Arnold Lecture notes on complex analysis [sharethefiles com]

COMPLEX ANALYSIS

Douglas N. Arnold

References:
John B. Conway,

Functions of One Complex Variable

, Springer-Verlag, 1978.

Lars V. Ahlfors,

Complex Analysis

, McGraw-Hill, 1966.

Raghavan Narasimhan,

Complex Analysis in One Variable

, Birkhauser, 1985.

CONTENTS

The Complex Number System

::::::::::::::::::::::::::::::::::::::

II.

Elementary Properties and Examples of Analytic Fns.

::::::::

Dierentiability and analyticity

::::::::::::::::::::::::::::::::::::

The Logarithm

:::::::::::::::::::::::::::::::::::::::::::::::::::::

Conformality

:::::::::::::::::::::::::::::::::::::::::::::::::::::::

Cauchy{Riemann Equations

::::::::::::::::::::::::::::::::::::::::

Mobius transformations

::::::::::::::::::::::::::::::::::::::::::::

III.

Complex Integration and Applications to Analytic Fns.

:::::

Local results and consequences

::::::::::::::::::::::::::::::::::::

Homotopy of paths and Cauchy's Theorem

::::::::::::::::::::::::

Winding numbers and Cauchy's Integral Formula

::::::::::::::::::

Zero counting Open Mapping Theorem

:::::::::::::::::::::::::::

Morera's Theorem and Goursat's Theorem

::::::::::::::::::::::::

IV.

Singularities of Analytic Functions

::::::::::::::::::::::::::::

Laurent series

:::::::::::::::::::::::::::::::::::::::::::::::::::::

Residue integrals

::::::::::::::::::::::::::::::::::::::::::::::::::

Further results on analytic functions

:::::::::::::::::::::::::

The theorems of Weierstrass, Hurwitz, and Montel

::::::::::::::::

Schwarz's Lemma

:::::::::::::::::::::::::::::::::::::::::::::::::

The Riemann Mapping Theorem

::::::::::::::::::::::::::::::::::

Complements on Conformal Mapping

:::::::::::::::::::::::::::::

VI.

Harmonic Functions

:::::::::::::::::::::::::::::::::::::::::::::::

The Poisson kernel

::::::::::::::::::::::::::::::::::::::::::::::::

Subharmonic functions and the solution of the Dirichlet Problem

The Schwarz Reection Principle

::::::::::::::::::::::::::::::::::

These lecture notes were prepared for the instructor's personal use in teaching a half-semester

course on complex analysis at the beginning graduate level at Penn State, in Spring 1997. They

are certainly not meant to replace a good text on the subject, such as those listed on this page.

Department of Mathematics, Penn State University, University Park, PA 16802. Email:

dna@math.psu.edu. Web: http://www.math.psu.edu/dna/.

Typeset by

-TEX

I. The Complex Number System

is a eld. For

n >

is a vectorspace over

, so is an additive group, but

doesn't have a multiplication on it. We can endow

with a multiplication by

(

)(

) = (

;

bdbc

)

Under this denition

becomes a eld, denoted

. Note that (

(

)

;

(

)) is the multiplicative inverse of (

). (Remark: it is not possible to endow

with a eld structure for

n >

2.) We denote (0

1) by

and identify

with

(

0), so

. Thus (

) =

. Note that

;

is generated

by adjoining

and closing under addition and multiplication. It is remarkable

that the addition of

lets us not only solve the equation

+ 1 = 0, but every

polynomial equation.

For

and

real and

we dene Re

, Im

;

, and

= (

)

. Then

= (

)

= (

;

)

w zw

z=w

The map

(cos

sin

) denes a 2

-periodic map of the real line onto the

unit circle in

. In complex notation this map is

cis

:= cos

sin

Every nonzero complex number can be written as

cis

where

r >

0 is uniquely

determined and

is uniquely determined modulo 2

. The number 0 is equal

cis

where

= 0 and

is arbitrary. The relation

cis

determines the

relations

which is simply the function

and

. The latter is

denoted

= arg

. Note that for

= 0, arg

is determined modulo 2

(while

arg0 is arbitrary). We can normalize arg by insisting that arg

(

;

]. Note

that if

cis

and

cis

then

cis(

). The latter

formula just encapsulates the formula for the sine and cosine of a sum, and gives

arg

= arg

+arg

. In particular,

cis

cis(

2), so multiplication by

is just the operation of rotation by

2 in the complex plane. Multiplication by an

arbitrary complex number

cis

is just rotation by arg

followed by (or preceded

by) dilation by a factor

. Further,

cis(

). Every nonzero

admits

distinct

th roots: the

th roots of

cis

are

cis(

+ 2

)

= 0

::: n

Lines and circles in the plane. Circles given by

;

where

is the

center and

r >

0 is the radius. If 0

then the line through the origin in the

direction

is the set of all points of the form

, or all

with Im(

z=b

) = 0.

and

c >

0 then (

)

cib

represents a point in the half plane to

the left of

determined by the line

, i.e.,

: Im(

z=b

)

is the equation of that

half-plane. Similarly,

: Im(

;

)

]

is the translation of that half-plane

, i.e., the half-plane determined by the line through

parallel to

and in the

direction to the left of

Stereographic projection determines a one-to-one correspondence between the

unit sphere in

minus the north-pole,

, and the complex plane via the corre-

spondence

;

= 2Re

1 +

= 2Im

1 +

;

+ 1

If we dene

f1g

, then we have a one-to-one correspondence between

and

. This allows us to dene a metric on

, which is given by

(

) =

;

(1 +

)(1 +

)

(

) =

1 +

II. Elementary Properties and Examples of Analytic Functions

For

= 1,

= (1

;

)

;

). Therefore the geometric series

converges (to 1

;

)) if

1. It clearly diverges, in fact its terms

become unbounded, if

Weierstrass M-Test.

Let

:::

be positive numbers with

and

suppose that

are functions on some set

satisfying

sup

(

)

. Then

(

)

is absolutely and uniformly convergent.

Theorem.

Let

be given and dene the number

= limsup

Then (1) for any

the power series

(

;

)

converges absolutely for

all

;

< R

and it converges absolutely and uniformly on the disk

;

for all

r < R

. (2) The sequence

(

;

)

is unbounded for all

;

> R

(and

hence the series is certainly divergent).

Thus we see that the set of points where a power series converges consists of a

disk

;

< R

and possibly a subset of its boundary.

is called the radius of

convergence of its series. The case

is allowed.

Proof of theorem.

For any

r < R

we show absolute uniform convergence on

;

. Choose ~

(

). Then, 1

r >

limsup

, so

for

all

suciently large. For such

and so

sup

(

;

)

(

)

Since

(

)

we get the absolute uniform convergence on

;

r > R

, take ~

(

). Then there exist

arbitrarily large such that

. Then,

(

;

)

(

)

, which can be arbitrarily large.

Theorem.

:::

and

lim

exists as a nite number or innity,

then this limit is the radius of convergence

(

;

)

Proof.

Without loss of generality we can suppose that

= 0. Suppose that

lim

. Then for all

suciently large

and

. Thus the series

has terms of increasing magnitude, and so

cannot be convergent. Thus

. This shows that lim

Similarly, suppose that

z <

lim

. Then for all

suciently large

and

. Thus the series has terms of decreas-

ing magnitude, and so, by the previous theorem,

. This shows that

lim

Remark.

On the circle of convergence, many dierent behaviors are possible.

diverges for all

= 1.

diverges for

= 1, else converges, but not absolutely

(this follows from the fact that the partial sums of

are bounded for

= 1 and

0).

converges absolutely on

1. Sierpinski gave a (complicated)

example of a function which diverges at every point of the unit circle except

= 1.

As an application, we see that the series

converges absolutely for all

and that the convergence is uniform on all

bounded sets. The sum is, by denition, exp

Now suppose that

(

;

)

has radius of convergence

, and consider

its formal derivative

(

;

)

(

+1)

(

;

)

. Now clearly

(

;

)

has the same radius of convergence as

(

;

)

since

(

;

)

(

;

)

(

;

)

;

and so the partial sums on the left and right either both diverge for a given

or both converge. This shows (in a roundabout way) that limsup

limsup

= 1

. Now lim(

+ 1)

= 1 as is easily seen by taking logs.

Moreover, it is easy to see that if limsup

and lim

c >

0, then

limsup

. Thus limsup

(

+1)

= 1

. This shows that the formal

derivative of a power series has the same radius of convergence as the original power

series.

Dierentiability and analyticity.

Denition of dierentiability at a point (as-

sumes function is dened in a neighborhood of the point).

Most of the consequences of dierentiability are quite dierent in the real and

complex case, but the simplest algebraic rules are the same, with the same proofs.

First of all, dierentiability at a point implies continuity there. If

and

are both

dierentiable at a point

, then so are

, and, if

(

)

= 0,

f=g

, and

the usual sum, product, and quotient rules hold. If

is dierentiable at

and

is dierentiable at

(

), then

is dierentiable at

and the chain rule holds.

Suppose that

is continuous at

is continous at

(

), and

(

)) =

for all

in a neighborhood of

. Then if

(

)) exists and is non-zero, then

(

) exists

and equals 1

(

)).

Denition.

Let

be a complex-valued function dened on an open set

Then

is said to be analytic on

exists and is continuous at every point of

Remark.

We shall prove later that if

is dierentiable at every point of an open set

it is automatically analytic in fact, it is automatically innitely dierentiable.

