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a |a n |
Now choose R large enough to insure that for each j пЂЅ 1, 2, пЃµ , n, we have znпЂїj пЂј 2n
j

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

a0
a a
|pпѓќzпѓћ| п‚і |a n | пЂї nпЂї1 пЂ« nпЂї2 пЂ«пЃµ пЂ« n |z| n
z z
2
z
п‚і |a n | пЂї a nпЂї1 пЂї a nпЂї2 пЂїпЃµ пЂї an |z| n
0
z z
2
z
|a | |a | |a |
пЂѕ |a n | пЂї n пЂї n пЂїпЃµ пЂї n |z| n
2n 2n 2n
|a |
пЂѕ n |z| n .
2

Hence, for |z| пЂѕ R,

1пЂј 2 2.
nп‚І
|a n |R n
|a n ||z|
pпѓќzпѓћ

1 1
Now suppose pпѓќzпѓћ п‚® 0 for all z. Then pпѓќzпѓћ is also bounded on the disk |z| п‚І R. Thus, pпѓќzпѓћ
is a bounded entire function, and hence, by LiouvilleвЂ™s Theorem, constant! Hence the
polynomial is constant if it has no zeros. In other words, if pпѓќzпѓћ is of degree at least one,
there must be at least one z 0 for which pпѓќz 0 пѓћ пЂЅ 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 fпѓќzпѓћ п‚І M for all
z. Prove that f is a constant function.

10. Suppose w is a solution of 5z 4 пЂ« z 3 пЂ« z 2 пЂї 7z пЂ« 14 пЂЅ 0. Prove that |w| п‚І 3.

11. Prove that if p is a polynomial of degree n, and if pпѓќaпѓћ пЂЅ 0, then pпѓќzпѓћ пЂЅ пѓќz пЂї aпѓћqпѓќzпѓћ,
where q is a polynomial of degree n пЂї 1.

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

pпѓќzпѓћ пЂЅ cпѓќz пЂї z 1 пѓћ k 1 пѓќz пЂї z 2 пѓћ k 2 пЃµ пѓќz пЂї z j пѓћ k j ,

where k 1 , k 2 , пЃµ , k j are positive integers such that n пЂЅ k 1 пЂ« k 2 пЂ«пЃµ пЂ«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, |fпѓќzпѓћ| must attain its maximum value somewhere in this domain. Suppose this
happens at an interior point. That is, suppose |fпѓќzпѓћ| п‚І M for all z пЂµ D and suppose that
|fпѓќz 0 пѓћ| пЂЅ 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 пЃѓ centered at z 0 having radius R is included in D. Let C be a
positively oriented circle of radius пЃџ п‚І R centered at z 0 . From CauchyвЂ™s formula, we
know

fпѓќsпѓћ
пЃ˜
1
fпѓќz 0 пѓћ пЂЅ s пЂї z 0 ds.
2пЃћi
C

Hence,

2пЃћ

пЃ˜ fпѓќz 0 пЂ« пЃџe it пѓћdt,
fпѓќz 0 пѓћ пЂЅ 1
2пЃћ
0

and so,

6.10
2пЃћ

пЃ˜ |fпѓќz 0 пЂ« пЃџe it пѓћ|dt п‚І M.
M пЂЅ |fпѓќz 0 пѓћ| п‚І 1
2пЃћ
0

since |fпѓќz 0 пЂ« пЃџe it пѓћ| п‚І M. This means

2пЃћ

пЃ˜ |fпѓќz 0 пЂ« пЃџe it пѓћ|dt.
MпЂЅ 1
2пЃћ
0

Thus,

2пЃћ 2пЃћ

пЃ˜ |fпѓќz 0 пЂ« пЃџe it пѓћ|dt пЂЅ пЃ˜ пѓџM пЂї |fпѓќz 0 пЂ« пЃџe it пѓћ| пѓ dt пЂЅ 0.
MпЂї 1 1
2пЃћ 2пЃћ
0 0

This integrand is continuous and non-negative, and so must be zero. In other words,
|fпѓќzпѓћ| пЂЅ M for all z пЂµ C. There was nothing special about C except its radius пЃџ п‚І R, and so
we have shown that f must be constant on the disk пЃѓ.

I hope it is easy to see that if D is a region (пЂЅconnected and open), then the only way in
which the modulus |fпѓќzпѓћ| 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 fпѓќzпѓћ п‚® 0 for all z пЂµ D.
Explain why |fпѓќzпѓћ| does not have a minimum in D.

15. Suppose fпѓќzпѓћ пЂЅ uпѓќx, yпѓћ пЂ« ivпѓќx, yпѓћ is analytic on a region D. Prove that if uпѓќx, yпѓћ 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

k пЃ˜пЃ˜ пЂґT пЂ¶ 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 conditionsвЂ”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 пЃ˜пЃ˜ пЂґT пЂ¶ dA пЂЅ 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

пЃ˜пЃ˜пЃ˜пѓќпЂґ пЂ¶ пЂґTпѓћdV пЂЅ 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
пЂґ пЂ¶ пЂґT пЂЅ пЂЇ T пЂ« пЂЇ T пЂ« пЂЇ T пЂЅ 0.
пЂЇx 2 пЂЇy 2 пЂЇz 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 problemвЂ”the temperature
depends only on x and y, and satisfies the two-dimensional version of the Laplace equation:

пЂЇ 2 T пЂ« пЂЇ 2 T пЂЅ 0.
пЂЇx 2 пЂЇy 2

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

LetвЂ™s look at another physical problem which leads to LaplaceвЂ™s equation. GaussвЂ™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

пЃ˜пЃ˜ E пЂ¶dA пЂЅ 0.
S

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

E пЂЅ пЂґпЃЄ.

