why there is no such sequence.

4. Give a sequence f n  of functions continuous on a set D with a limit that is not
continuous.

5. Give a sequence of real functions differentiable on an interval which converges
uniformly to a nondifferentiable function.


8.2 Series. A series is simply a sequence s n  in which s n  a 1 « a 2 «µ «a n . In other
words, there is sequence a n  so that s n  s n1 « a n . The s n are usually called the partial
n
 a j has a limit, then it must be
sums. Recall from Mrs. Turner™s class that if the series
j1
true that lim a n   0.
n‚


n
 f j z of functions. Chances are this series will converge for some
Consider a series
j1
values of z and not converge for others. A useful result is the celebrated Weierstrass
M-test: Suppose M j  is a sequence of real numbers such that M j ‚ 0 for all j  J, where
n
 M j converges. If for all zD, we
J is some number., and suppose also that the series
j1
n
 f j z converges uniformly on D.
have |f j z| ‚ M j for all j  J, then the series
j1


To prove this, begin by letting   0 and choosing N  J so that

n

 Mj  
jm



for all n, m  N. (We can do this because of the famous Cauchy criterion.) Next, observe
that




8.3
n n n

 f j z |f j z| ‚  M j  .

jm jm jm


n
 f j z converges. To see the uniform convergence, observe that
This shows that
j1


n n m1

 f j z  f j z   f j z
 
jm j0 j0



for all zD and n  m  N. Thus,


n m1 m1

 f j z   f j z  f j z   f j z
 ‚
lim
n‚
j0 j0 j0 j0



n
 a j is almost always written as  a j .)
for m  N.(The limit of a series
j0 j0




Exercises

zn
6. Find the set D of all z for which the sequence has a limit. Find the limit.
n 3 n
z


n n
 aj  Re a j
7. Prove that the series convegres if and only if both the series and
j1 j1
n
 Im a j converge.
j1


n
 1  j
8. Explain how you know that the series converges uniformly on the set
z
j1
|z| ‚ 5.


8.3 Power series. We are particularly interested in series of functions in which the partial
sums are polynomials of increasing degree:

s n z  c 0 « c 1 z  z 0  « c 2 z  z 0  2 «µ «c n z  z 0  n .


8.4
(We start with n  0 for esthetic reasons.) These are the so-called power series. Thus,
n
 c j z  z 0  j .
a power series is a series of functions of the form
j0


Let™s look first as a very special power series, the so-called Geometric series:

n

 zj .
j0



Here
s n  1 « z « z 2 «µ «z n , and
zs n  z « z 2 « z 3 «µ «z n«1 .

Subtracting the second of these from the first gives us


1  zs n  1  z n«1 .

If z  1, then we can™t go any further with this, but I hope it™s clear that the series does not
have a limit in case z  1. Suppose now z ‚® 1. Then we have

1  z n«1 .
sn 
1z 1z

Now if |z|  1, it should be clear that limz n«1   0, and so

n

 zj 1.
 lim s n 
lim
1z
j0



Or,



 zj  1 , for |z|  1.
1z
j0



There is a bit more to the story. First, note that if |z|  1, then the Geometric series does
not have a limit (why?). Next, note that if |z| ‚   1, then the Geometric series converges


8.5
1
uniformly to . To see this, note that
1z


n

 j
j0



has a limit and appeal to the Weierstrass M-test.

Clearly a power series will have a limit for some values of z and perhaps not for others.
First, note that any power series has a limit when z  z 0 . Let™s see what else we can say.
n
 c j z  z 0  j . Let
Consider a power series
j0

–  lim sup |c j | .
j




1
(Recall from 6 th grade that lim supa k   limsupa k : k ‚ n. ) Now let R  – . (We
shall say R  0 if –  , and R   if –  0. ) We are going to show that the series
converges uniformly for all |z  z 0 | ‚   R and diverges for all |z  z 0 |  R.

