Exercises

12. Is the collection of all values of logi 1/2 the same as the collection of all values of

1

log i ? Explain.

2

13. Is the collection of all values of logi 2 the same as the collection of all values of

2 log i ? Explain.

3.8

14. Find all values of logz 1/2 . (in rectangular form)

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

16. Find the principal value of

a) i i . b) 1 i 4i

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

3.9

Chapter Four

Integration

4.1. Introduction. If : D ‚ C is simply a function on a real interval D , , then the

integral ˜ tdt is, of course, simply an ordered pair of everyday 3 rd grade calculus

integrals:

˜ tdt ˜ xtdt « i ˜ ytdt,

where t xt « iyt. Thus, for example,

1

˜t 2 « 1 « it 3 dt 4 « i.

3 4

0

Nothing really new here. The excitement begins when we consider the idea of an integral

of an honest-to-goodness complex function f : D ‚ C, where D is a subset of the complex

plane. Let™s define the integral of such things; it is pretty much a straight-forward extension

to two dimensions of what we did in one dimension back in Mrs. Turner™s class.

Suppose f is a complex-valued function on a subset of the complex plane and suppose a

and b are complex numbers in the domain of f. In one dimension, there is just one way to

get from one number to the other; here we must also specify a path from a to b. Let C be a

path from a to b, and we must also require that C be a subset of the domain of f.

4.1

(Note we do not even require that a ‚® b; but in case a b, we must specify an orientation

for the closed path C.) Next, let P be a partition of the curve; that is, P z 0 , z 1 , z 2 , µ , z n

is a finite subset of C, such that a z 0 , b z n , and such that z j comes immediately after

z j1 as we travel along C from a to b.

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

n

fz „ z j ,

SP j

j1

where z „ is a point on the arc between z j1 and z j , and z j z j z j1 . (Note that for a

j

given partition P, there are many SP”depending on how the points z „ are chosen.) If

j

there is a number L so that given any 0, there is a partition P of C such that

|SP L|

whenever P ‘ P , then f is said to be integrable on C and the number L is called the

integral of f on C. This number L is usually written ˜ fzdz.

C

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

˜ cfzdz c ˜ fzdz

C C

for any complex constant c.

4.2

˜fz « gzdz ˜ fzdz « ˜ gzdz

C C C

4.2 Evaluating integrals. Now, how on Earth do we ever find such an integral? Let

: , ‚ C be a complex description of the curve C. We partition C by partitioning the

, t 0 t 1 t 2 µ t n .

interval in the usual way: Then

a , t 1 , t 2 , µ , b is partition of C. (Recall we assume that t ‚® 0

for a complex description of a curve C.) A corresponding Riemann sum looks like

n

ft „ t j t j1 .

SP j

j1

We have chosen the points z „ t „ , where t j1 ‚ t „ ‚ t j . Next, multiply each term in the

j j j

sum by 1 in disguise:

n

ft „ t jtj tt j1 t j t j1 .

SP j

j1

j1

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

˜ fzdz ˜ ft tdt.

C

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

Example

We shall find the integral of fz x 2 « y « ixy from a 0 to b 1 « i along three

different paths, or contours, as some call them.

First, let C 1 be the part of the parabola y x 2 connecting the two points. A complex

description of C 1 is 1 t t « it 2 , 0 ‚ t ‚ 1:

4.3

1

0.8

0.6

0.4

0.2

0 0.2 0.4 0.6 0.8 1

x

Now, 1 t 1 « 2ti, and f 1 t t 2 « t 2 « itt 2 2t 2 « it 3 . Hence,

1

˜ fzdz ˜ f 1 t 1 tdt

C1 0

1

˜2t 2 « it 3 1 « 2tidt

0

1

˜2t 2 2t 4 « 5t 3 idt

0

4 « 5i

15 4

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

1

0.8

0.6

0.4

0.2

0 0.2 0.4 0.6 0.8 1

x

Here we have 2 t t « it, 0 ‚ t ‚ 1. Thus, 2 t 1 « i, and our integral looks like

4.4

1

˜ fzdz ˜ f 2 t 2 tdt

C2 0

1

˜t 2 « t « it 2 1 « idt

0

1

˜ t « it « 2t 2 dt

0

1 « 7i

2 6

Finally, let™s integrate along C 3 , the path consisting of the line segment from 0 to 1

together with the segment from 1 to 1 « i.

1

0.8

0.6

0.4

0.2

0 0.2 0.4 0.6 0.8 1

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

Then we have

˜ fzdz ˜ fzdz « ˜ fzdz.

C3 C 31 C 32

For C 31 we have t t, 0 ‚ t ‚ 1. Hence,

1

˜ fzdz ˜ dt 1.