Thus,

пЃ˜пЃ˜ E пЂ¶dA пЂЅ пЃ˜пЃ˜ пЂґпЃЄ пЂ¶dA.
S S

Again, we call on the Divergence Theorem to conclude that пЃЄ 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 equationвЂ”You probably already know
of some others. Let D be a domain and let пЃЎ be a given function continuous on the
boundary of D. The problem of finding a function пЃЄ harmonic on the interior of D and
which agrees with пЃЎ 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 uпѓќx, yпѓћ
having continuous second partial derivatives is said to be harmonic on D if it satisfies
LaplaceвЂ™s equation on D :

пЂЇ 2 u пЂ« пЂЇ 2 u пЂЅ 0.
пЂЇx 2 пЂЇy 2

There is an intimate relationship between harmonic functions and analytic functions.
Suppose f is analytic on D, and let fпѓќzпѓћ пЂЅ uпѓќx, yпѓћ пЂ« ivпѓќx, yпѓћ. Now, from the Cauchy-Riemann
equations, we know

пЂЇu пЂЅ пЂЇv , and
пЂЇx пЂЇy
пЂЇu пЂЅ пЂї пЂЇv .
пЂЇy пЂЇx

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

пЂЇ 2 u пЂ« пЂЇ 2 u пЂЅ пЂЇ 2 v пЂї пЂЇ 2 v пЂЅ 0.
пЂЇxпЂЇy пЂЇyпЂЇx
пЂЇx 2 пЂЇy 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

пЂЇ 2 v пЂ« пЂЇ 2 v пЂЅ 0,
пЂЇx 2 пЂЇy 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 пЃЄ harmonic in a simply
connected region D. Then there is a function f analytic on D which is such that Re f пЂЅ пЃЄ.
LetвЂ™s see why this is so. First, define g by

7.3
пЂЇпЃЄ пЂЇпЃЄ
gпѓќzпѓћ пЂЅ пЂїi .
пЂЇx пЂЇy

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

пЂЇ2пЃЄ пЂЇ2пЃЄ
пЂЇпЃЄ пЂЇпЃЄ
пЂЇ пЂЅпЂї 2 пЂЅ пЂЇ пЂї
пЂЅ ,
пЂЇx пЂЇx пЂЇy пЂЇy
пЂЇx 2 пЂЇy

since пЃЄ is harmonic. Next,

пЂЇ2пЃЄ пЂЇ2пЃЄ
пЂЇпЃЄ пЂЇпЃЄ
пЂЇ пЂЅпЂї пЂЇ пЂї
пЂЅ пЂЅ .
пЂЇy пЂЇx пЂЇyпЂЇx пЂЇxпЂЇy пЂЇx пЂЇy

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 Gпѓќzпѓћ пЂЅ u пЂ« iv. This antiderivative
is, of course, analytic on D, and we know that

пЂЇпЃЄ пЂЇпЃЄ
G пЃ¶ пѓќzпѓћ пЂЅ пЂЇu пЂї i пЂЇu пЂЅ пЂїi .
пЂЇx пЂЇy пЂЇx пЂЇy

Thus, uпѓќx, yпѓћ пЂЅ пЃЄпѓќx, yпѓћ пЂ« hпѓќyпѓћ. From this,

пЂЇu пЂЅ пЂЇпЃЄ пЂ« h пЃ¶ пѓќyпѓћ,
пЂЇy пЂЇy

and so h пЃ¶ пѓќyпѓћ пЂЅ 0, or h пЂЅ constant, from which it follows that uпѓќx, yпѓћ пЂЅ пЃЄпѓќx, yпѓћ пЂ« c. In other
words, Re G пЂЅ u, as we promised to show.

Example

The function пЃЄпѓќx, yпѓћ пЂЅ x 3 пЂї 3xy 2 is harmonic everywhere. We shall find an analytic
function G so that Re G пЂЅ пЃЄ. We know that Gпѓќzпѓћ пЂЅ пѓќx 3 пЂї 3xy 2 пѓћ пЂ« iv, and so from the
Cauchy-Riemann equations:

пЂЇv пЂЅ пЂї пЂЇu пЂЅ 6xy
пЂЇx пЂЇy

7.4
Hence,

vпѓќx, yпѓћ пЂЅ 3x 2 y пЂ« kпѓќyпѓћ.

To find kпѓќyпѓћ differentiate with respect to y :

пЂЇv пЂЅ 3x 2 пЂ« k пЃ¶ пѓќyпѓћ пЂЅ пЂЇu пЂЅ 3x 2 пЂї 3y 2 ,
пЂЇy пЂЇx

and so,

k пЃ¶ пѓќyпѓћ пЂЅ пЂї3y 2 , or
kпѓќyпѓћ пЂЅ пЂїy 3 пЂ« any constant.