First, let™s show the series does not converge for |z  z 0 |  R. To begin, let k be so that

1  k  1  –.
R
|z  z 0 |

There are an infinite number of c j for which j |c j |  k, otherwise lim sup ‚ k. For
|c j |
j


each of these c j we have

j
 k|z  z 0 | j  1.
|c j z  z 0  j |  |c j | |z  z 0 |
j




It is thus not possible for lim |c n z  z 0  n |  0, and so the series does not converge.
n‚


Next, we show that the series does converge uniformly for |z  z 0 | ‚   R. Let k be so
that


–  1  k  .
1
R

Now, for j large enough, we have j |c j |  k. Thus for |z  z 0 | ‚ , we have



8.6
j
|c j z  z 0  j |   k|z  z 0 | j  k j .
|c j | |z  z 0 |
j




n
k j converges because k  1 and the uniform convergence
The geometric series
j0
n
 c j z  z 0  j follows from the M-test.
of
j0



Example

n
 j! z j . Let™s compute R  1/ lim sup
1
 lim sup j j! . Let
Consider the series |c j |
j

j0
K 2K
K be any positive integer and choose an integer m large enough to insure that 2 m  .
2K!
n!
Now consider K n , where n  2K « m:


n!  2K « m!  2K « m2K « m  1µ 2K « 12K!
Kn K 2K«m K m K 2K
2K!
 2 m 2K  1
K

Thus n n!  K. Reflect on what we have just shown: given any number K, there is a
number n such that n n! is bigger than it. In other words, R  lim sup j j!   , and so the
n
 j! z j converges for all z.
1
series
j0


n
 c j z  z 0  j , there is a number
Let™s summarize what we have. For any power series
j0
1
R such that the series converges uniformly for |z  z 0 | ‚   R and does not
lim sup j |c j |
converge for |z  z 0 |  R. (Note that we may have R  0 or R  .) The number R is
called the radius of convergence of the series, and the set |z  z 0 |  R is called the circle
of convergence. Observe also that the limit of a power series is a function analytic inside
the circle of convergence (why?).

Exercises

9. Suppose the sequence of real numbers  j  has a limit. Prove that




8.7
lim sup j   lim j .

For each of the following, find the set D of points at which the series converges:

n
 j!z j .
10.
j0


n
 jz j .
11.
j0


n
j2
 zj .
12. 3j
j0


n
1 j
 z 2j
13. 2 2j j! 2
j0




8.4 Integration of power series. Inside the circle of convergence, the limit



 c j z  z 0  j
Sz 
j0



is an analytic function. We shall show that this series may be integrated
”term-by-term””that is, the integral of the limit is the limit of the integrals. Specifically, if
C is any contour inside the circle of convergence, and the function g is continuous on C,
then


˜ gzSzdz   c j ˜ gzz  z 0  j dz.
j0
C C



Let™s see why this. First, let   0. Let M be the maximum of |gz| on C and let L be the
length of C. Then there is an integer N so that



 c j z  z 0  j 
ML
jn




8.8
for all n  N. Thus,


˜ gz  c j z  z 0  j dz  ML   ,
ML
jn
C



Hence,


n1
˜ gzSzdz   c j ˜ gzz  z 0  j dz ˜ gz  c j z  z 0  j dz

j0 jn
C C C

 ,

and we have shown what we promised.


8.5 Differentiation of power series. Again, let




 c j z  z 0  j .
Sz 
j0



Now we are ready to show that inside the circle of convergence,



 jc j z  z 0  j1 .
S  z 
j1



Let z be a point inside the circle of convergence and let C be a positive oriented circle
centered at z and inside the circle of convergence. Define

1
gs  ,
2is  z 2

and apply the result of the previous section to conclude that




8.9

˜ gsSsds   c j ˜ gss  z 0  j ds, or
j0
C C

s  z 0  j
Ss
˜  cj ˜
1 ds  ds. Thus
2i s  z 2 s  z 2
j0
C C


 jc j z  z 0  j1 ,
S  z 
j0



as promised!

Exercises

14. Find the limit of

n

j « 1z j .
j0

For what values of z does the series converge?