3

C 31 0

For C 32 we have t 1 « it, 0 ‚ t ‚ 1. Hence,

1

˜ fzdz ˜1 « t « ititdt 1 « 5 i.

3 6

C 32 0

4.5

Thus,

˜ fzdz ˜ fzdz « ˜ fzdz

C3 C 31 C 32

5 i.

6

Suppose there is a number M so that |fz| ‚ M for all zC. Then

˜ fzdz ˜ ft tdt

C

‚ ˜|ft t|dt

‚ M ˜| t|dt ML,

where L ˜| t|dt is the length of C.

Exercises

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

C

2. Evaluate ˜ 1

dz, where C is the circle of radius 2 centered at 0 oriented

z

C

counterclockwise.

4. Evaluate ˜ fzdz, where C is the curve y x 3 from 1 i to 1 « i , and

C

for y 0

1

fz .

4y for y ‚ 0

5. Let C be the part of the circle t e it in the first quadrant from a 1 to b i. Find as

small an upper bound as you can for ˜ z 2 z 4 « 5dz .

C

4.6

6. Evaluate ˜ fzdz where fz z « 2 z and C is the path from z 0 to z 1 « 2i

C

consisting of the line segment from 0 to 1 together with the segment from 1 to 1 « 2i.

4.3 Antiderivatives. Suppose D is a subset of the reals and : D ‚ C is differentiable at t.

Suppose further that g is differentiable at t. Then let™s see about the derivative of the

composition gt. It is, in fact, exactly what one would guess. First,

gt uxt, yt « ivxt, yt,

where gz ux, y « ivx, y and t xt « iyt. Then,

d gt u dx « u dy « i v dx « v dy .

x dt y dt x dt y dt

dt

The places at which the functions on the right-hand side of the equation are evaluated are

obvious. Now, apply the Cauchy-Riemann equations:

d gt u dx v dy « i v dx « u dy

x dt x dt x dt x dt

dt

dx « i dy

u « i v

x x dt dt

g t t.

The nicest result in the world!

Now, back to integrals. Let F : D ‚ C and suppose F z fz in D. Suppose moreover

that a and b are in D and that C D is a contour from a to b. Then

˜ fzdz ˜ ft tdt,

C

where : , ‚ C describes C. From our introductory discussion, we know that

d

Ft F t t ft t. Hence,

dt

4.7

˜ fzdz ˜ ft tdt

C

˜ d Ftdt F F

dt

Fb Fa.

This is very pleasing. Note that integral depends only on the points a and b and not at all

on the path C. We say the integral is path independent. Observe that this is equivalent to

saying that the integral of f around any closed path is 0. We have thus shown that if in D

the integrand f is the derivative of a function F, then any integral ˜ fzdz for C D is path

C

independent.

Example

1 i

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

9

x2

˜ z 2 dz.

C

1

This is easy”we know that F z z 2 , where Fz z 3 . Thus,

3

3

˜ z 2 dz 1 1 « i 3 3 « 9

i

3

C

260 728 i

27 2187

Now, instead of assuming f has an antiderivative, let us suppose that the integral of f

between any two points in the domain is independent of path and that f is continuous.

Assume also that every point in the domain D is an interior point of D and that D is

connected. We shall see that in this case, f has an antiderivative. To do so, let z 0 be any

point in D, and define the function F by

˜ fzdz,

Fz

Cz

where C z is any path in D from z 0 to z. Here is important that the integral is path

independent, otherwise Fz would not be well-defined. Note also we need the assumption

that D is connected in order to be sure there always is at least one such path.

4.8

Now, for the computation of the derivative of F:

˜ fsds,

Fz « z Fz

L z

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

Next, observe that ˜ ds z. Thus, fz ˜ fzds, and we have

1

z

L z L z

Fz « z Fz

˜ fs fz ds.

fz 1

z z

L z

Now then,

˜ fs fz ds

1 1 |z| max|fs fz| : sL z

‚

z z

L z

‚ max|fs fz| : sL z .

We know f is continuous at z, and so lim max|fs fz| : sL z 0. Hence,

z‚0

Fz « z Fz

˜ fs fz ds

1

fz lim

lim

z z

z‚0 z‚0

L z

0.

4.9

In other words, F z fz, and so, just as promised, f has an antiderivative! Let™s

summarize what we have shown in this section:

Suppose f : D ‚ C is continuous, where D is connected and every point of D is an interior

point. Then f has an antiderivative if and only if the integral between any two points of D is

path independent.

Exercises

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

˜ cos z dz.