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

v пЂЅ 3x 2 y пЂ« kпѓќyпѓћ пЂЅ 3x 2 y пЂї y 3 ,

and finally,

Gпѓќzпѓћ пЂЅ u пЂ« iv пЂЅ пѓќx 3 пЂї 3xy 2 пѓћ пЂ« iпѓќ3x 2 y пЂї y 3 пѓћ.

Exercises

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

2. Suppose пЃЄ and пЃЎ are harmonic in a simply connected region D bounded by the curve C.
Suppose moreover that пЃЄпѓќx, yпѓћ пЂЅ пЃЎпѓќx, yпѓћ for all пѓќx, yпѓћ пЂµ C. Explain how you know that
пЃЄ пЂЅ пЃЎ everywhere in D.

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

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

7.5
7.3. PoissonвЂ™s integral formula. Let пЃѓ be the disk bounded by the circle
C пЃџ пЂЅ пѓЎz : |z| пЂЅ пЃџпѓў. Suppose пЃЄ is harmonic on пЃѓ and let f be a function analytic on пЃѓ and
such that Re f пЂЅ пЃЄ. Now then, for fixed z with |z| пЂј пЃџ, the function

fпѓќsпѓћ z
gпѓќsпѓћ пЂЅ
пЃџ2 пЂї s z

is analyic on пЃѓ. Thus from CauchyвЂ™s Theorem

fпѓќsпѓћ z
пЃ˜ gпѓќsпѓћds пЂЅ пЃ˜ ds пЂЅ 0.
пЃџ2 пЂї s z
CпЃџ CпЃџ

We know also that

fпѓќsпѓћ
пЃ˜
1
fпѓќzпѓћ пЂЅ s пЂї z ds.
2пЃћi
CпЃџ

Adding these two equations gives us

пЃ˜
1 1 z
fпѓќzпѓћ пЂЅ s пЂї z пЂ« пЃџ2 пЂї s z fпѓќsпѓћds
2пЃћi
CпЃџ

пЃџ 2 пЂї |z| 2
пЃ˜
пЂЅ1 fпѓќsпѓћds.
2пЃћi пѓќs пЂї zпѓћпѓќпЃџ 2 пЂї s z пѓћ
CпЃџ

Next, let пЃЊпѓќtпѓћ пЂЅ пЃџe it , and our integral becomes

2пЃћ
пЃџ 2 пЂї |z| 2
пЃ˜
fпѓќzпѓћ пЂЅ 1 fпѓќпЃџe it пѓћiпЃџe it dt
2пЃћi it 2 it
пѓќпЃџe пЂї zпѓћпѓќпЃџ пЂї пЃџe z пѓћ
0
2пЃћ
2
2
fпѓќпЃџe it пѓћ
пЃџ пЂї |z|
пЃ˜
пЂЅ dt
2пЃћ пѓќпЃџe it пЂї zпѓћпѓќпЃџe пЂїit пЂї z пѓћ
0
2пЃћ
2
2
fпѓќпЃџe it пѓћ
пЃџ пЂї |z|
пЃ˜
пЂЅ dt
2пЃћ 2
it
|пЃџe пЂї z|
0

Now,

7.6
2пЃћ
2
2
пЃЄпѓќпЃџe it пѓћ
пЃџ пЂї |z|
пЃ˜
пЃЄпѓќx, yпѓћ пЂЅ Re f пЂЅ dt.
2пЃћ 2
it
|пЃџe пЂї z|
0

Next, use polar coordinates: z пЂЅ re iпЃ“ :

2пЃћ
пЃџ2 пЂї r2 пЃЄпѓќпЃџe it пѓћ
пЃ˜
пЃЄпѓќr, пЃ“пѓћ пЂЅ dt.
2пЃћ пЃџe it пЂї re iпЃ“ | 2
|
0

Now,
2
|пЃџe it пЂї re iпЃ“ | пЂЅ пѓќпЃџe it пЂї re iпЃ“ пѓћпѓќпЃџe пЂїit пЂї re пЂїiпЃ“ пѓћ пЂЅ пЃџ 2 пЂ« r 2 пЂї rпЃџпѓќe iпѓќtпЂїпЃ“пѓћ пЂ« e пЂїiпѓќtпЂїпЃ“пѓћ пѓћ
пЂЅ пЃџ 2 пЂ« r 2 пЂї 2rпЃџ cosпѓќt пЂї пЃ“пѓћ.

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

2пЃћ
2 2
пЃЄпѓќпЃџe it пѓћ
пЃџ пЂїr
пЃ˜
пЃЄпѓќr, пЃ“пѓћ пЂЅ dt
2пЃћ пЃџ 2 пЂ« r 2 пЂї 2rпЃџ cosпѓќt пЂї пЃ“пѓћ
0

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

Exercises

2пЃћ
5. Evaluate пЃ˜ 1
dt. [Hint: This is easy.]
пЃџ 2 пЂ«r 2 пЂї2rпЃџ cosпѓќtпЂїпЃ“пѓћ
0

6. Suppose пЃЄ is harmonic in a region D. If пѓќx 0 , y 0 пѓћ пЂµ D and if C пѓђ D is a circle centered at
пѓќx 0 , y 0 пѓћ, the inside of which is also in D, then пЃЄпѓќx 0 , y 0 пѓћ is the average value of пЃЄ on the
circle C.

