The sum of the limits of these two series is frequently written

c j z z 0 j ,

fz

j

where

fs

˜

1

cj , j 0, ‚±1, ‚±2, µ .

2i s z 0 j«1

C

This recipe for fz is called a Laurent series, although it is important to keep in mind that

it is really two series.

9.4

Okay, now let™s derive the above formula. First, let r 1 and r 2 be so that

R 1 r 1 ‚ |z z 0 | ‚ r 2 R 2 and so that the point z and the curve C are included in the

region r 1 ‚ |z z 0 | ‚ r 2 . Also, let be a circle centered at z and such that is included in

this region.

Then fs is an analytic function (of s) on the region bounded by C 1 , C 2 , and , where C 1 is

sz

the circle |z| r 1 and C 2 is the circle |z| r 2 . Thus,

fs fs fs

˜ ˜ ds « ˜ s z ds.

s z ds sz

C2 C1

(All three circles are positively oriented, of course.) But ˜ fs

ds 2ifz, and so we have

sz

fs fs

˜ ˜

2ifz s z ds s z ds.

C2 C1

Look at the first of the two integrals on the right-hand side of this equation. For sC 2 , we

have |z z 0 | |s z 0 |, and so

1 1

s z s z 0 z z 0

1 1

s z0

1 zz 0

sz 0

z z0 j

1

s z0 s z0

j0

1 z z 0 j .

j«1

s z 0

j0

9.5

Hence,

fs fs

˜ ˜ ds z z 0 j

s z ds j«1

s z 0

j0

C2 C2

fs

˜

. ds z z 0 j

j«1

s z 0

j0 C

For the second of these two integrals, note that for sC 1 we have |s z 0 | |z z 0 |, and so

1 1

1 1

s z z z 0 s z 0 z z 0 1 sz 0

zz 0

s z0 j

z 1 0 s z 0 j 1

z z z0 z z 0 j«1

j0 j0

s z 0 j1

1 1 1

s z 0 j«1

j

z z 0 j

z z 0

j1 j1

As before,

fs fs

˜ ˜

s z ds

1

ds

s z 0 j«1 z z 0 j

j1

C1 C2

fs

˜

1

ds

s z 0 j«1 z z 0 j

j1 C

Putting this altogether, we have the Laurent series:

fs fs

˜ ˜

1 1

fz s z ds 2i s z ds

2i

C2 C1

fs fs

˜ ˜

ds z z 0 j «

1 1 1

ds .

2i 2i s z 0 j«1

j«1

z z 0 j

s z 0

j0 j1

C C

Example

9.6

Let f be defined by

1

fz .

zz 1

First, observe that f is analytic in the region 0 |z| 1. Let™s find the Laurent series for f

valid in this region. First,

1 1 « 1 .

fz z z1

zz 1

From our vast knowledge of the Geometric series, we have

fz 1 z j .

z

j0

Now let™s find another Laurent series for f, the one valid for the region 1 |z| .

First,

1 1 1 .

z

z1 1

1 z

Now since | 1 | 1, we have

z

z j z j ,

1 1 1 1

z z

z1 1

1 z j0 j1

and so

fz 1 « 1 1 « z j

z z

z1

j1

z j .

fz

j2

Exercises

8. Find two Laurent series in powers of z for the function f defined by

9.7

1

fz 2

z 1 z

and specify the regions in which the series converge to fz.

9. Find two Laurent series in powers of z for the function f defined by

1

fz

z1 « z 2

and specify the regions in which the series converge to fz.

1

10. Find the Laurent series in powers of z 1 for fz in the region 1 |z 1| .

z

9.8

Chapter Ten

Poles, Residues, and All That

10.1. Residues. A point z 0 is a singular point of a function f if f not analytic at z 0 , but is

analytic at some point of each neighborhood of z 0 . A singular point z 0 of f is said to be

isolated if there is a neighborhood of z 0 which contains no singular points of f save z 0 . In

other words, f is analytic on some region 0 |z z 0 | .

Examples

The function f given by

1

fz 2

zz « 4

has isolated singular points at z 0, z 2i, and z 2i.

Every point on the negative real axis and the origin is a singular point of Log z , but there

are no isolated singular points.

Suppose now that z 0 is an isolated singular point of f . Then there is a Laurent series

c j z z 0 j

fz

j

valid for 0 |z z 0 | R, for some positive R. The coefficient c 1 of z z 0 1 is called the

residue of f at z 0 , and is frequently written

Res f.

zz 0

Now, why do we care enough about c 1 to give it a special name? Well, observe that if C is

any positively oriented simple closed curve in 0 |z z 0 | R and which contains z 0

inside, then

˜ fzdz.

1

c 1

2i

C

10.1

This provides the key to evaluating many complex integrals.