This is of course vastly dierent from the real case.

is an arbitrary non-empty subset of

we say

is analytic on

if it is

dened and analytic on an open set containing

We now show that a power series is dierentiable at every point in its disk

of convergence and that its derivative is given by the formal derivative obtained

by dierentiating term-by-term. Since we know that that power series has the

same radius of convergence, it follows that a power series is analytic and innitely

dierentiable in its convergence disk. For simplicity, and without loss of generality

we consider a power series centered at zero:

(

) =

. Suppose that the

radius of convergence is

and that

< R

. We must show that for any

the inequality

(

)

;

(

)

;

(

)

is satised for all

suciently close to

, where

(

) =

. Let

(

) =

(

) =

. Then

(

)

;

(

)

;

(

)

(

)

;

(

)

;

(

)

(

)

;

(

)

(

)

;

(

)

;

Now

(

) is just a partial sum for

(

), so for

suciently large (and all

3. Also,

(

)

;

(

)

;

Now

< r < R

for some

, and if we restrict to

< r

, we have

;

Since

is convergent, we have for

suciently large and all

< r

then

< =

3. Now x a value of

which is suciently large by both criteria. Then

the dierentiability of the polynomial

shows that

3 for all

suciently

close to

We thus know that if

(

) =

, then, within the disk of convergence,

(

) =

, and by induction,

0 0

(

) =

(

;

, etc. Thus

(0),

(0)

3!, etc. This shows that any

convergent power series is the sum of its Taylor series in the disk of convergence:

(

) =

(

)

! (

;

)

In particular, exp

= exp.

Lemma.

A function with vanishing derivative on a region (connected open set) is

constant.

Indeed, it is constant on disks, since we can restrict to segments and use the real

result. Then we can use connectedness to see that the set of points where it takes

a given value is open and closed.

Use this to show that exp(

)exp(

;

)

exp

. Dene cos

and sin

, get

cos

+sin

1, exp

= cos

sin

, cis

= exp(

), exp

= exp(Re

)cis(Im

The Logarithm.

= 0, the most general solution of exp

= log

arg

+ 2

. There is no solution to exp

= 0.

Denition.

is an open set and

is a continuous function satisfying

exp(

(

)) =

, then

is called a branch of the logarithm on

Examples:

;

(principal branch),

, a spiral strip.

By the formula for the derivative of an inverse, a branch of the logarithm is

analytic with derivative 1

A branch of the logarithm gives a branch of

:= exp(

log

) (understood to be

the principal branch if not otherwise noted). Note that

= exp

Note that dierent branches of the logarithm assign values to log

that dier

by addition of 2

. Consequently dierent branches of

dier by factors of

exp(2

inb

). If

is an integer, all values agree. If

is a rational number with

denominator

there are

values.

Conformality.

Consider the angle between two line segments with a common

vertex in the complex plane. We wish to consider how this angle is transformed

under a complex map. The image will be two smooth curves meeting at a common

point, so we have to dene the angle between two such curves.

Dene a path in the complex plane as a continuous map

]

. If

(

)

exists at some point and is not zero, then it is a vector tangent to the

curve at

(

), and hence its argument is the angle between the (tangent to the)

curve and the horizontal. (Note: if

(

) = 0, the curve may not have a tangent

e.g.,

(

) =

;

.) Let

]

and ~

: ~

]

be two smooth

curves, such that

(

) = ~

) =

, and

(

)

= 0. Then arg ~

)

;

arg

(

)

measures the angle between the curves. Now consider the images under

, e.g., of

(

) =

(

)). Then

(

) =

(

)

(

), ~

) =

(

). Therefore arg ~

)

;

arg

(

) = arg ~

) + arg

(

)

;

arg

(

)

;

arg

(

), i.e., the angle is invariant

in both magnitude and sign. This says that an analytic mapping is conformal,

whenever its derivative is not zero. (Not true if the derivative is zero. E.g.,

doubles angles at the origin.)

The conformality of an analytic map arises from the fact that in a neighborhood

of a point

such a map behaves to rst order like multiplication by

(

). Since

(

) =

exp

, near

simply behave like rotation by

followed by dilation.

By contrast, a smooth map from

can behave like an arbitrary linear

operator in a neighborhood of a point.

At this point, show the computer graphics square1.mps and square2.mps.

Cauchy{Riemann Equations.

Let

with

open. Abuse notation

slightly by writing

(

) as an alternative for

(

). If

(

) exists for some

, then

(

) = lim

(

)

;

(

)

= lim

(

)

;

(

)

and

(

) = lim

(

)

;

(

)

= lim

;

(

)

;

(

)

;

i@f

@y :

Thus complex-dierentiability of

implies not only that the partial derivatives

exist there, but also that they satisfy the Cauchy{Riemann equation

;

i@f

@y :

, then this equation is equivalent to the system

;

@y :

Another convenient notation is to introduce

:= 12

;

i@f

:= 12

i@f

(These are motivated by the equations

= (

)

= (

;

)

), which, if

and

were independent variables, would give

@x=@z

= 1

@y=@z

;

2, etc.)

In terms of these, the Cauchy{Riemann equations are exactly equivalent to

= 0

which is also equivalent to

@x:

Thus, if

is analytic on , then the partial derivaties of

exist and are contin-

uous on , and the Cauchy-Riemann equations are satised there. The converse is

true as well:

Theorem.

If the partial derivatives of

exist and are continuous on

, and the

Cauchy-Riemann equations are satised there, then

is analytic on

Proof.

Let

. We must show that

(

) exists. Let

be small

enough that the disk of radius

around

belongs to and choose

with

< r

. Then, by the mean value theorem,

(

)

;

(

)

(

)

;

(

)

(

)

;

(

)

(

)

;

(

)

(

)

(

)

where

< s

and

< t

. Note that

(

)

;

(

)

0, and similarly for the second partial derivative. Moreover

(

)

stays

bounded (by 1). Thus

(

)

;

(

)

(

)

(

)

(

)

where lim

(

) = 0. Now if the Cauchy{Riemann equations hold, then the the

rst two terms on the right hand side sum to

@f=@x

(

), which is independent of

, so the limit exists and is equal to

@f=@x

(

Remark.

The theorem can be weakened to say that if

is continuous on and the

partial derivatives exist and satisfy the Cauchy{Riemann equations there (without

assuming that the partial derivatives are continuous), then the complex derivative

exists on (which is equivalent to

being analytic on . This is the Looman{

Mencho Theorem. See Narasimhan,

Complex Analysis in One Variable

, for a

proof. We do need at least continuity, since otherwise we could take

to be the

characteristic function of the coordinate axes.

Note that

= 14

This shows that any analytic function is harmonic (equivalently, its real and

imaginary parts are harmonic). It also shows that the conjugate of an analytic

function, while not analytic, is harmonic.

(

) = Re

(

), then

is harmonic. Conversely, if

is harmonic on a

region , does there exist a

harmonic conjugate

, i.e., a function

on such that

is analytic? Clearly, if there exists

it is determined up to addition of

a real constant (or

is determined up to addition of an imaginary constant). The

book gives an elementary construction of the harmonic conjugate in a disk and in

the whole plane. We sketch the idea of a more general proof. We require that the

domain satisfy the condition that if

is any piecewise simple smooth closed curve

in , then

is the boundary of a subset

of . In other words, has no holes

(is simply-connected). Our proof follows directly from the following result, which

holds on simply-connected domains: let

be a

vectoreld. Then there

exists

such that

if and only if

=@y

=@x

. The \only

if" part is obvious. To prove the \if" part, x a point (

) in and for any

(

)

, let

(

)

be a piecewise smooth path in from (

) to (

), and

let

be the unit tangent to the path pointing from (

) towards (

). Dene

(

) =

y )

ds:

It is essential that this quantity doesn't depend on the choice of path, or, equiva-

lently, that

= 0

for all piecewise smooth simple closed paths

. But

(

;

)

nds

(

;

)

dxdy

= 0

by the divergence theorem. It is easy to check that

For a non-simply connected region, there may exist no harmonic conugate. E.g.,

= log

, then

arg

is analytic on

, but cannot be

extended to an analytic function on

Mobius transformations.
Denition.

Let

be complex numbers with

. Then the mapping

(

) =

;

d=c

is a Mobius transformation.

Remarks.

1) Note that if

the same expression would yield a constant.

2) The coecents aren't unique, since we can multiply them all by any nonzero

complex constant. To each MT we associate the nonsingular matrix

a b

c d

of its

coecients, which is determined up to a non-zero multiple. 3) The linear (but non-

constant) polynomials are MTs, namely the ones for which

= 0. 4) If

;

d=c

then

(

)

and

(

) = (

;

)

(

)

= 0. 5) We can dene

(

;

d=c

) =

and

(

) =

a=c

, so

can be viewed of as a map from

into itself (which is, as

we will now see, 1-1 and onto).

and

are MTs, then so is

, its coecient matrix being the product of the

coecient matrices of

and

. Any MTs admits an inverse, namely the transform

with the inverse coecient matrix. With the understanding that

(

) =

a=c

and

(

;

d=c

) =

(or, if

= 0, then

(

) =

, the MT maps

1-1 onto itself. This

is evident from the existence of its inverse

(

) = (

;

)

(

;

). Moreover,

the composition of Mobius transformations is again one, so they form a group under

composition. If

is a non-linear MT we can write it as

;

d :

This shows that any MT can be written as a composition of translations, rotations

and dilations (multiplication by a complex number), and inversion (reciprocals).