7. Suppose пЃЄ is harmonic on the disk пЃѓ пЂЅ пѓЎz : |z| п‚І пЃџпѓў. Prove that

пЃ˜пЃ˜ пЃЄdA.
1
пЃЄпѓќ0, 0пѓћ пЂЅ
пЃћпЃџ 2
пЃѓ

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 пЂ« п‚ё 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 gпѓќnпѓћ п‚Ї z n and an explicit name for the sequence
is seldom used; we write simply пѓќz n пѓћ to stand for the sequence g which is such that
i i
gпѓќnпѓћ пЂЅ z n . For example, пѓќ n пѓћ is the sequence g for which gпѓќnпѓћ пЂЅ n .

The number L is a limit of the sequence пѓќz n пѓћ if given an пЃђ пЂѕ 0, there is an integer N пЃђ such
that |z n пЂї L| пЂј пЃђ for all n п‚і N пЃђ . If L is a limit of пѓќz n пѓћ, we sometimes say that пѓќz n пѓћ
converges to L. We frequently write limпѓќz n пѓћ пЂЅ L. It is relatively easy to see that if the
complex sequence пѓќz n пѓћ пЂЅ пѓќu n пЂ« iv n пѓћ converges to L, then the two real sequences пѓќu n пѓћ and
пѓќv n пѓћ each have a limit: пѓќu n пѓћ converges to Re L and пѓќv n пѓћ converges to Im L. Conversely, if
the two real sequences пѓќu n пѓћ and пѓќv n пѓћ each have a limit, then so also does the complex
sequence пѓќu n пЂ« iv n пѓћ. All the usual nice properties of limits of sequences are thus true:

limпѓќz n п‚± w n пѓћ пЂЅ limпѓќz n пѓћ п‚± limпѓќw n пѓћ;
limпѓќz n w n пѓћ пЂЅ limпѓќz n пѓћ limпѓќw n пѓћ; and
limпѓќz n пѓћ
z
lim wnn пЂЅ .
limпѓќw n пѓћ

provided that limпѓќz n пѓћ and limпѓќw n пѓћ exist. (And in the last equation, we must, of course,
insist that limпѓќw n пѓћ п‚® 0.)

A necessary and sufficient condition for the convergence of a sequence пѓќa n пѓћ is the
celebrated Cauchy criterion: given пЃђ пЂѕ 0, there is an integer N пЃђ so that |a n пЂї a m | пЂј пЃђ
whenever n, m пЂѕ N пЃђ .

A sequence пѓќf n пѓћ 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 zпЃЏD, we have an
ordinary sequence пѓќf n пѓќzпѓћпѓћ. If each of the sequences пѓќf n пѓќzпѓћпѓћ converges, then we say the
sequence of functions пѓќf n пѓћ converges to the function f defined by fпѓќzпѓћ пЂЅ limпѓќf n пѓќzпѓћпѓћ. This
pretty obvious stuff. The sequence пѓќf n пѓћ is said to converge to f uniformly on a set S if
given an пЃђ пЂѕ 0, there is an integer N пЃђ so that |f n пѓќzпѓћ пЂї fпѓќzпѓћ| пЂј пЃђ for all n п‚і N пЃђ and all z пЂµ 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 пѓќf n пѓћ of continuous functions converges uniformly to f on a domain D, let
z 0 пЃЏD, and let пЃђ пЂѕ 0. We need to show there is a пЃЋ so that |fпѓќz 0 пѓћ пЂї fпѓќzпѓћ| пЂј пЃђ whenever

8.1
пЃђ
|z 0 пЂї z| пЂј пЃЋ. LetвЂ™s do it. First, choose N so that |f N пѓќzпѓћ пЂї fпѓќzпѓћ| пЂј 3 . We can do this because
of the uniform convergence of the sequence пѓќf n пѓћ. Next, choose пЃЋ so that
пЃђ
|f N пѓќz 0 пѓћ пЂї f N пѓќzпѓћ| пЂј 3 whenever |z 0 пЂї z| пЂј пЃЋ. This is possible because f N is continuous.
Now then, when |z 0 пЂї z| пЂј пЃЋ, we have

|fпѓќz 0 пѓћ пЂї fпѓќzпѓћ| пЂЅ |fпѓќz 0 пѓћ пЂї f N пѓќz 0 пѓћ пЂ« f N пѓќz 0 пѓћ пЂї f N пѓќzпѓћ пЂ« f N пѓќzпѓћ пЂї fпѓќzпѓћ|
п‚І |fпѓќz 0 пѓћ пЂї f N пѓќz 0 пѓћ| пЂ« |f N пѓќz 0 пѓћ пЂї f N пѓќzпѓћ| пЂ« |f N пѓќzпѓћ пЂї fпѓќzпѓћ|
пЂј пЃђ пЂ« пЃђ пЂ« пЃђ пЂЅ пЃђ,
3 3 3

and we have done it!