Example

We shall evaluate the integral

˜ e 1/z dz

C

where C is the circle |z| 1 with the usual positive orientation. Observe that the integrand

has an isolated singularity at z 0. We know then that the value of the integral is simply

2i times the residue of e 1/z at 0. Let™s find the Laurent series about 0. We already know

that

1 zj

ez

j!

j0

for all z . Thus,

1 z j 1 « 1 « 1 1 «µ

e 1/z z 2! z 2

j!

j0

The residue c 1 1, and so the value of the integral is simply 2i.

Now suppose we have a function f which is analytic everywhere except for isolated

singularities, and let C be a simple closed curve (positively oriented) on which f is analytic.

Then there will be only a finite number of singularities of f inside C (why?). Call them z 1 ,

z 2 , µ , z n . For each k 1, 2, µ , n, let C k be a positively oriented circle centered at z k and

with radius small enough to insure that it is inside C and has no other singular points inside

it.

10.2

Then,

˜ fzdz ˜ fzdz « ˜ fzdz «µ « ˜ fzdz

C C1 C2 Cn

2i Res f « 2i Res f «µ «2i Res f

zz 1 zz 2 zz n

n

2i Res f.

zz k

k1

This is the celebrated Residue Theorem. It says that the integral of f is simply 2i times

the sum of the residues at the singular points enclosed by the contour C.

Exercises

Evaluate the integrals. In each case, C is the positively oriented circle |z| 2.

˜ e 1/z 2 dz.

1.

C

2. ˜ sin 1 dz.

z

C

3. ˜ cos 1 dz.

z

C

4. ˜ 1

sin 1 dz.

z z

C

5. ˜ 1

cos 1 dz.

z z

C

10.3

10.2. Poles and other singularities. In order for the Residue Theorem to be of much help

in evaluating integrals, there needs to be some better way of computing the

residue”finding the Laurent expansion about each isolated singular point is a chore. We

shall now see that in the case of a special but commonly occurring type of singularity the

residue is easy to find. Suppose z 0 is an isolated singularity of f and suppose that the

Laurent series of f at z 0 contains only a finite number of terms involving negative powers

of z z 0 . Thus,

c n c n«1 «µ « c 1 « c 0 « c 1 z z 0 «µ .

fz «

z z 0 n z z 0

z z 0 n1

Multiply this expression by z z 0 n :

¤z z z 0 n fz c n « c n«1 z z 0 «µ «c 1 z z 0 n1 «µ .

What we see is the Taylor series at z 0 for the function ¤z z z 0 n fz. The coefficient

of z z 0 n1 is what we seek, and we know that this is

¤ n1 z 0

n 1!

.

The sought after residue c 1 is thus

¤ n1 z 0

Res f

c 1 ,

n 1!

zz 0

where ¤z z z 0 n fz.

Example

We shall find all the residues of the function

ez

fz .

z 2 z 2 « 1

First, observe that f has isolated singularities at 0, and ‚±i. Let™s see about the residue at 0.

Here we have

10.4

ez .

¤z z 2 fz

z 2 « 1

The residue is simply ¤ 0 :

z 2 « 1e z 2ze z

¤ z .

z 2 « 1 2

Hence,

Res f ¤ 0 1.

z0

Next, let™s see what we have at z i:

ez

¤z z ifz ,

z 2 z « i

and so

i

Res fz ¤i e .

2i

zi

In the same way, we see that

i

Res f e .

2i

zi

Let™s find the integral ˜ ez

dz , where C is the contour pictured:

z 2 z 2 «1

C

10.5

This is now easy. The contour is positive oriented and encloses two singularities of f; viz, i

and i. Hence,

˜ ez dz 2i Res f «Res f

z 2 z 2 « 1 zi zi

C

i

i

2i e « e

2i 2i

2i sin 1.

Miraculously easy!

There is some jargon that goes with all this. An isolated singular point z 0 of f such that the

Laurent series at z 0 includes only a finite number of terms involving negative powers of

z z 0 is called a pole. Thus, if z 0 is a pole, there is an integer n so that ¤z z z 0 n fz

is analytic at z 0 , and fz 0 ‚® 0. The number n is called the order of the pole. Thus, in the

preceding example, 0 is a pole of order 2, while i and i are poles of order 1. (A pole of

order 1 is frequently called a simple pole.) We must hedge just a bit here. If z 0 is an

isolated singularity of f and there are no Laurent series terms involving negative powers of

z z 0 , then we say z 0 is a removable singularity.

Example

Let

fz sin z ;

z

then the singularity z 0 is a removable singularity:

3 5

fz 1 sin z 1 z z « z µ

z z 3! 5!

2 4

1 z « z µ

3! 5!

and we see that in some sense f is ”really” analytic at z 0 if we would just define it to be

the right thing there.