Since the geometric action of translation, dilation, and rotation are quite evident,

let's consider inversion. It takes the unit disk

. We now show that

inversion maps lines and circles to other lines and circles.

Consider the equation

;

where

k >

. This is

(

;

)

+ (

;

)

(

;

)

+ (

;

)

]

which is clearly the equation of a circle if

= 1 and the equation of a line if

= 1. In either case we have a circle in the Riemann sphere, which we call

Substituting 1

for

and doing some simple manipulations gives the equation

;

= (

)

;

, which is another such circle,

. Thus we have

(

)

, showing that inversion maps circles in the sphere onto

other circles in the sphere. Since the same property is evident for translation,

dilation, and rotation, we conclude that this property holds for all MT.

Some further analysis, which will be omitted, establishes two further properties.

1) A MT is orientiation preserving in the sense that, if we traverse a circle in the

order of three distinct points on it,

, the region to the left of the circle

will map to the region to the left of the image circle, with respect to the image

orientation. 2) A MT preserves the property of two points being symmetric with

respect to a circle, i.e., lying on the same ray from the center, and such that the

geometric mean of their distances from the center equals the radius.

Let

be distinct points in

. Then it is easy to see that there is a unique

MT taking these points to 1, 0, and

, respectively, namely

;

We can also handle, as a special case, the possibility that one of the

. We

infer from this and the invertibility of the MTs that given two ordered triples of

distinct points in

, there is a unique MT taking the rst onto the second.

If we have any two disks or halfplanes, or one of each, we can map one to the

other by a MT, and can arrange that the image of any three points on the boundary

of one go to three points on the boundary of the other.

The book makes heavy use of the the notation (

) (cross ratio) for the

image of

under the transformation taking

to 1, 0, and

, respectively,

(

) =

;

From the denition, (

) = (

) for any MT

Example:

;

)

(1 +

) takes the right half plane onto the unit disk.

III. Complex Integration and Applications to Analytic Functions

]

is piecewise smooth, and

is a continuous complex-valued

function dened on the image of

, dene

(

)

(

))

(

)

dt:

This is simply the line integral of

along

. It can be dened analogously to the

Riemann integral as the limit of sums of the form

(

)(

)

(

)

;

(

)], so

is the

Riemann{Stieltjes integral

with respect to

. Using this denition

it can be extended to rectiable paths, i.e., ones for which

is only of bounded

variation.

is a piecewise smooth increasing function from

] onto

] and we let

(so

is the same path as

but with a dierent parametrization), then

(

)

(

)

From the fundamental theorem of calculus we see that if

is any analytic func-

tion on a neighborhood of

, then

(

)

(

)

;

(

) where

(

). If

is a closed path, this is zero.

As a consequence, if

, then

= 0

for all

;

By direct calculation we have

= 2

More generally, let

< r

and consider

;

. Let

be the point on the

circle

;

with arg(

;

) =

(

;

). Clearly, as

, the

point on the circle with the same imaginary part as

and negative real part. Now

let

be the curve running from

;

on the circle. In a neighborhood of

the function Log(

;

) is a primitive of 1

(

;

). So

;

= Log(

;

)

;

Log(

;

) = log

(

;

)

;

log

(

;

)

Taking the limit as

0 gives

;

= 2

Local results and consequences.
Theorem (Cauchy's integral formula for a disk).

Let

be analytic on a

closed disk and let

be a path around the boundary of the disk. Then

(

) = 1

(

)

;

z dw

for all

in the open disk.

Proof.

Consider the function

(

) =

(

;

))

;

dw:

Then

(

) =

(

;

))

and, since

(

) =

(

;

))

is a primitive for

(

;

)),

(

)

Therefore,

(

)

;

z dw

(1) =

(0) =

(

)

;

z dw

= 2

(

)

As an easy corollary, we have:

Cauchy's Theorem for a disk.

Let

be analytic on a closed disk and let

be a

path around the boundary of the disk. Then

(

)

= 0

Proof.

Fix

in the open disk and apply Cauchy's formula to the function

(

)(

;

We can now show that an analytic (i.e., continuously complex dierentiable)

function has a power series expansion around each point in its domain. (And we

already know the converse is true.)

Theorem.

Let

be analytic in

(

)

. Then

(

) =

(

;

)

for

;

< R

for some

Proof.

Fix

and choose

r < R

with

;

< r

. Let

be the path around

;

Then

(

;

)

(

;

)

1, so the Weierstrass M-test implies that the series

(

;

)

(

;

)

= 1

;

converges absolutely and uniformly in

for

;

. Since

is bounded on the

circle, the same sort of convergence holds for

(

)

(

;

)

(

;

)

(

)

;

Therefore we can interchange summation and integraton to get

(

) = 1

(

)

;

z dw

(

;

)

where

(

)

(

;

)

dw:

Corollary (of proof).

Under the same hypotheses

(

)

(

) =

(

)

(

;

)

dw:

Corollary (Cauchy's estimate).

is analytic in

(

)

, then

(

)

(

)

!max

(

)

Proof.

Evident if

is replaced by

r < R

. Then pass to limit.

Corollary (Maximum Modulus Principle).

An analytic function on a region

which achieves a maximum modulus in the region is constant.
Proof.

By the Cauchy estimate for

= 0, if the function achieves its maximum

modulus at a point, it achieves it everywhere on a disk containing the point. Thus,

the set of points were the maximum modulus is achieved is open. It is also clearly

closed. Hence the function has constant modulus. It follows from the Cauchy{

Riemann equations that the function itself is constant.

Corollary (Liouville's Theorem).

A bounded entire function is constant.

Proof.

Cauchy's estimate for

= 1 implies

Corollary (Fundamental Theorem of Algebra).

Every nonconstant polyno-

mial has a root in

Theorem.

is analytic on a region then there does not exist a point

such that

(

)

(

) = 0

for all

, unless

is identically zero on the region.

Proof.

The set of such

is closed, and using the Taylor expansion, is open.

Corollary.

is analytic on a region

and not identically zero, then for each

there exists a unique integer

and a function

, analytic in a neighborhood

and nonzero at

, such that

(

) = (

;

)

(

)

Proof.

(

) =

(

;

)

with

(

)

(

)

!. Let

be the least integer with

= 0 (which exists by the preceeding theorem), and

(

) =

(

;

)

;

The number

is the

multiplicity

as a root of

Corollary.

is analytic on a region and not identically zero, then the roots of

are isolated.

Corollary.

An analytic function on a region is completely determined by its value

on any set of points containing a limit point.

Homotopy of paths and Cauchy's Theorem.
Denition.

Let

: 0

be two closed paths in a region

. We

say that the paths are

homotopic in

if there exists a continuous function ; :

such that for each

;(

) is a closed path with ;(

0) =

and ;(

1) =

This is an equivalence relation, written

Theorem.

Suppose that

and

are two homotopic piecewise smooth closed

curves in

and that

is analytic on

. Then

(

)

(

)

dz:

If we assume that the homotopy ; is

, the result is easy: let

(

) =

(;(

))

;

(

)

dt:

Then

(

) =

(;(

))

;

(

)]

(;(

))

;

(

)]

= 0

since both integrands equal

(;(

))

;

(;(

))

;

@t@s:

The proof without assuming smoothness is based on Cauchy's theorem for a disk

and a polygonal approximation argument.

Sketch of proof.

Since ; is continuouson the compact set 0

1], it is uniformly

continuous. Moreover, there exists

0 so that

contains a closed disk of radius

about each point in the range of ;. Therefore, we can nd an integer

, so that ;

maps each of the squares

j=n

(

+1)

]

k=n

(

+1)

] into the disk of radius

about ;(

j=nk=n

). Let

be the closed polygonal path determined by the points

;(

j=nk=n

= 0

::: n

. It is easy to conclude from Cauchy's theorem for a

disk that

(

)

k+1

(

)

and also that

(

)

(

)

(

)

(

)

dz:

Denition.

A closed path in

homotopic to 0

if it is homotopic to a

constant path.

Corollary.

is a piecewise smooth path which is homotopic to

, and

is analytic in

, then

(

)

= 0

Denition.

A region

is simply-connected if every closed path in

is homotopic

to 0.

As an immediate consequence of the theorem above, we have Cauchy's theorem

(previously proved only for

a disk).

Cauchy's Theorem.

is is a simply-connected region in the complex plane

then

(

)

= 0

for every function

analytic in

and every piecewise smooth closed curve

Winding numbers and Cauchy's Integral Formula.

Intuitive meaning of

winding number: if a pole is planted in the plane at a point

and a closed curve is

drawn in the plane not meeting

, if the curve were made of well-lubricated rubber

and were to be contracted as small as possible, if it contracts to a point other than

its winding number is 0. If it contracts around the pole planted at

, its winding

number is the number of times it circles the pole (signed by orientation).

Denition.

Let

be a piecewise smooth closed curve in

and

. Then

the

index

with respect to

, or the

winding number

about

is dened to

(

) = 1

;

adz:

Demonstrate how the winding number can be calculated by drawing a negatively

directed ray from

and integrating over pieces of the curve between successive

intersections of the curve with the ray.

Proposition.

The index

(

)

is an integer for all

. It is constant on

each connected component of

Proof.