2

C

8. a)Let Fz log z, 0 arg z 2. Show that the derivative F z 1 .

z

7

b)Let Gz log z, 4 arg z 4 . Show that the derivative G z 1 .

z

c)Let C 1 be a curve in the right-half plane D 1 z : Re z ‚ 0 from i to i that does not

pass through the origin. Find the integral

˜ 1 dz.

z

C1

d)Let C 2 be a curve in the left-half plane D 2 z : Re z ‚ 0 from i to i that does not

pass through the origin. Find the integral.

˜ 1 dz.

z

C2

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

˜ 1 dz.

z

C

10. a)Let Hz z c , arg z . Find the derivative H z.

b)Let Kz z c , arg z 7 . What is the largest subset of the plane on which

4 4

Hz Kz?

c)Let C be any path from 1 to 1 that lies completely in the upper half-plane. (Upper

4.10

half-plane z : Im z ‚ 0.) Find

˜ Fzdz,

C

where Fz z i , arg z ‚ .

11. Suppose P is a polynomial and C is a closed curve. Explain how you know that

˜ Pzdz 0.

C

4.11

Chapter Five

Cauchy™s Theorem

5.1. Homotopy. Suppose D is a connected subset of the plane such that every point of D is

an interior point”we call such a set a region”and let C 1 and C 2 be oriented closed curves

in D. We say C 1 is homotopic to C 2 in D if there is a continuous function H : S ‚ D,

where S is the square S t, s : 0 ‚ s, t ‚ 1, such that Ht, 0 describes C 1 and Ht, 1

describes C 2 , and for each fixed s, the function Ht, s describes a closed curve C s in D.

The function H is called a homotopy between C 1 and C 2 . Note that if C 1 is homotopic to

C 2 in D, then C 2 is homotopic to C 1 in D. Just observe that the function

Kt, s Ht, 1 s is a homotopy.

It is convenient to consider a point to be a closed curve. The point c is a described by a

constant function t c. We thus speak of a closed curve C being homotopic to a

constant”we sometimes say C is contractible to a point.

Emotionally, the fact that two closed curves are homotopic in D means that one can be

continuously deformed into the other in D.

Example

Let D be the annular region D z : 1 |z| 5. Suppose C 1 is the circle described by

1 t 2e i2t , 0 ‚ t ‚ 1; and C 2 is the circle described by 2 t 4e i2t , 0 ‚ t ‚ 1. Then

Ht, s 2 « 2se i2t is a homotopy in D between C 1 and C 2 . Suppose C 3 is the same

circle as C 2 but with the opposite orientation; that is, a description is given by

3 t 4e i2t , 0 ‚ t ‚ 1. A homotopy between C 1 and C 3 is not too easy to construct”in

fact, it is not possible! The moral: orientation counts. From now on, the term ”closed

curve” will mean an oriented closed curve.

5.1

Another Example

Let D be the set obtained by removing the point z 0 from the plane. Take a look at the

picture. Meditate on it and convince yourself that C and K are homotopic in D, but and

are homotopic in D, while K and are not homotopic in D.

Exercises

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

homotopic to C 3 in D.

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

3. A region D is said to be simply connected if every closed curve in D is contractible to a

point in D. Prove that any two closed curves in a simply connected region are homotopic in

D.

5.2 Cauchy™s Theorem. Suppose C 1 and C 2 are closed curves in a region D that are

homotopic in D, and suppose f is a function analytic on D. Let Ht, s be a homotopy

between C 1 and C 2 . For each s, the function s t describes a closed curve C s in D. Let

Is be given by

˜ fzdz.

Is

Cs

Then,

5.2

1

˜ fHt, s Ht, s dt.

Is

t

0

Now let™s look at the derivative of Is. We assume everything is nice enough to allow us

to differentiate under the integral:

1

˜ fHt, s Ht, s dt

I s d

t

ds

0

1

2 Ht, s

Ht, s Ht, s

˜

f Ht, s « fHt, s dt

s t st

0

1

2 Ht, s

Ht, s Ht, s

˜

f Ht, s « fHt, s dt

s t ts

0

1

fHt, s Ht, s dt

˜

t s

0

H1, s H0, s

fH1, s fH0, s .

s s

But we know each Ht, s describes a closed curve, and so H0, s H1, s for all s. Thus,

H1, s H0, s

I s fH1, s fH0, s 0,

s s

which means Is is constant! In particular, I0 I1, or

˜ fzdz ˜ fzdz.

C1 C2

This is a big deal. We have shown that if C 1 and C 2 are closed curves in a region D that are

homotopic in D, and f is analytic on D, then ˜ fzdz ˜ fzdz.