15. Find the limit of

n
zj
 .
j
j1

For what values of z does the series converge?

n
 c j z  1 j such that
16. Find a power series
j0




 c j z  1 j , for |z  1|  1.
1
z
j0


n
 c j z  1 j such that
17. Find a power series
j0




8.10


 c j z  1 j , for |z  1|  1.
Log z 
j0




8.11
Chapter Nine

Taylor and Laurent Series

9.1. Taylor series. Suppose f is analytic on the open disk |z  z 0 |  r. Let z be any point in
this disk and choose C to be the positively oriented circle of radius , where
|z  z 0 |    r. Then for sC we have


z  z 0  j

1 1 1 1
s  z  s  z 0   z  z 0   s  z 0 
zz 0
1 s  z 0  j«1
sz 0 j0



since | zz 0 |  1. The convergence is uniform, so we may integrate
sz 0



fs fs
˜ ˜ ds z  z 0  j , or
s  z ds  j«1
s  z 0 
j0
C C


fs fs
˜ ˜

1 1 ds z  z 0  j .
fz  s  z ds 
2i 2i j«1
s  z 0 
j0
C C



We have thus produced a power series having the given analytic function as a limit:



 c j z  z 0  j , |z  z 0 |  r,
fz 
j0



where

fs
˜
1
cj  ds.
2i s  z 0  j«1
C



This is the celebrated Taylor Series for f at z  z 0 .

We know we may differentiate the series to get



 jc j z  z 0  j1
f  z 
j1




9.1
and this one converges uniformly where the series for f does. We can thus differentiate
again and again to obtain



 jj  1j  2µ j  n « 1c j z  z 0  jn .
f n z 
jn



Hence,

f n z 0   n!c n , or
f n z 0 
cn  .
n!

But we also know that

fs
˜
1
cn  ds.
2i s  z 0  n«1
C

This gives us

fs
˜
f n z 0   n! ds, for n  0, 1, 2, µ .
2i s  z 0  n«1
C



This is the famous Generalized Cauchy Integral Formula. Recall that we previously
derived this formula for n  0 and 1.

What does all this tell us about the radius of convergence of a power series? Suppose we
have



 c j z  z 0  j ,
fz 
j0



and the radius of convergence is R. Then we know, of course, that the limit function f is
analytic for |z  z 0 |  R. We showed that if f is analytic in |z  z 0 |  r, then the series
converges for |z  z 0 |  r. Thus r ‚ R, and so f cannot be analytic at any point z for which
|z  z 0 |  R. In other words, the circle of convergence is the largest circle centered at z 0
inside of which the limit f is analytic.




9.2
Example

Let fz  expz  e z . Then f0  f  0 µ  f n 0 µ  1, and the Taylor series for f
at z 0  0 is



 1 zj
ez 
j!
j0



and this is valid for all values of z since f is entire. (We also showed earlier that this
particular series has an infinite radius of convergence.)


Exercises

1. Show that for all z,


e z  e  1 z  1 j .
j!
j0


n
 c j z j for tanh z ?
2. What is the radius of convergence of the Taylor series
j0


3. Show that


z  i j

1
1z 1  i j«1
j0



for |z  i|  2.

1
, what is f 10 i ?
4. If fz  1z


5. Suppose f is analytic at z  0 and f0  f  0  f  0  0. Prove there is a function g
analytic at 0 such that fz  z 3 gz in a neighborhood of 0.

6. Find the Taylor series for fz  sin z at z 0  0.

7. Show that the function f defined by


9.3
sin z
for z ‚® 0
z
fz 
for z  0
1


is analytic at z  0, and find f  0.

9.2. Laurent series. Suppose f is analytic in the region R 1  |z  z 0 |  R 2 , and let C be a
positively oriented simple closed curve around z 0 in this region. (Note: we include the
possiblites that R 1 can be 0, and R 2  .) We shall show that for z  C in this region

 
bj
 a j z  z 0  j « 
fz  ,
j
z  z 0 
j0 j1



where
fs
˜
1
aj  ds, for j  0, 1, 2, µ
2i s  z 0  j«1
C



and

fs
˜
1
bj  ds, for j  1, 2, µ .
2i s  z 0  j«1

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