Now suppose we have a sequence пѓќf n пѓћ of continuous functions which converges uniformly
пЃ˜ f n пѓќzпѓћdz converges to пЃ˜ fпѓќzпѓћdz. This
on a contour C to the function f. Then the sequence
C C
пЃђ
is easy to see. Let пЃђ пЂѕ 0. Now let N be so that |f n пѓќzпѓћ пЂї fпѓќzпѓћ| пЂј for n пЂѕ N, where A is the
A
length of C. Then,

пЃ˜ f n пѓќzпѓћdz пЂї пЃ˜ fпѓќzпѓћdz пЃ˜пѓќf n пѓќzпѓћ пЂї fпѓќzпѓћпѓћdz
пЂЅ
C C C

пЂј пЃђAпЂЅпЃђ
A

whenever n пЂѕ N.

Now suppose пѓќf n пѓћ 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 functionsвЂ”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 пЃЏD, and let S пЂЅ пѓЎz : |z пЂї z 0 | пЂј rпѓў пѓђ D . Now
consider any simple closed curve C пѓђ S. Each f n is analytic, and so пЃ˜ f n пѓќzпѓћdz пЂЅ 0 for every
C
n. From the uniform convergence of пѓќf n пѓћ , we know that пЃ˜ fпѓќzпѓћdz is the limit of the sequence
C

пЃ˜ f n пѓќzпѓћdz , and so пЃ˜ fпѓќzпѓћdz пЂЅ 0. MoreraвЂ™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 пѓќf n пѓћ 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 пѓќs n пѓћ in which s n пЂЅ a 1 пЂ« a 2 пЂ«пЃµ пЂ«a n . In other
words, there is sequence пѓќa n пѓћ so that s n пЂЅ s nпЂї1 пЂ« a n . The s n are usually called the partial
n
пЂѕ a j has a limit, then it must be
sums. Recall from Mrs. TurnerвЂ™s class that if the series
jпЂЅ1
true that lim пѓќa n пѓћ пЂЅ 0.
nп‚ёпЃ‹

n
пЂѕ f j пѓќzпѓћ of functions. Chances are this series will converge for some
Consider a series
jпЂЅ1
values of z and not converge for others. A useful result is the celebrated Weierstrass
M-test: Suppose пѓќM j пѓћ is a sequence of real numbers such that M j п‚і 0 for all j пЂѕ J, where
n
пЂѕ M j converges. If for all zпЃЏD, we
J is some number., and suppose also that the series
jпЂЅ1
n
пЂѕ f j пѓќzпѓћ converges uniformly on D.
have |f j пѓќzпѓћ| п‚І M j for all j пЂѕ J, then the series
jпЂЅ1

To prove this, begin by letting пЃђ пЂѕ 0 and choosing N пЂѕ J so that

n

пЂѕ Mj пЂј пЃђ
jпЂЅm

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

8.3
n n n

пЂѕ f j пѓќzпѓћ пЂѕ|f j пѓќzпѓћ| п‚І пЂѕ M j пЂј пЃђ.
п‚І
jпЂЅm jпЂЅm jпЂЅm

n
пЂѕ f j пѓќzпѓћ converges. To see the uniform convergence, observe that
This shows that
jпЂЅ1

n n mпЂї1

пЂѕ f j пѓќzпѓћ пЂѕ f j пѓќzпѓћ пЂї пЂѕ f j пѓќzпѓћ
пЂЅ пЂјпЃђ
jпЂЅm jпЂЅ0 jпЂЅ0

for all zпЃЏD and n пЂѕ m пЂѕ N. Thus,

пЃ‹
n mпЂї1 mпЂї1

пЂѕ f j пѓќzпѓћ пЂї пЂѕ f j пѓќzпѓћ пЂѕ f j пѓќzпѓћ пЂї пЂѕ f j пѓќzпѓћ
пЂЅ п‚ІпЃђ
lim
nп‚ёпЃ‹
jпЂЅ0 jпЂЅ0 jпЂЅ0 jпЂЅ0

пЃ‹
n
пЂѕ a j is almost always written as пЂѕ a j .)
for m пЂѕ N.(The limit of a series
jпЂЅ0 jпЂЅ0

Exercises

zn
6. Find the set D of all z for which the sequence has a limit. Find the limit.
n пЂї3 n
z

n n
пЂѕ aj пЂѕ Re a j
7. Prove that the series convegres if and only if both the series and
jпЂЅ1 jпЂЅ1
n
пЂѕ Im a j converge.
jпЂЅ1

n
пЂѕпѓќ 1 пѓћ j
8. Explain how you know that the series converges uniformly on the set
z
jпЂЅ1
|z| п‚і 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 пѓќzпѓћ пЂЅ c 0 пЂ« c 1 пѓќz пЂї z 0 пѓћ пЂ« c 2 пѓќz пЂї z 0 пѓћ 2 пЂ«пЃµ пЂ«c n пѓќz пЂї z 0 пѓћ n .

8.4
(We start with n пЂЅ 0 for esthetic reasons.) These are the so-called power series. Thus,
n
пЂѕ c j пѓќz пЂї z 0 пѓћ j .
a power series is a series of functions of the form
jпЂЅ0

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

n

пЂѕ zj .
jпЂЅ0

Here
s n пЂЅ 1 пЂ« z пЂ« z 2 пЂ«пЃµ пЂ«z n , and
zs n пЂЅ z пЂ« z 2 пЂ« z 3 пЂ«пЃµ пЂ«z nпЂ«1 .