C1 C2

An easy corollary of this result is the celebrated Cauchy™s Theorem, which says that if f is

analytic on a simply connected region D, then for any closed curve C in D,

5.3

˜ fzdz 0.

C

In court testimony, one is admonished to tell the truth, the whole truth, and nothing but the

truth. Well, so far in this chapter, we have told the truth and nothing but the truth, but we

have not quite told the whole truth. We assumed all sorts of continuous derivatives in the

preceding discussion. These are not always necessary”specifically, the results can be

proved true without all our smoothness assumptions”think about approximation.

Example

Look at the picture below and convince your self that the path C is homotopic to the closed

path consisting of the two curves C 1 and C 2 together with the line L. We traverse the line

twice, once from C 1 to C 2 and once from C 2 to C 1 .

Observe then that an integral over this closed path is simply the sum of the integrals over

C 1 and C 2 , since the two integrals along L , being in opposite directions, would sum to

zero. Thus, if f is analytic in the region bounded by these curves (the region with two holes

in it), then we know that

˜ fzdz ˜ fzdz « ˜ fzdz.

C C1 C2

Exercises

4. Prove Cauchy™s Theorem.

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

orientation. Find

5.4

˜ 1 dz.

z

S

6. a)Find ˜ 1

dz, where C is any circle centered at z 1 with the usual counterclockwise

z1

C

orientation: t 1 « Ae 2it , 0 ‚ t ‚ 1.

b)Find ˜ z«1 dz, where C is any circle centered at z 1 with the usual counterclockwise

1

C

orientation.

c)Find ˜ 1

dz, where C is the ellipse 4x 2 « y 2 100 with the counterclockwise

z 2 1

C

orientation. [Hint: partial fractions]

d)Find ˜ 1

dz, where C is the circle x 2 10x « y 2 0 with the counterclockwise

z 2 1

C

orientation.

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

C

9. Evaluate ˜ 1

dz where C is the circle described by t e 2it , 0 ‚ t ‚ 1, and n is an

zn

C

integer ‚® 1.

10. a)Does the function fz 1 have an antiderivative on the set of all z ‚® 0? Explain.

z

1

b)How about fz z n , n an integer ‚® 1 ?

2

11. Find as large a set D as you can so that the function fz e z have an antiderivative

on D.

12. Explain how you know that every function analytic in a simply connected (cf. Exercise

3) region D is the derivative of a function analytic in D.

5.5

Chapter Six

More Integration

6.1. Cauchy™s Integral Formula. Suppose f is analytic in a region containing a simple

closed contour C with the usual positive orientation and its inside , and suppose z 0 is inside

C. Then it turns out that

fz

˜

1

fz 0 z z 0 dz.

2i

C

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

Let 0 be any positive number. We know that f is continuous at z 0 and so there is a

number such that |fz fz 0 | whenever |z z 0 | . Now let 0 be a number

such that and the circle C 0 z : |z z 0 | is also inside C. Now, the function

fz

zz 0 is analytic in the region between C and C 0 ; thus

fz fz

˜ ˜

z z 0 dz z z 0 dz.

C C0

We know that ˜ 1

dz 2i, so we can write

zz 0

C0

fz fz

˜ ˜ ˜ 1

z z 0 dz 2ifz 0 z z 0 dz fz 0 z z 0 dz

C0 C0 C0

fz fz 0

˜

z z 0 dz.

C0

For zC 0 we have

|fz fz 0 |

fz fz 0

z z0 |z z 0 |

‚ .

Thus,

6.1

fz fz 0

fz

˜ ˜

z z 0 dz 2ifz 0 z z 0 dz

C0 C0

‚ 2 2.

But is any positive number, and so

fz

˜ z z 0 dz 2ifz 0 0,

C0

or,

fz fz

˜ ˜

1 dz 1

fz 0 z z 0 dz,

z z0

2i 2i

C0 C

which is exactly what we set out to show.

Meditate on this result. It says that if f is analytic on and inside a simple closed curve and

we know the values fz for every z on the simple closed curve, then we know the value for

the function at every point inside the curve”quite remarkable indeed.

Example

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

the integral

˜ cos z dz.

2

z 6z « 5

C

We simply write the integrand as

fz

cos z cos z

,

z1

z 5z 1

z 2 6z « 5

where

fz cos z .

z5

Observe that f is analytic on and inside C, and so,

6.2

fz

˜ ˜

cos z dz dz 2if1

z1

2

z 6z « 5

C C

2i cos 1 i cos 1

15 2

Exercises

1. Suppose f and g are analytic on and inside the simple closed curve C, and suppose

moreover that fz gz for all z on C. Prove that fz gz for all z inside C.