Assuming

is parametrized by 0

1], dene

(

) =

(

)

(

)

;

a ds:

An easy direct calculation shows that

exp

;

(

)]

(

)

;

] = 0

so, indeed, exp

;

(

)]

(

)

;

] =

(0)

;

. Taking

= 1,

exp

;

(

)]

(1)

;

] =

(0)

;

or, exp

;

(

)] = 1, which implies that

(

) is indeed integral.

The function is obviously continuous, and being integer, it is constant on con-

nected components. Clearly also, it tends to 0 as

tends to

, so it is identically

0 on the unbounded component.

The next result is an immediate corollary of the invariance of path integrals of

analytic functions under homotopy.

Proposition.

and

are paths which are homotopic in

for some point

, then

(

) =

(

)

Cauchy's Integral Formula.

Let

a piecewise smooth curve in a region

which

is null homotopic there, and let

be an analytic function on

. Then

(

)

(

) = 1

(

)

;

a dw a

Proof.

For

xed, let

(

) =

(

)

;

(

)

;

a z

(

)

It is easy to show that

is continuous (even analytic!) on

(show the continuity

as a homework exercise). Applying Cauchy's theorem to

gives the desired result

immediately.

Remark.

This presentation is quite a bit dierent from the book's. The book

proves Cauchy's formula under the assumption that

homologous to 0

, i.e., that

(

) = 0 for all

. It is easy to see that homotopic to 0 implies homologous

to 0, but the reverse is not true, so the book's result is a bit stronger. It is true

that all piecewise smooth closed curves in region are homologous to zero i

simply-connected (see Theorem 2.2 of Chapter 8).

The book also goes on to consider chains, which are unions of curves and proves

that

(

)

(

) = 1

(

)

;

z dz

if the chain

is homologous to zero, i.e.,

(

) = 0 for all

. This sort of result is useful, e.g., in the case of an annulus. However we

can deal with this solution in other ways (by adding a line connecting the circles).

Zero counting Open Mapping Theorem.
Theorem.

Let

be a region,

a curve homotopic to

and

an analytic

function on

with zeros

:::

repeated according to multiplicity. Then

(

)

(

)

(

)

Proof.

We can write

(

) = (

;

)(

;

)

:::

(

;

)

(

) where

(

) is analytic

and nonzero on

. Then

(

)

(

) =

;

+ 1

;

(

)

(

)

and the theorem follows.

If the

are the solutions of the equation

(

) =

, then

(

)

(

)

;

(

)

Note that we may as well write the integral as

;

dw:

Theorem.

is analytic at

and

(

) =

with nite multiplicity

, then

there exists

such that for all

;

, the equation

(

) =

has

precisely

roots in

;

and all are simple.

Proof.

That the number of roots is

comes from the continuity of the integral as

long as

remains in the same connected component of

, and the fact

that it is integer valued. We can insure that the roots are simple by taking

small

enough to avoid a root of

Open Mapping Theorem.

A non-constant analytic function is open.

Proof.

This is an immediate corollary.

Morera's Theorem and Goursat's Theorem.
Morera's Theorem.

is a region and

a continuous function such

that

f dz

= 0

for each triangular path

, then

is analytic.

Proof.

Since it is enough to show

is analytic on disks, we can assume

is a disk.

Let

be the center of the disk, and dene

(

) =

]

. By hypothesis

(

)

;

(

)

;

= 1

;

]

(

)

(

;

))

dt:

Continuity then implies that

(

) =

(

), so

has a primitive, so is analytic.

Before proceeding to Goursat's Theorem, we consider another result that uses

an argument similar to Morera's Theorem.

Theorem.

Let

be a region and

a continuous function. Then

admits

a primitive on

if and only if

= 0

for all piecewise smooth closed curves

Proof.

Only if is clear. For the if direction, x

and dene

(

) =

(

)

where

(

) is any piecewise smooth path from

. Reasoning as in Morera's

theorem

In particular, if

is a simply-connected region then every analytic function

admits a primitive. If 0

, then 1

admits a primitive. Adjusting the primitive

by a constant we can assure that it agrees with a logarithm of

at some point

It is then easy to check by dierentiation that exp(

(

))

1. Thus

Proposition.

is a simply connected domain,

, then a branch of the

logarithm may be dened there.

(As an interesting example, take

to be the complement of the spiral

We can at long last show that a function which is complex dierentiable in a

region (but without assuming it is continuously dierentiable) is analytic.

Goursat's Theorem.

Let

be a region and

complex dierentiable at

each point of

. Then

is analytic.

Proof.

We will show that the integral of

vanishes over every triangular path. Let

be a closed triangle with boundary

. Using repeated quadrisection we obtain

triangles

:::

with boundaries

such that

Let

= diam

= perim

. Then diam

, perim

. By

compactness

. Let

be in the intersection (it's unique).

From the dierentiability of

we have: given

(

)

;

(

)

;

(

)(

;

)

;

for all

in some disk around

, therefore for all

, some

. Since

(

)

z dz

= 0,

(

)

;

(

)

;

(

)(

;

)]

diam

perim

dp=

Therefore

, and, since

was arbitrary, it must vanish.

IV. Singularities of Analytic Functions

Denition.

is analytic in

(

)

, for some

R >

0, but not in

(

). We say

has an

isolated singularity

. If there exists

analytic on

some disk centered at

which agrees with

except at

, we say

has a

removable

singularity.

Theorem.

Suppose

has an isolated singularity at

. It is removable if and only

lim

(

;

)

(

) = 0

Proof.

\Only if" is clear. For \if" let

(

) = (

;

)

(

) = 0. Using

Morera's theorem (separating the cases

inside the triangle, on the boundary, or

in the interior), it is easy to see that

is analytic. Therefore

(

) = (

;

)

(

) for

analytic, and

is the desired extension of

Corollary.

lim

(

)

, or even

(

)

is bounded in a punctured neighbor-

hood of

, or even

(

)

;

;(1;

)

for some

, some

, then

has a

removable singularity.

Now either lim

(

)

exists as a nite real number, equals +

, or does not

exist as a nite number or innity. The rst case is that of a removable singularity.

The second is called a

pole

, the third, an

essential singularity

Note that if

has a pole at

, then 1

(

) has a removable singularity there and

extends to

with value 0. Using this we see that

has a pole at

if and only if

(

) = (

;

)

;

(

) for some integer

m >

0 and

analytic in a neighborhood of

. Thus

(

) =

(

;

)

in a punctured neighborhood of

The function

has an essential singularity since

lim

= +

lim

= 0

Theorem (Casorati-Weierstrass).

has an essential singularity at

, then

the image under

of any punctured disk around

is dense in

Proof.

WLOG, assume

= 0. If the conclusion is false, there is a punctured disk

around 0 in which

(

) stays a xed positive distance

away from some number

. Consider the function

(

) =

(

)

;

]

. It tends to

0, so it has

a pole at 0. Therefore

(

)

;

]

0 as

0 for suciently large

. Therefore,

(

)

0, which is not possible if

has an essential singularity.

Laurent series.
Denition.

Suppose

is analytic on

> R

for some

. Then we say

that

has an isolated singularity at

. It is removable if

) has a removable

singularity at 0.

We easily see that

has a removable singularity at

(

)

lim

(

)

(

)

is bounded in a neighborhood of innity.

Note that if

has a removable singularity at

then we can expand

in a power

series in 1

, convergent for

> R

uniformly and absolutely for

> R

We say that a series of the form

=;1

(

;

)

converges if both the series

;

(

;

)

;

and

(

;

)

converge (and their sum is the value of

the doubly innite summation).

Lemma.

Let

be a piecewise smooth curve and

a continuous complex-valued

function on

. Dene

(

) =

(

)

;

wdw

z =

The

is analytic on

Proof.

Fix

z =

. Then

(

)

;

(

)

;

(

)

(

;

)(

;

)

;

(

)

(

;

)

with the convergence uniform in

. Thus

is dierentiable at each

Let

be dened and analytic in an annulus

< R

. Dene

(

) = 1

(

)

;

zdw

where

> r >

(it doesn't matter which such

we take, by Cauchy's theorem).

Then

is analytic in

< R

and so has a convergent power series there:

(

) =

Also, let

(

) =

;

(

)

;

zdw

where now

is taken with

< r <

. Then this is analytic in

> R

and tends

to 0 at

. Thus

(

) =

;

Now, it is easy to use Cauchy's integral formula to see that

in the

annulus. This shows that

(

) =

=;1

with the convergence absolute and uniform on compact subsets of the annulus.

Note that such an expansion is unique, since we can easily recover the

= 1

(

)

where

is any number in (

) (result is independent of choice of

since two

dierent circles are homotopic in the annulus).

Summarizing:

Theorem (Laurent Expansion).

is an analytic function on the annulus

= ann(0

)

for some

< R

, then

(

) =

=;1

for all

in the annulus. The convergence is uniform and absolute on compact

subsets of the annulus and the coecients

are uniquely determined by the formula

above.

has an isolated singularity at

, it has a Laurent expansion in a punctured

disk at

(

) =

=;1

(

;

)

;

< R:

We dene

ord

(

) = inf

= 0

Thus

has a removable singularity

(

)

ord

(

)

0, and it extend to

as 0

(

)

ord

(

)

0, in which case this is the order of the 0. ord

(

) = +

(

)

is identically 0 near

. ord

(

) =

(

)

has an essential singularity at

ord

(

)

0 but nite

(

)

has a pole, in which case

;

ord

(

) is the order of

the pole.