Subtracting the second of these from the first gives us

пѓќ1 пЂї zпѓћs n пЂЅ 1 пЂї z nпЂ«1 .

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

1 пЂї z nпЂ«1 .
sn пЂЅ
1пЂїz 1пЂїz

Now if |z| пЂј 1, it should be clear that limпѓќz nпЂ«1 пѓћ пЂЅ 0, and so

n

пЂѕ zj 1.
пЂЅ lim s n пЂЅ
lim
1пЂїz
jпЂЅ0

Or,

пЃ‹

пЂѕ zj пЂЅ 1 , for |z| пЂј 1.
1пЂїz
jпЂЅ0

There is a bit more to the story. First, note that if |z| пЂѕ 1, then the Geometric series does
not have a limit (why?). Next, note that if |z| п‚І пЃџ пЂј 1, then the Geometric series converges

8.5
1
uniformly to . To see this, note that
1пЂїz

n

пЂѕ пЃџj
jпЂЅ0

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 пЂЅ z 0 . LetвЂ™s see what else we can say.
n
пЂѕ c j пѓќz пЂї z 0 пѓћ j . Let
Consider a power series
jпЂЅ0

пЃ– пЂЅ lim sup |c j | .
j

1
(Recall from 6 th grade that lim supпѓќa k пѓћ пЂЅ limпѓќsupпѓЎa k : k п‚і nпѓў. ) Now let R пЂЅ пЃ– . (We
shall say R пЂЅ 0 if пЃ– пЂЅ пЃ‹, and R пЂЅ пЃ‹ if пЃ– пЂЅ 0. ) We are going to show that the series
converges uniformly for all |z пЂї z 0 | п‚І пЃџ пЂј R and diverges for all |z пЂї z 0 | пЂѕ R.

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

1 пЂј k пЂј 1 пЂЅ пЃ–.
R
|z пЂї z 0 |

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

each of these c j we have

j
пЂѕ пѓќk|z пЂї z 0 |пѓћ j пЂѕ 1.
|c j пѓќz пЂї z 0 пѓћ j | пЂЅ |c j | |z пЂї z 0 |
j

It is thus not possible for lim |c n пѓќz пЂї z 0 пѓћ n | пЂЅ 0, and so the series does not converge.
nп‚ёпЃ‹

Next, we show that the series does converge uniformly for |z пЂї z 0 | п‚І пЃџ пЂј R. Let k be so
that

пЃ– пЂЅ 1 пЂј k пЂј пЃџ.
1
R

Now, for j large enough, we have j |c j | пЂј k. Thus for |z пЂї z 0 | п‚І пЃџ, we have

8.6
j
|c j пѓќz пЂї z 0 пѓћ j | пЂЅ пЂј пѓќk|z пЂї z 0 |пѓћ j пЂј пѓќkпЃџпѓћ j .
|c j | |z пЂї z 0 |
j

n
пЂѕпѓќkпЃџпѓћ j converges because kпЃџ пЂј 1 and the uniform convergence
The geometric series
jпЂЅ0
n
пЂѕ c j пѓќz пЂї z 0 пѓћ j follows from the M-test.
of
jпЂЅ0

Example

n
пЂѕ j! z j . LetвЂ™s compute R пЂЅ 1/ lim sup
1
пЂЅ lim supпѓќ j j! пѓћ. Let
Consider the series |c j |
j

jпЂЅ0
K 2K
K be any positive integer and choose an integer m large enough to insure that 2 m пЂѕ .
пѓќ2Kпѓћ!
n!
Now consider K n , where n пЂЅ 2K пЂ« m:

n! пЂЅ пѓќ2K пЂ« mпѓћ! пЂЅ пѓќ2K пЂ« mпѓћпѓќ2K пЂ« m пЂї 1пѓћпЃµ пѓќ2K пЂ« 1пѓћпѓќ2Kпѓћ!
Kn K 2KпЂ«m K m K 2K
пѓќ2Kпѓћ!
пЂѕ 2 m 2K пЂѕ 1
K

Thus n n! пЂѕ 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 пЂЅ lim supпѓќ j j! пѓћ пЂЅ пЃ‹, and so the
n
пЂѕ j! z j converges for all z.
1
series
jпЂЅ0

n
пЂѕ c j пѓќz пЂї z 0 пѓћ j , there is a number
LetвЂ™s summarize what we have. For any power series
jпЂЅ0
1
RпЂЅ such that the series converges uniformly for |z пЂї z 0 | п‚І пЃџ пЂј R and does not
lim supпѓќ j |c j |
converge for |z пЂї z 0 | пЂѕ R. (Note that we may have R пЂЅ 0 or R пЂЅ пЃ‹.) The number R is
called the radius of convergence of the series, and the set |z пЂї z 0 | пЂЅ 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 пѓќпЃЉ j пѓћ has a limit. Prove that

8.7
lim supпѓќпЃЉ j пѓћ пЂЅ limпѓќпЃЉ j пѓћ.