A singularity that is neither a pole or removable is called an essential singularity.

Let™s look at one more labor-saving trick”or technique, if you prefer. Suppose f is a

function:

10.6

pz

fz ,

qz

where p and q are analytic at z 0 , and we have qz 0 0, while q z 0 ‚® 0, and pz 0 ‚® 0.

Then

pz 0 « p z 0 z z 0 «µ

pz

fz ,

qz q z 0

2

q z 0 z z 0 « 2 z z 0 µ

and so

pz 0 « p z 0 z z 0 «µ

¤z z z 0 fz .

q z 0

q z 0 « 2 z z 0 «µ

Thus z 0 is a simple pole and

pz 0

Res f ¤z 0 .

q z 0

zz 0

Example

Find the integral

˜ cos z dz,

ez 1

C

where C is the rectangle with sides x ‚±1, y , and y 3.

The singularities of the integrand are all the places at which e z 1, or in other words, the

points z 0, ‚±2i, ‚±4i, µ . The singularities enclosed by C are 0 and 2i. Thus,

˜ cos z dz 2i Res f «Res f ,

ez 1 z0 z2i

C

where

fz cos z .

ez 1

10.7

pz

Observe this is precisely the situation just discussed: fz , where p and q are

qz

analytic, etc.,etc. Now,

pz

cos z .

ez

q z

Thus,

Res f cos 0 1, and

1

z0

2 2

Res f cos2i e « e cosh 2.

2i

2

e

z2i

Finally,

˜ cos z dz 2i Res f «Res f

ez 1 z0 z2i

C

2i1 « cosh 2

Exercises

6. Suppose f has an isolated singularity at z 0 . Then, of course, the derivative f also has an

isolated singularity at z 0 . Find the residue Res f .

zz 0

7. Given an example of a function f with a simple pole at z 0 such that Res f 0, or explain

zz 0

carefully why there is no such function.

8. Given an example of a function f with a pole of order 2 at z 0 such that Res f 0, or

zz 0

explain carefully why there is no such function.

9. Suppose g is analytic and has a zero of order n at z 0 (That is, gz z z 0 n hz, where

hz 0 ‚® 0.). Show that the function f given by

1

fz

gz

10.8

has a pole of order n at z 0 . What is Res f ?

zz 0

10. Suppose g is analytic and has a zero of order n at z 0 . Show that the function f given by

g z

fz

gz

has a simple pole at z 0 , and Res f n.

zz 0

11. Find

˜ cos z dz,

z2 4

C

where C is the positively oriented circle |z| 6.

12. Find

˜ tan zdz,

C

where C is the positively oriented circle |z| 2.

13. Find

˜ 1 dz,

2

z «z«1

C

where C is the positively oriented circle |z| 10.

10.9

Chapter Eleven

Argument Principle

11.1. Argument principle. Let C be a simple closed curve, and suppose f is analytic on C.

Suppose moreover that the only singularities of f inside C are poles. If fz ‚® 0 for all zC,

then fC is a closed curve which does not pass through the origin. If

t, ‚ t ‚

is a complex description of C, then

‘t ft, ‚ t ‚

is a complex description of . Now, let™s compute

f t

f z

˜ ˜

dz tdt.

fz ft

C

But notice that ‘ t f t t. Hence,

‘ t

f t

f z

˜ ˜ ˜

dz tdt dt

‘t

fz ft

C

˜ 1 dz n2i,

z

where |n| is the number of times ”winds around” the origin. The integer n is positive in

case is traversed in the positive direction, and negative in case the traversal is in the

negative direction.

f z

Next, we shall use the Residue Theorem to evaluate the integral ˜ dz. The singularities

fz

C

f z

of the integrand are the poles of f together with the zeros of f. Let™s find the residues at

fz

these points. First, let Z z 1 , z 2 , µ , z K be set of all zeros of f. Suppose the order of the

zero z j is n j . Then fz z z j n j hz and hz j ‚® 0. Thus,

11.1

z z j n j h z « n j z z j n j 1 hz

f z

z z j n j hz

fz

nj

h z

« .

z z j

hz

Then

f z h z

¤z z z j z z j « n j,

fz hz

and

Res f n j .

f

zz j

The sum of all these residues is thus

N n 1 « n 2 «µ «n K .