We call

the

residue

, denoted Res(

). Note that, if

has an

isolated singularity at

, then for

suciently small,

Res(

) = 1

;

(

)

dz:

Give pictoral proof that if

is simply-connected domain,

analytic on

except

for isolated singularies, and

is a simple closed curve that encloses singularities at

::: a

(necessarily nite), then

(

)

Res(

)

a simple closed curve in

More precisely:

Theorem (Residue Theorem).

Let

be an open set,

a discrete subset of

and

a null-homotopic piecewise smooth closed curve in

which doesn't intersect

. Then

(

)

= 0

is nite and

(

)

(

)Res(

)

for all functions

which are analytic in

Proof.

Let

: 0

be a homotopy from

to a constant. The image

is compact, so has a nite intersection with

. If

does not belong

to this intersection, the homotopy takes place in the the complement of

, showing

that the

(

) = 0.

Let

:::a

be the points in the intersection, and let

be the singular part of

. The

;

is analytic on

(after removing the removable singularities

at the

), so

;

= 0. The result follows easily.

We will proceed to two easy but useful consequences of the residue theorem,

the argument principle and Rouch%e's theorem. First, however, we consider the

application of the residue theorem to the computation of denite integrals.

Residue integrals.

First, how to practically compute residues. For essential singularities it can be

very dicult, but for poles it is usually easy. If

has a simple pole at

, then

(

) =

(

)

(

;

) with

analytic at

and Res(

) =

(

) (and thus Cauchy's

therem may be viewed as the special case of the residue theorem where

has a

simple pole). If

has a double pole at

, then

(

) =

(

)

(

;

)

(

)

(

;

)

(

)

;

so Res(

) =

(

). Similarly, if

has a pole of order

with

(

) =

(

)

(

;

)

, we expand

as a Taylor series around

and can read o Res(

). (It is

(

;1)!

(

;1)

(

), but it is easier to remember the procedure than the formula.)

Example 1: evaluate

(

)(

)

< a < b:

Use a semicircle centered at the origin in upper half plane and let radius tend to

innity to get the answer of

(

)]. Same technique works for all rational

functions of degree

;

Example 2: evaluate

sin

(

)

dx:

Write as

, with

the unit circle, to get the answer 3

4. This

approach applies to rational functions in sin and cos.

Example 3: evaluate

dt:

Use a branch of the

with the

positive

real axis as the branch cut, and integrate

around this cut and over a large circle centered at the origin (avoid origin with a

small circle). The answer is

. This approach applies to integrals of the form

(

)

where

1) and

(

) is a rational function of degree

;

2 with at

worst a simple pole at 0.

Example 4: evaluate

sin

x dx

We will evaluate

lim

!+1

;

z dz:

The imaginary part of the quantity inside the limit is 2

(sin

)

=xdx

. Note: it

is essential that when taking the limit we excise a symmetric interval about zero.

(The ordinary integral doesn't exist:

has a non-integrable singularity near

the origin, even though its imaginary part, (sin

)

doesn't.) On the other hand,

it is

not

necessary that we cut o the integrand at symmetric points

;

and

(we could have used

;

and

and let them go independently to zero).

We use a somewhat fancy path: starting at

;

we follow the

-axis to

;

, then

a semicircle in the upper-half plane to

, then along the

-axis to

, then along

a vertical segment to

, then back along a horizontal segment to

;

and then down a vertical segment back to

;

. The integrand

has no poles

inside this curve, so its contour integral is 0. It is easy to see that the integral is

bounded by 1

on each of the vertical segments and by 2

;

on the horizontal

segment. On the other hand,

= 1

+ analytic and the integral of 1

over

the semicircle (integrated clockwise) is easily seen to be

;

. We conclude that the

limit above equals

, whence

sin

x dx

Example 5: evaluate

=;1

(

;

)

Use a rectangular path with opposite corners

(

+ 1

2 +

a positive integer

and evaluate

(

;

)

cot

z dz

a =

. The poles of the integrand are at

and

the integers and the residue is

;

sin

(

) and 1

(

;

)

]. Since sin

sin

cosh

cos

sinh

and cos

= cos

cosh

;

sin

sinh

, we get

cot

= cos

+ sinh

sin

+ sinh

so the

cot

equals 1 on the vertical segments and, if

is large, is bounded by 2 on

the horizontal segments. It follows easily that the integral goes to 0 with

, so

=;1

(

;

)

sin

To state the argument principle we dene:

Denition.

A complex function on an open set

is called

meromorphic

if it is

analytic on

except for a set of poles.

Note that the pole set of a meromorphicfunction is discrete (but may be innite).

A meromorphic function is continuous when viewed as taking values in

Suppose

is meromorphic at

(i.e., on a neighborhood of

). Then

Res(

f=f

) = ord

(

). Indeed,

(

) = (

;

)

(

) with

= ord

(

) and

(

)

= 0, and

(

;

) +

. The residue theorem then immediately

gives

Theorem (Argument Principle).

Let

be an open set,

a meromorphic func-

tion on

, and

a null-homotopic piecewise smooth closed curve in

which doesn't

intersect either the zero set

or the pole set

. Then

(

)ord

(

)

where the sum is over

and contains only nitely many terms.

Note that the left hand side is nothing other than

(

0), which is behind

the name (how much does the argument change as we traverse this curve). We saw

this theorem previously in the case where

was analytic, in which case the integral

counts the zeros of

. For meromorphic

we count the poles as negative zeros.

Our nal application of the residue theorem is Rouch%e's theorem.

Theorem (Rouche).

Let

be an open set,

and

meromorphic functions on

and

a null-homotopic piecewise smooth closed curve in

which doesn't intersect

. If

(

) +

(

)

(

)

(

)

then

(

)ord

(

) =

(

)ord

(

)

Proof.

The hypothesis means that on

the quotient

(

)

(

) is not a non-negative

real number. Let

be a branch of the logarithm on

). Then

(

f=g

) is a

primitive of (

f=g

)

(

f=g

) =

;

which is analytic in a neighborhood of

Hence,

and the theorem follows from the argument principle.

Corollary (Classical statement of Rouche's theorem).

Suppose

and

are

analytic on a simply-connected region

and that

is a simple closed piecewise

smooth curve in

. If

, then

has the same number of zeros as

in the region bounded by

Proof.

(

) + (

;

)

;

Example: If

1, then the equation

exp(

;

) = 1 has exactly one solution

in the unit disk (take

exp(

;

1).

V. Further results on analytic functions

The theorems of Weierstrass, Hurwitz, and Montel.

Consider the space of

continuous functions on a region in the complex plane, endowed with the topology

of uniform convergence on compact subsets. (In fact this is a metric space topology:

we can dene a metric such that a sequence of continuous functions converge to

another continuous function in this metric, if and only if the sequence converges

uniformly on compact subsets.) Weierstrass's theorems states that the space of

analytic functions on the region is a closed subspace and dierentiation acts con-

tinuously on it. Montel's theorem identies the precompact subsets of the space of

analytic functions.

Theorem (Weierstrass).

Let

be open,

= 1

:::

, analytic,

. If

uniformly on compact subsets of

, then

is analytic.

Moreover

converges to

uniformly on compact subsets of

Proof.

is a triangular path,

= lim

= 0

is analytic by Morera's theorem.

Let

be the boundary of a disk around

contained in

lim

(

) = lim 1

(

)

(

;

)

= 1

(

)

(

;

)

(

)

Combining Weierstrass's Theorem and the Argument Principle we get:

Theorem (Hurwitz).

Let

be a domain in

analytic functions on

with

uniformly on compact subsets. If each of the

is nowhere-vanishing,

then

is either nowhere-vanishing or identically zero.

Proof.

is not identically zero, but

(

) = 0, we can choose

r >

0 so that

(

)

, and

is the only zero of

on this disk. Letting

be the boundary

of the disk, the argument principle then tells us that

= 0, but

ord

(

)

= 0. But, in view of Weierstrass's Theorem the hypothesis tells us that

converges to

uniformly on

, so we have a contradiction.

Corollary.

Let

be a domain in

analytic functions on

with

uniformly on compact subsets. If each of the

is injective, then

is either injective

or identically constant.
Proof.

is not identically constant, but there exists distinct

with

(

) =

(

then we can choose

r >

0 so that

(

)

(

)

, and

(

)

(

) =

. Now

(

)

;

(

) converges to

(

)

;

(

) uniformly on compact subsets of

(

) and the latter function has a zero but is not identically zero there, so for all

suciently large

(

) =

(

) for some

(

), and, by the same argument,

for some

(

). This contradicts the injectivity of the

Theorem (Montel).

Let

be open,

a family of analytic functions on

. Suppose that the functions

:::

are uniformly bounded on each compact

subset of

. Then there exists a subsequence which converges uniformly on compact

subsets.
Proof.

Fix

and let

(

) = dist(

a@G

)

2. First we show that we can nd a

subsequence that converges uniformly on

(

)

2). Now

(

) =

(

)(

;

)

(

))

where

(

) =

(

)

(

)

!. By the Cauchy estimates

(

)

M=d

(

)

where

is the uniform bound for the

(

)).

Since

(

) is uniformly bounded in

, we can nd a subsequence

(

)

for

which

(

)

) converges as

. Then we can nd a further subsequence so

that

(

)

) converges, as well, etc. For the diagonal sequence

(

)

(

)

converges for all

Claim:

converges uniformly on

(

)

2). To prove this, we must show

that for any

;

(

)

2).