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

n
пЂѕ j!z j .
10.
jпЂЅ0

n
пЂѕ jz j .
11.
jпЂЅ0

n
j2
пЂѕ zj .
12. 3j
jпЂЅ0

n
пѓќпЂї1пѓћ j
пЂѕ z 2j
13. 2 2j пѓќj!пѓћ 2
jпЂЅ0

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

пЃ‹

пЂѕ c j пѓќz пЂї z 0 пѓћ j
Sпѓќzпѓћ пЂЅ
jпЂЅ0

is an analytic function. We shall show that this series may be integrated
вЂќterm-by-termвЂќвЂ”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

пЃ‹
пЃ˜ gпѓќzпѓћSпѓќzпѓћdz пЂЅ пЂѕ c j пЃ˜ gпѓќzпѓћпѓќz пЂї z 0 пѓћ j dz.
jпЂЅ0
C C

LetвЂ™s see why this. First, let пЃђ пЂѕ 0. Let M be the maximum of |gпѓќzпѓћ| on C and let L be the
length of C. Then there is an integer N so that

пЃ‹
пЃђ
пЂѕ c j пѓќz пЂї z 0 пѓћ j пЂј
ML
jпЂЅn

8.8
for all n пЂѕ N. Thus,

пЃ‹
пЃ˜ gпѓќzпѓћ пЂѕ c j пѓќz пЂї z 0 пѓћ j dz пЂј ML пЃђ пЂЅ пЃђ,
ML
jпЂЅn
C

Hence,

пЃ‹
nпЂї1
пЃ˜ gпѓќzпѓћSпѓќzпѓћdz пЂї пЂѕ c j пЃ˜ gпѓќzпѓћпѓќz пЂї z 0 пѓћ j dz пЃ˜ gпѓќzпѓћ пЂѕ c j пѓќz пЂї z 0 пѓћ j dz
пЂЅ
jпЂЅ0 jпЂЅn
C C C

пЂј пЃђ,

and we have shown what we promised.

8.5 Differentiation of power series. Again, let

пЃ‹

пЂѕ c j пѓќz пЂї z 0 пѓћ j .
Sпѓќzпѓћ пЂЅ
jпЂЅ0

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

пЃ‹

пЂѕ jc j пѓќz пЂї z 0 пѓћ jпЂї1 .
S пЃ¶ пѓќzпѓћ пЂЅ
jпЂЅ1

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
gпѓќsпѓћ пЂЅ ,
2пЃћiпѓќs пЂї zпѓћ 2

and apply the result of the previous section to conclude that

8.9
пЃ‹
пЃ˜ gпѓќsпѓћSпѓќsпѓћds пЂЅ пЂѕ c j пЃ˜ gпѓќsпѓћпѓќs пЂї z 0 пѓћ j ds, or
jпЂЅ0
C C
пЃ‹
пѓќs пЂї z 0 пѓћ j
Sпѓќsпѓћ
пЃ˜ пЂѕ cj пЃ˜
1 ds пЂЅ ds. Thus
2пЃћi пѓќs пЂї zпѓћ 2 пѓќs пЂї zпѓћ 2
jпЂЅ0
C C
пЃ‹

пЂѕ jc j пѓќz пЂї z 0 пѓћ jпЂї1 ,
S пЃ¶ пѓќzпѓћ пЂЅ
jпЂЅ0

as promised!

Exercises

14. Find the limit of

n

пЂѕпѓќj пЂ« 1пѓћz j .
jпЂЅ0

For what values of z does the series converge?

15. Find the limit of

n
zj
пЂѕ .
j
jпЂЅ1

For what values of z does the series converge?

n
пЂѕ c j пѓќz пЂї 1пѓћ j such that
16. Find a power series
jпЂЅ0

пЃ‹

пЂѕ c j пѓќz пЂї 1пѓћ j , for |z пЂї 1| пЂј 1.
1пЂЅ
z
jпЂЅ0

n
пЂѕ c j пѓќz пЂї 1пѓћ j such that
17. Find a power series
jпЂЅ0

8.10
пЃ‹

пЂѕ c j пѓќz пЂї 1пѓћ j , for |z пЂї 1| пЂј 1.
Log z пЂЅ
jпЂЅ0

8.11
Chapter Nine

Taylor and Laurent Series

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

пЃ‹
пѓќz пЂї z 0 пѓћ j
пЂѕ
1 1 1 1
s пЂї z пЂЅ пѓќs пЂї z 0 пѓћ пЂї пѓќz пЂї z 0 пѓћ пЂЅ s пЂї z 0 пЂЅ
zпЂїz 0
1пЂї пѓќs пЂї z 0 пѓћ jпЂ«1
sпЂїz 0 jпЂЅ0

since | zпЂїz 0 | пЂј 1. The convergence is uniform, so we may integrate
sпЂїz 0

пЃ‹
fпѓќsпѓћ fпѓќsпѓћ
пЃ˜ пЂѕпЃ˜ ds пѓќz пЂї z 0 пѓћ j , or
s пЂї z ds пЂЅ jпЂ«1
пѓќs пЂї z 0 пѓћ
jпЂЅ0
C C

пЃ‹
fпѓќsпѓћ fпѓќsпѓћ
пЃ˜ пЃ˜
пЂѕ
1 1 ds пѓќz пЂї z 0 пѓћ j .
fпѓќzпѓћ пЂЅ s пЂї z ds пЂЅ
2пЃћi 2пЃћi jпЂ«1
пѓќs пЂї z 0 пѓћ
jпЂЅ0
C C