Next, we go after the residues at the poles of f. Let the set of poles of f be

P p 1 , p 2 , µ , p J . Suppose p j is a pole of order m j . Then

hz z p j m j fz

is analytic at p j . In other words,

hz

fz .

z p j m j

Hence,

z p j m j h z m j z p j m j 1 hz z p j m j

f z

fz hz

z p j 2m j

mj

h z

.

z p j m j

hz

Now then,

11.2

f z h z

¤z z p j m j z p j m j mj,

fz hz

and so

Res f ¤p j m j .

zp j f

The sum of all these residues is

P m 1 m 2 µ m J

Then,

f z

˜ dz 2iN P ;

fz

C

and we already found that

f z

˜ dz n2i,

fz

C

where n is the ”winding number”, or the number of times winds around the

origin”n 0 means winds in the positive sense, and n negative means it winds in the

negative sense. Finally, we have

n N P,

where N n 1 « n 2 «µ «n K is the number of zeros inside C, counting multiplicity, or the

order of the zeros, and P m 1 « m 2 «µ «m J is the number of poles, counting the order.

This result is the celebrated argument principle.

Exercises

1. Let C be the unit circle |z| 1 positively oriented, and let f be given by

11.3

fz z 3 .

How many times does the curve fC wind around the origin? Explain.

2. Let C be the unit circle |z| 1 positively oriented, and let f be given by

2

fz z « 2 .

z3

How many times does the curve fC wind around the origin? Explain.

3. Let pz a n z n « a n1 z n1 «µ «a 1 z « a 0 , with a n ‚® 0. Prove there is an R 0 so that if

C is the circle |z| R positively oriented, then

p z

˜ dz 2ni.

pz

C

4. How many solutions of 3e z z 0 are in the disk |z| ‚ 1? Explain.

5. Suppose f is entire and fz is real if and only if z is real. Explain how you know that f

has at most one zero.

11.2 Rouche™s Theorem. Suppose f and g are analytic on and inside a simple closed

contour C. Suppose moreover that |fz| |gz| for all zC. Then we shall see that f and

f « g have the same number of zeros inside C. This result is Rouche™s Theorem. To see

why it is so, start by defining the function t on the interval 0 ‚ t ‚ 1 :

f z « tg t

˜

t 1 dz.

2i fz « tgz

C

Observe that this is okay”that is, the denominator of the integrand is never zero:

|fz « tgz| ‚ ||ft| t|gt|| ‚ ||ft| |gt|| 0.

Observe that is continuous on the interval 0, 1 and is integer-valued”t is the

11.4

number of zeros of f « tg inside C. Being continuous and integer-valued on the connected

set 0, 1, it must be constant. In particular, 0 1. This does the job!

f z

˜

0 1 dz

2i fz

C

is the number of zeros of f inside C, and

f z « g z

˜

1 1 dz

2i fz « gz

C

is the number of zeros of f « g inside C.

Example

How many solutions of the equation z 6 5z 5 « z 3 2 0 are inside the circle |z| 1?

Rouche™s Theorem makes it quite easy to answer this. Simply let fz 5z 5 and let

gz z 6 « z 3 2. Then |fz| 5 and |gz| ‚ |z| 6 « |z| 3 « 2 4 for all |z| 1. Hence

|fz| |gz| on the unit circle. From Rouche™s Theorem we know then that f and f « g

have the same number of zeros inside |z| 1. Thus, there are 5 such solutions.

The following nice result follows easily from Rouche™s Theorem. Suppose U is an open set

(i.e., every point of U is an interior point) and suppose that a sequence f n of functions

analytic on U converges uniformly to the function f. Suppose further that f is not zero on

the circle C z : |z z 0 | R U. Then there is an integer N so that for all n ‚ N, the

functions f n and f have the same number of zeros inside C.

This result, called Hurwitz™s Theorem, is an easy consequence of Rouche™s Theorem.

Simply observe that for zC, we have |fz| 0 for some . Now let N be large enough

to insure that |f n z fz| on C. It follows from Rouche™s Theorem that f and

f « f n f f n have the same number of zeros inside C.

Example

2 n

On any bounded set, the sequence f n , where f n z 1 « z « z2 «µ « z , converges

n!

uniformly to fz e z , and fz ‚® 0 for all z. Thus for any R, there is an N so that for

2 n

n N, every zero of 1 « z « z2 «µ « z has modulus R. Or to put it another way, given

n!

2 n

an R there is an N so that for n N no polynomial 1 « z « z2 «µ « z has a zero inside the

n!

11.5

circle of radius R.

Exercises

6. Show that the polynomial z 6 « 4z 2 1 has exactly two zeros inside the circle |z| 1.

7. How many solutions of 2z 4 2z 3 « 2z 2 2z « 9 0 lie inside the circle |z| 1?

8. Use Rouche™s Theorem to prove that every polynomial of degree n has exactly n zeros

(counting multiplicity, of course).

9. Let C be the closed unit disk |z| ‚ 1. Suppose the function f analytic on C maps C into

the open unit disk |z| 1”that is, |fz| 1 for all zC. Prove there is exactly one wC

such that fw w. (The point w is called a fixed point of f .)

11.6