Now, for any

(

)

2),

(

)

;

(

)

(

)

;

(

)

;

(

)

;

(

)

;

(

)

(

)

+ 2

Given

0 we can choose

large enough that

and then choose

and

large

enough that

(

)

;

(

)

+ 1)

(

)

] for 0

. This proves

the claim.

Fix

and let

(

) = dist(

a@G

)

2. First we show that we can nd a

subsequence that converges uniformly on

(

)

2). Now

(

) =

(

)

(

)

! (

;

)

(

))

By the Cauchy estimates

(

)

(

)

(

)

where

is the uniform bound for the

(

)). In particular,

sup

(

)

(

)

for each

Thus we can nd a subsequence

(

)

for which

(

)

(

) converges as

Then we can nd a further subsequence so that

(

)

(

) converges as well, etc.

For the diagonal sequence

(

)

(

)

(

) converges for all

Claim:

converges uniformly on

(

)

2). To prove this, we must show

that for any

;

(

)

2).

Now, for any

(

)

2),

(

)

;

(

)

(

)

(

)

;

(

)

(

)

;

! +

(

)

;

(

)

(

)

;

(

)

(

)

;

! + 2

Given

0 we can choose

large enough that 2

2 and can then

choose

and

large enough that

(

)

(

)

;

(

)

(

)

+1)

(

)

for

. This proves the claim and shows that we can nd a subsequence which

converges uniformly on

(

)

2).

Now we may choose a countable sequence of points

such that

(

)

2) (e.g., all points in

with rational real and imaginary parts).

Given the sequence of functions

uniformly bounded on compact sets, we may nd

a subsequence

(

)

which converges uniformly on

(

)

2). Then we may

take a subsequence of that sequence which converges uniformly on

(

)

2),

etc. The diagonal subsequence

converges uniformly on each

(

)

2).

We complete the proof by showing that

converges uniformly on all compact

subsets. Indeed this is immediate, because any compact subset of

is contained

in a nite union of the

(

)

2).

Schwarz's Lemma.

Schwarz's lemma is an easy, but very useful, consequence

of the maximum modulus principle. The most dicult part is to remember that

the German analyst Herman Schwarz (actually Karl Herman Amandus Schwarz,

1843-1921), inventor of the \Schwarz inequality," \Schwarz lemma," and \Schwarz

alternating method," spells his name without a \t", while the French analyst, Lau-

rent Schwartz, 1915{, inventor of the theory of distributions, uses a \t".

Suppose that an analytic map

maps a disk of radius

about

into the disk

of radius

about

(

). We can use this to estimate the derivative

(

) and the

stretching

(

)

;

(

)

;

. To keep the statement simple, we translate in the

domain and range space so that

(

) = 0 and dilate in the domain and range

so that

= 1. We then get:

Theorem (Schwarz's Lemma).

Suppose

maps the open unit disk into itself

leaving the origin xed. Then

(

)

for all

in the disk, and if equality holds

for any nonzero point, then

(

) =

for some

of modulus 1. Also

(0)

and if equality holds, then

(

) =

for some

of modulus 1.

Proof.

The function

(

) =

(

)

has a removable singularity at the origin, and

extends analytically to the disk with

(0) =

(0). It satises

(

)

the circle of radius

r <

1, so by the maximum modulus theorem it satises this

condition on the disk of radius

. Letting

tend to one gives the result. Equality

implies

is constant.

From a homework exercise (

III .3, no. 10), we know that the most general

Mobius transformations of the open unit disk

into itself is given by

c z

;

with

in the disk and

= 1. We can use Schwarz's lemma to show that this is

most general one-to-one analytic map of the disk onto itself.

First consider the special case

(0) = 0. Then Schwarz shows that

(0)

with equality if and only if

(

) =

some

= 1. Applying Schwarz to

gives

(0)

(

)

(0)

1. Thus equality does indeed hold.

Returning to the general case, say

(

) = 0. Then we can apply the special case

above to

;

to get

(

;

(

)) =

(

) =

(

The Riemann Mapping Theorem.

We begin by deriving some easy conse-

quences of simple-connectivity. Recall that every analytic function on a simply-

connected region admits a primitive. It follows that if

is a nowhere-vanishing

function on a simply-connected domain, then there exists an analytic function

on that domain such that

= exp(

). Indeed, let

be primitive of

. It fol-

lows immediately that

exp(

;

) is constant, and so adjusting

by an additive

constant,

= exp(

). Another consequence is the

square-root property

: if

is a

nowhere-vanishing analytic function on the domain, then there exists an analytic

function

with

(

)]

(

). Indeed, we can just take

= exp(

2). For future

use we remark that if there exists an analytic isomorphism of a region

onto a

region

, then

has the square-root property if and only if

does.

We now turn to a lemma of Koebe.

Lemma.

Let

and let

(

be a region containing the origin and

having the square root property. Then there exists an injective analytic map

such that

(0) = 0

(

)

Proof.

Let

and let

be one of the square roots of

;

. Dene

(

) =

;

(

;

(

)]

)

(where

(

) := (

;

)

;

)). Clearly

(0) = 0, and

is not just multiplication

by a constant (in fact, it is not 1-1 on

), so Schwarz's lemma implies

(

)

Next, note that

is an analytic map of

into

which never vanishes, so admits

an analytic square root

(

), which we may determine uniquely by insisting that

(0) =

. Dening

we have

(0) = 0, and

(

))

. Thus if 0

(

)

= 0 and

(

))

(

)

Two regions in the complex plane are called conformally equivalent if they are

analytically isomorphic, that is, if there exists a one-to-one analytic mapping of the

rst domain onto the second (whose inverse is then automatically analytic).

Riemann Mapping Theorem.

Let

(

be simply-connected. Then

is con-

formally equivalent to the open unit disk.
Proof.

The only consequence of simple-connectivity that we shall use in the proof

is that

has the square-root property.

The structure of the proof is as follows: rst we use the hypotheses to exhibit an

injective analytic map of

onto a domain

which satises the hypotheses

of Koebe's lemma. Then we use an extremal problem to dene an injective analytic

map

onto a domain

contained in the disk. If

lls the entire disk we

are done otherwise,

satises the hypotheses of Koebe's lemma and we use the

lemma to contradict the extremality of

Let

and let

be an analytic square root of

;

. It is easy to

see that

is injective and that if

(

) then

;

w =

(

). Pick

(

)

and then

r >

0 such that

(

)

(

) (Open Mapping Theorem). Therefore

(

;

)

(

) =

. Set

(

) =

(

) +

]

so that

) is

analytic and injective. Then set

(

) =

(

)

;

(

)]

3 where

. This map

is an analytic isomorphism of

onto a connected open set

(

containing 0.

Let

be the set of all injective analytic functions

with

(0) = 0.

Fix 0

, and let

= sup

(

)

. Note

1]. We claim that

the supremum is achieved. Indeed, let

(

)

. Since the

are

uniformly bounded (by 1), Montel's theorem assures us that there is a subsequence

which converges uniformly on compact subsets to some analytic function

. Obviously

(0) = 0 and

(

)

0, so

is not constant, and since the

are injective, so is

. Clearly

1 on

, and by the maximum modulus

principle,

1. Thus

and achieves the supremum.

We complete the theorem by showing that

is an isomorphism of

onto

(so

is the desired analytic isomorphism of

onto

). We already know that

analytic and injective, so we need only show that

maps onto

. If, to the contrary,

(

)

(

then (since

inherits the square-root property from

which

inherits it from

) Koebe's lemma gives an injective analytic function

with

(0) = 0,

(

)

for

= 0. Then

, but

(

))

(

)

, a

contradiction.

We now note that the square-root property is equivalent to simple-connectivity.

Indeed we have seen that simple-connectivity implies the square-root property. On

the other hand, if

satises the square root property, then either

, which

is simply-connected, or

is isomorphic to the disk (by the proof of the Riemann

mapping theorem above), and so is simply-connected.

(

is a simply-connected region and

is arbitrary we may take

any analytic isomorphism of

onto the disk and follow it with a suitably chosen

Mobius transformation (

(

;

)

;

= 1, to obtain an analytic

isomorphism

onto

satisfying

(

) = 0

(

)

and

were two such analytic isomorphisms, then

is an analytic

isomorphism of the disk onto itself with

(0) = 0,

(0)

0, and it easily follows

that

is the identity and

Complements on Conformal Mapping.

There are many interesting questions

in conformal mapping which we will not have time to investigate, but I will quickly

state some key results.

Extension to boundary.

If the boundary of a simply-connected region is a simple

closed curve, then it can be proved that a conformal map of the region onto the

open disk extends to a topological homeomorphism of the closure of the region on

to the closed disk.

Constructive conformal mapping.

Conformal is used to obtain explicit solutions

to problems involving analytic and harmonic functions. For example, Joukowski

used the conformal mapping

where

(

) = 21 +

;

(

) =

;

+ 1

applied to a circle passing through

;

1 and containing 1 in its interior. The result-

ing region, called a Joukowski airfoil, looks like a cross-section of a wing, and by

adjusting the circle, one can adjust the exact shape of the airfoil. Using the explicit

conformal map to the disk and explicit formulas for the solution of Laplace's equa-

tion on (the exterior of) a disk (similar to those that will be derived in the next

section, one can calculate things like the drag and lift of the airfoil, and so this was

a useful approach to airfoil design.