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

пЃ‹

пЂѕ c j пѓќz пЂї z 0 пѓћ j , |z пЂї z 0 | пЂј r,
fпѓќzпѓћ пЂЅ
jпЂЅ0

where

fпѓќsпѓћ
пЃ˜
1
cj пЂЅ ds.
2пЃћi пѓќs пЂї z 0 пѓћ jпЂ«1
C

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

We know we may differentiate the series to get

пЃ‹

пЂѕ jc j пѓќz пЂї z 0 пѓћ jпЂї1
f пЃ¶ пѓќzпѓћ пЂЅ
jпЂЅ1

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

пЃ‹

пЂѕ jпѓќj пЂї 1пѓћпѓќj пЂї 2пѓћпЃµ пѓќj пЂї n пЂ« 1пѓћc j пѓќz пЂї z 0 пѓћ jпЂїn .
f пѓќnпѓћ пѓќzпѓћ пЂЅ
jпЂЅn

Hence,

f пѓќnпѓћ пѓќz 0 пѓћ пЂЅ n!c n , or
f пѓќnпѓћ пѓќz 0 пѓћ
cn пЂЅ .
n!

But we also know that

fпѓќsпѓћ
пЃ˜
1
cn пЂЅ ds.
2пЃћi пѓќs пЂї z 0 пѓћ nпЂ«1
C

This gives us

fпѓќsпѓћ
пЃ˜
f пѓќnпѓћ пѓќz 0 пѓћ пЂЅ n! ds, for n пЂЅ 0, 1, 2, пЃµ .
2пЃћi пѓќs пЂї z 0 пѓћ nпЂ«1
C

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

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

пЃ‹

пЂѕ c j пѓќz пЂї z 0 пѓћ j ,
fпѓќzпѓћ пЂЅ
jпЂЅ0

and the radius of convergence is R. Then we know, of course, that the limit function f is
analytic for |z пЂї z 0 | пЂј R. We showed that if f is analytic in |z пЂї z 0 | пЂј r, then the series
converges for |z пЂї z 0 | пЂј r. Thus r п‚І R, and so f cannot be analytic at any point z for which
|z пЂї z 0 | пЂѕ 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 fпѓќzпѓћ пЂЅ expпѓќzпѓћ пЂЅ e z . Then fпѓќ0пѓћ пЂЅ f пЃ¶ пѓќ0пѓћ пЂЅпЃµ пЂЅ f пѓќnпѓћ пѓќ0пѓћ пЂЅпЃµ пЂЅ 1, and the Taylor series for f
at z 0 пЂЅ 0 is

пЃ‹

пЂѕ 1 zj
ez пЂЅ
j!
jпЂЅ0

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,

пЃ‹
e z пЂЅ e пЂѕ 1 пѓќz пЂї 1пѓћ j .
j!
jпЂЅ0

n
пЂѕ c j z j for tanh z ?
2. What is the radius of convergence of the Taylor series
jпЂЅ0

3. Show that

пЃ‹
пѓќz пЂї iпѓћ j
пЂѕ
1пЂЅ
1пЂїz пѓќ1 пЂї iпѓћ jпЂ«1
jпЂЅ0

for |z пЂї i| пЂј 2.

1
, what is f пѓќ10пѓћ пѓќiпѓћ ?
4. If fпѓќzпѓћ пЂЅ 1пЂїz

5. Suppose f is analytic at z пЂЅ 0 and fпѓќ0пѓћ пЂЅ f пЃ¶ пѓќ0пѓћ пЂЅ f пЃ¶пЃ¶ пѓќ0пѓћ пЂЅ 0. Prove there is a function g
analytic at 0 such that fпѓќzпѓћ пЂЅ z 3 gпѓќzпѓћ in a neighborhood of 0.

6. Find the Taylor series for fпѓќzпѓћ пЂЅ sin z at z 0 пЂЅ 0.

7. Show that the function f defined by

9.3
sin z
for z п‚® 0
z
fпѓќzпѓћ пЂЅ
for z пЂЅ 0
1

is analytic at z пЂЅ 0, and find f пЃ¶ пѓќ0пѓћ.

9.2. Laurent series. Suppose f is analytic in the region R 1 пЂј |z пЂї z 0 | пЂј 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 пЂЅ пЃ‹.) We shall show that for z пЂ¶ C in this region

пЃ‹ пЃ‹
bj
пЂѕ a j пѓќz пЂї z 0 пѓћ j пЂ« пЂѕ
fпѓќzпѓћ пЂЅ ,
j
пѓќz пЂї z 0 пѓћ
jпЂЅ0 jпЂЅ1

where
fпѓќsпѓћ
пЃ˜
1
aj пЂЅ ds, for j пЂЅ 0, 1, 2, пЃµ
2пЃћi пѓќs пЂї z 0 пѓћ jпЂ«1
C

and

fпѓќsпѓћ
пЃ˜
1
bj пЂЅ ds, for j пЂЅ 1, 2, пЃµ .
2пЃћi пѓќs пЂї z 0 пѓћ пЂїjпЂ«1
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