An important class of regions are the polygons. The Schwarz{Christoel formula

is an explicit formula for the conformal map of a given polygon onto the disk. The

formula involves a few complex parameters which are determined by the vertices

of the polygon. Given the vertices, one cannot usually compute the parameters

analytically, but they are not dicult to compute numerically (with a computer).

Multiply connected regions

. For lack of time we will not study conformal mapping

of multiply connected regions, but we just state, without proof, the result for doubly

connected regions (regions whose complement in the Riemann sphere consists of two

connected components, i.e., roughly regions in the plane with one hole). It can be

shown (with basically the tools at our disposal) that any such region is conformally

equivalent to the annulus 1

< R

for some

R >

1. The value of

is uniquely

determined. In particular, two such annuli with dierent values of

are

not

conformally equivalent.

VI. Harmonic Functions

A real-valued

function on an open subset of

is called

harmonic

if its Lapla-

cian vanishes. We have seen previously that the real part of an analytic function

is harmonic and that if the domain is simply-connected every real-valued harmonic

function

admits a harmonic conjugate, i.e., a real-valued function

, for which

is analytic. In particular, we see that every harmonic function is

Suppose

is harmonic on

. If

r >

0 is such that

(

), then we can

choose a disk

with

(

)

, and since

is simply-connected

= Re

for some analytic function

. Cauchy's integral formula then gives

(

) = 1

;

(

)

;

adz

= 12

(

)

Taking real parts we get:

Mean Value Theorem.

is harmonic and

(

)

, then

(MVP)

(

) = 12

(

)

A harmonic function satises the

Maximum Principle

: it does not assume its

maximum on a region unless it is constant there. Indeed the conclusion is true

for any continuous function with the mean-value property (MVP), or even just

satisfying the inequality

(

)

(

)

(since this implies that the

set of points were the maximum is achieved is open, and it is obviously closed). The

reverse inequality similarly shows that a harmonic function satises the Minimum

Principle.

Given a bounded domain

and a function

, the

Dirichlet problem

consists of nding a function

such that

is harmonic on

, continuous

, and coincident with

. An important consequence of the maximum

and minimum principles is uniqueness for the Dirichlet problem. If

and

are

both harmonic on

, continuous on

, and agree on

, then

;

is harmonic

and vanishes on the boundary, but it takes its maximum and minimum on the

boundary, so it is identically zero, i.e.,

When applying the maximum principle to functions which are not known to

extend continuously to the boundary this easy-to-prove consequence is useful: If

satises the maximum principle on a bounded open set and there is a number

such that limsup

(

)

, then

. Proof: Let

= sup

(

)

and take

with

(

)

. Pass to a subsequence with

. If

, then

takes its maximum there, so is constant and obviously

Otherwise

, so

limsup

(

)

lim

(

) =

The Poisson kernel.

For 0

r <

let

(

) = Re

1 +

;

cos

dene the Poisson kernel. Note that

(

) = Re(1 + 2

) =

=;1

Lemma.

For

r <

(

)

is a smooth, positive, even,

-periodic function of

with mean value

. Moreover if

, then

lim

(

) = 0

and if

, then the convergence is uniform over

such that

;

for all

Proof.

Elementary (use the power series expansion for the integral and the formula

(

) =

;

)

+ 2

;

cos

)

for the limits).

Now let

be the open unit disk and

its boundary. The next theorem shows

how to solve the Dirichlet problem for

using the Poisson kernel:

Theorem.

Given

continuous, dene

(

) =

(

)

(

) =

(

;

)

(

)

r <

The

is continuous on

and harmonic on

Proof.

(

) = Re

(

)

1 +

(

;

)

;

(

;

)

= Re

(

)

;

(

) = Re

(

)

;

The integral denes an analytic function of

(since the integrand is continuous

and

and analytic in

), so

, being the real part of an analytic function, is

harmonic.

It remains to show that lim

(

) =

(

) for

. Now

(

)

;

(

) = 12

;

(

;

)

(

)

;

(

)]

so it suces to show that the last integral can be made arbitrarily small by taking

suciently close to

and

r <

1 suciently close to 1. Now given

0 we

can choose

0 so that

(

)

;

(

)

;

. Split the interval of

integration as

;

, the integrand is bounded by

(

;

)

, so the integral is bounded by

(

;

)

(

;

)

Now suppose that

is taken suciently near

, namely

;

2. Then for

;

Thus

(

;

)

0 uniformly for

1. Since

is bounded, the entire

integrand does so as well. Thus if

;

2 we may take

suciently close to

1 that the integral over

is bounded by

It is now easy to deduce that the mean value property characterizes harmonic

functions.

Theorem.

is a continuous function satisfying the mean value prop-

erty, then it is harmonic.
Proof.

It is enough to show that

is harmonic on each disk with closure in

From the Poisson kernel construction we can construct a harmonic function with

the same boundary values as

on this disk. The dierence between this function

and

satises the mean value property and is zero on the boundary of the disk, so

it vanishes on the disk.

Corollary.

If a sequence of harmonic functions converges uniformly on compact

subsets, then the limit function is harmonic.

From the expressions

(

) =

;

(1 +

)

;

(1 + cos

) =

;

)

+ 2

;

cos

)

we have

;

1 +

(

)

1 +

;

Now suppose

is harmonic on

) and non-negative, and

is an arbitrary

point of

). Choose

0 small enough that

r < R

;

, and set

(

) =

((

;

)

is harmonic on a neighborhood of

1), so

(

) =

(

;

)]

) = 12

(

;

)

(

;

)

(

)

dt:

Using the non-negativity of

, the mean value property, and the upper bound for

, we get

(

)

1 +

(

;

)

;

(

;

)

(0)

Letting

0 this gives

(

)

;

(0)

We obtain a lower bound in a similar fashion. Of course we can translate the disk

so it is centered at an arbitrary point

. Thus we have proven:

Theorem (Harnack's Inequalities).

is non-negative and harmonic on

(

)

, then

;

(

)

(

)

;

(

)

for

r < R

For example, if

is harmonic and non-negative on the disk of radius

about

then

(

)

(

)

(

)] on the disk of radius

2 about

Theorem (Harnack).

Let

:::

be a sequence of harmonic functions

on a region

. Then either

lim

(

) = +

uniformly on compact subsets of

or there is an harmonic function

with

lim

(

) =

(

)

uniformly on compact

subsets of

Proof.

For any point

let

be a disk centered at

contained in

. Then

there exists a constant

C >

1 with

(

)

(

)

(

)

(

)

Consequently, the numbers

(

) remain bounded as

if and only if

(

)

remains bounded for all

. This shows that the set of points where

stays

bounded is both open and closed, so it is either the empty set or all of

. In

the rst case,

(

)

for all

, and the above estimate shows that the

convergence is uniform on a covering set of discs, hence on all compact subsets.

In the second case, lim

(

)

for all

, and since

(

)

;

(

)

(

)

;

(

)]

the sequence is uniformly Cauchy on

so converges uniformly on

. Again,

uniform convergence on compact sets follows.

Subharmonic functions and the solution of the Dirichlet problem.
Denition.

Let

be a continuous real-valued function on a region

. If

(

)

(

)

whenever

(

)

, we say that

subharmonic

As mentioned earlier, a subharmonic function satises the maximum principle.

Other simple properties are:

harmonic functions are subharmonic

the sum of two subharmonic functions is subharmonic

the pointwise maximum of two subharmonic functions is subharmonic

The next lemma states that subharmonicity is a local property:

Lemma.

Suppose that

is subharmonic in a neighborhood of each point of a region

. Then

is subharmonic on

Proof.

(

)

, let ~

be the continuous function on

which agrees

with

and is harmonic on

. From the Poisson kernel representation, we

have

(

) = 12

(

)

Now

;

is subharmonic in a neighborhood of each point of

, and it follows by

the usual connectedness argument that the points where it achieves its maximum

is either empty or all of

. Hence its maximum on

is achieved on the

boundary, so is 0. This shows that

(

)

(

) as desired.

One more property we shall use:

Lemma.

Suppose that

is subharmonic in a region

and

(

)

Dene

as the continous function on

which agrees with

and is

harmonic on

, and set

. Then

is subharmonic on

Proof.

Clearly it is continuous and subharmonic in

and

. So it suces to

show that for each point

, ~

is bounded by its mean on any circle around

. Now

;

is subharmonic on

and 0 on

, so

, and therefore on

. Thus

(

) =

(

)

(

)

(

)

Note that

Let

be a bounded region. If

is a harmonic function on

continuous up to the

boundary, and

is any subharmonic function that is less than

on the boundary,

then the maximum principle implies that

. Hence it is reasonable to

try to solve the Dirichlet problem by seeking the largest subharmonic function that

doesn't exceed the given boundary values. This is the approach of O. Perron.

Let

be a bounded (at this point not necessarily continuous)real-valued function

. Dene the Perron family

(

) as the set of subharmonic functions

for which

(

)

limsup

(

)

(

) for all

@G:

Note that

(

) contains all constants not exceeding min

, so it is not empty.

From the maximum principle we obtain:

Lemma.

(

)

then

sup

Now dene the

Perron function

(

) =

(

sup

(

)

(

)

(

)

@G:

We will show that (1) the Perron function is harmonic on

, and (2) under mild

restrictions on

, it extends continuously to

with value

. Thus we will

have exhibited a solution to the Dirichlet problem.

Proof that the Perron function is harmonic.

Let

be any disk with closure in

and

let

be any point in the disk. Choose functions

(

) with

(

)

(

Let

= max(

::: v

), and then dene ~

to be

and to be harmonic

and continuous on

. We know that

and ~

are subharmonic, and clearly

the

and thus the ~

are non-decreasing. All belong to

(

) so are bounded

, but

, so ~

(

)

(

). By Harnack's Theorem, ~

converges to

an harmonic function

. We know that

(

) =

(

). We will now show that

is any other point of

then

(

) =

(

). This will show that

is harmonic in

as desired.

Choose functions

(

) with

(

)

(

), and set

= max(

)

(

). Construct

and ~

from

in analogy with

and ~

. Then

, ~

, and

converges to a harmonic function

with

(

) =

(

). Also

(

) =

(

) =

(

), so

;

is a non-positive harmonic

function on

which achieves its maximum at

. Therefore

, so indeed

(

) =

(

We now turn to the continuity of the Perron solution at the boundary.

Denition.

Let

be a bounded region and

. A continuous function

which is harmonic on

is called a

barrier function

for

if it is positive

and zero at

For example, suppose that there is a line through

such that

lies inside

one of the open half-planes determined by the line. We may write the half-plane as

Im(

;

)

]

0 for some

, and so Im(

;

)

] is a barrier function for

Next consider the function

;

where we use the principal branch of the

square root, dened on

(

0]. This function is analytic on

1] and

extends continuously to 1 with value 0, and its real part is everywhere positive on

1]. Thus if

= 1 belongs to

, and is the only point of intersection of the

interval 0

1] with

, we can use Re

;

as a barrier function for

Similarly, if

is any boundary point for which there exists a line segment which

intersects

only at

, then there is a barrier function for

. Thus barrier

functions exist at all boundary points of smooth domains, polygonal domains, even

on a disk minus a segment, and on many other domains. A standard example of a

domain without a barrier at some boundary point is a punctured disk.

Theorem.

Let

be a bounded domain,

a bounded function on

, and

the

corresponding Perron function. If

is a point in

possessing a barrier function

and

is continuous at

, then

lim

(

) =

(

)

Proof.

First we show that for any

0, there is a harmonic function

continuous on

, which satises

w > f

and

(

) =

(

) +

. Indeed, let

be a disk about

such that

(

)

;

(

)

for

. Let

denote

the maximum of

and

denote to the minimum of the barrier function

, so

m >

0. Set

(

) =

(

)

;

(

)] +

(

) +

(

)

;

(

)] +

Then for

(

)

(

) +

> f

(

)

while for

(

)

> M

(

)

w > f

on the boundary as claimed, and obviously

(

) =

(

) +

It follows that if

(

), then

. Since this is true of all such

we have

, and so limsup

(

)

(

) =

(

) +

. Since

was

arbitrary, limsup

(

)

(

Next let

(

) =

(

)

;

(

)

(

)] =

;

(

)

(

)]

;

Now we get

v < f

and

(

) =

(

)

;

. Since

(

, and

so liminf

(

)

(

)

;

. Letting

tend to zero we have liminf

(

)

(

Corollary.

Let

be a bounded region in

which possesses a barrier at each point

of its boundary, and let

be a continuous function on

. Then the Perron function

for

solves the Dirichlet problem.

The Schwarz Reection Principle.

To state the reection principles, we intro-

duce the following terminology: a region

is symmetric with respect to the real

axis if

implies

Reection Principle for Harmonic Functions.

Let

be a region which is

symmetric with respect to the real axis and dene

;

, and

as the intersec-

tion of

with the upper half-plane, lower half-plane, and real axis, respectively. If

is a continuous real-valued function on

, which is harmonic on

and

zero on

, then

admits a unique extension to a harmonic function on all of

The extension is given by

(

) =

;

(

)

for

;

Proof.

If such an extension exists it is certainly unique, so it suces to show that the

stated extension denes a harmonic function in

. It certainly denes a continuous

extension which is harmonic in

;

, so it suces to show that it is harmonic

in a neighborhood of

. By translating and extending we can assume that

= 0 and that

contains the closed unit disk,

. Dene

as the solution

of the Dirichlet problem with boundary data

(where

has been extended to

;

by the formula above). From the Poisson kernel representation we have

= 0

on the real interval (

;

1). Thus

on the entire boundary of the upper

half-disk, and the entire boundary of the lower half-disk. Since both

and

are

harmonic in the half-disks, they coincide, and thus

, like

is harmonic in the

whole disk.

Reection Principle for Analytic Functions.

Let

;

, and

be as

above. If

is a continuous complex-valued function on

, which is analytic

and real on

, then

admits a unique extension to a harmonic function

on all of

. The extension is given by

(

) =

(

)

for

;

Proof.

We could base a proof on the previous result applied to Im

and harmonic

conjugates, but it is also easy to verify this directly using Morera's theorem.

Of course the line of symmetry could be any line, not just the real axis. Thus

is symmetric with respect to any line, dened and analytic on one side of the

line, and real on the line, it can be extended to be analytic on the whole domain.

The proof can be found by translating and rotating the domain to the standard

case.

We can also translate and rotate the image, so it is not necessary that

be real on

the symmetry line, it is sucient that it map it into some other line. An important

application is to analytic continuation. For example, if

maps a rectangle into

itself analytically and takes the edges into the edges, we can apply the reection

principle repeatedly to extend

to an entire function.

The reection principles can be applied for symmetries with respect to circles as

well as for lines.

Theorem.

Let

be analytic on the unit disk, continuous on its closure, and real on

the boundary. Then

admits a unique extension to the entire plane. The extension

is given by

(

) =

)

for

Proof.

Let

(

) =

((1 +

)

;

)), so

maps the upper half plane into

and

is real on the real axis. Applying the reection principle to

gives an extension of

with

1 +

;

1 +

;

for

in the lower half plane, or, equivalently,

(

) =

)

for

in the exterior of the disk.

As an application of the reection principle consider an analytic function on

the unit disk which extends continuously to the closure taking the unit circle into

itself. We may extend this function to

(

) = 1

) for

1. (We

obtain this by applying the standard reection principle to

where

(

) = (1 +

)

;

).) The extended function is analytic everywhere except at

the points symmetric to the zeros of

, where it has a pole. Innity is either a

removable singularity or a pole according to whether

is 0 at 0 or not. Thus the

extended function is meromorphic on

with a nite number of poles. It follows

that

is rational.

MATH 502

Midterm

March 7, 1997

Professor Arnold

Do 3 problems, omit 1. State which problem you omit. Partial credit will be awarded, but

errors and unacknowledged omissions count negatively.

1. Prove or provide a counterexample (with justication): if

is analytic on

(0 1) and

continuous on its closure, then

extends to an analytic function on

) for some

2. (a) Suppose that

is analytic and nowhere zero on some bounded region of the complex

plane, continuous on its closure, and of constant modulus on its boundary. Prove that

is constant on the region.

(b) Can the hypothesis that

is nowhere zero be dropped?

unbounded?
3. How many roots (counting multiplicity) does the function

(

) =

;

2 have

in the annulus 1

4. Find

sin(exp

)

MATH 502

Midterm

March 7, 1997

Professor Arnold

Do 3 problems, omit 1. State which problem you omit. Partial credit will be awarded, but

errors and unacknowledged omissions count negatively.

1. Prove or provide a counterexample (with justication): if

is analytic on

(0 1) and

continuous on its closure, then

extends to an analytic function on

) for some

It's false. One counterexample is given by

(

) =

n=0

. The radius of convergence

is given by

= limsup

n!1

2=n

= limsup

;2n

log

= 1

so this function is not analytic on any disk strictly containing

(0 1). However

, so the Weierstrass M-test implies that the series converges absolutely and uniformly

on the closed disk

(0 1), and hence

is continuous there. Another counterexample is

simply

(

) =

;

2. (a) Suppose that

is analytic and nowhere zero on some bounded region of the complex

plane, continuous on its closure, and of constant modulus on its boundary. Prove that

is constant on the region.

(b) Can the hypothesis that

is nowhere zero be dropped?

unbounded?
(a) Let

denote the constant value of

on the boundary. Then, by the maximum

modulus principle,

on the region and

is constant if equality holds at any point.

= 0 this shows that

0 which is a contradiction to the hypotheses. Thus

and, since

is nowhere zero, the function 1

is analytic, continuous on the closure, and

of constant modulus 1

on the boundary. Thus

, or

. Thus

, and

is constant.

(b) No. Let

(

) =

on the unit disk.

(

) =

on the right half plane.

3. How many roots (counting multiplicity) does the function

(

) =

;

2 have

in the annulus 1

On the circle

= 1,

= 5 and

;

4, so that

(

) has no zero on

= 1

and, by Rouche's theorem, it has the same number of zeros as 5

inside the circle, namely

2. On

= 2,

= 64 and

+ 5

;

30, so again

has no root on the circle and

Rouche's theorem implies that it has 6 roots inside the circle. Thus

has 4 = 6

;

2 roots

in the annulus.

4. Find

sin(exp

)

, so the integral is

sin(exp

)

where

the unit circle. The integrand is analytic except for the pole at zero, where the residue is

sin(exp0) = sin1. By the Residue Theorem the integral is

sin1 = 2

sin1.