2. Let C be the ellipse 9x 2 « 4y 2 36 traversed once in the counterclockwise direction.

Define the function g by

˜ s 2 « s « 1 ds.

gz sz

C

Find a) gi b) g4i

3. Find

˜ e 2z dz,

z2 4

C

where C is the closed curve in the picture:

4. Find ˜ e 2z

dz, where is the contour in the picture:

z 2 4

6.3

6.2. Functions defined by integrals. Suppose C is a curve (not necessarily a simple closed

curve, just a curve) and suppose the function g is continuous on C (not necessarily analytic,

just continuous). Let the function G be defined by

gs

˜

Gz s z ds

C

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

Consider,

Gz « z Gz

˜

1 1 1

s z gsds

z z s z z

C

gs

˜

ds.

s z zs z

C

Next,

Gz « z Gz gs

˜ ˜ 1 1

ds gsds

z s z zs z

s z 2 s z 2

C C

s z s z z

˜

gsds

s z zs z 2

C

gs

z ˜ ds.

s z zs z 2

C

Now we want to show that

6.4

gs

z ˜ 0.

lim ds

s z zs z 2

z‚0

C

To that end, let M max|gs| : s µ C, and let d be the shortest distance from z to C.

Thus, for s µ C, we have |s z| ‚ d 0 and also

|s z z| ‚ |s z| |z| ‚ d |z|.

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

gs M

‚

s z zs z 2 d |z|d 2

for all s µ C. Finally,

gs

z ˜ M

ds ‚ |z| lengthC,

2

d |z|d 2

s z zs z

C

and it is clear that

gs

z ˜ 0,

lim ds

s z zs z 2

z‚0

C

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

gs

˜

G z ds.

s z 2

C

Truly a miracle!

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

6.5

G z « z G z

˜

1 1 1

gsds

z z s z z 2 s z 2

C

s z 2 s z z 2

˜

1 gsds

z s z z 2 s z 2

C

2s zz z 2

˜

1 gsds

z s z z 2 s z 2

C

2s z z

˜

gsds

s z z 2 s z 2

C

Next,

G z « z G z gs

2˜ ds

z s z 3

C

2s z z

˜ 2

gsds

2 2

s z 3

s z z s z

C

2s z 2 zs z 2s z z 2

˜

gsds

s z z 2 s z 3

C

2s z 2 zs z 2s z 2 « 4zs z 2z 2

˜

gsds

s z z 2 s z 3

C

3zs z 2z 2

˜

gsds

s z z 2 s z 3

C

Hence,

G z « z G z 3zs z 2z 2

gs

2˜ ˜

ds gsds

z s z 3 s z z 2 s z 3

C C

|3m| « 2|z|M

‚ |z| ,

d z 2 d 3

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

G z « z G z gs

2˜ ds 0,

lim

z s z 3

z‚0

C

or in other words,

6.6

gs

G z 2 ˜ ds.

s z 3

C

Suppose f is analytic in a region D and suppose C is a positively oriented simple closed

curve in D. Suppose also the inside of C is in D. Then from the Cauchy Integral formula,

we know that

fs

˜

2ifz s z ds

C

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

fs fs

˜ ˜

1 ds, and f z 2

f z ds

2i 2i

2

s z 3

s z

C C

for all z inside the closed curve C. Meditate on these results. They say that the derivative

of an analytic function is also analytic. Now suppose f is continuous on a domain D in

which every point of D is an interior point and suppose that ˜ fzdz 0 for every closed

C

curve in D. Then we know that f has an antiderivative in D”in other words f is the

derivative of an analytic function. We now know this means that f is itself analytic. We

thus have the celebrated Morera™s Theorem:

If f:D ‚ C is continuous and such that ˜ fzdz 0 for every closed curve in D, then f is

C

analytic in D.

Example

Let™s evaluate the integral

˜ e z dz,

z3

C

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

equation

fs

˜

2

f z ds

2i s z 3

C

6.7

with z 0 and fs e s . Thus,

˜ e z dz.

ie 0 i

z3

C

Exercises

5. Evaluate

˜ sin z dz

z2

C

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

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

a) ˜ b) ˜

1 1

dz dz

z 2 «4 z 2 «4 2

C C

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

f z fz

˜ ˜

z w dz dz

z w 2

C C

for every w C.

8. a) Let be a real constant, and let C be the circle t e it , ‚ t ‚ . Evaluate

e z dz.

˜ z

C

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

˜ e cos t cos sin tdt .

0

6.3. Liouville™s Theorem. Suppose f is entire and bounded; that is, f is analytic in the

entire plane and there is a constant M such that |fz| ‚ M for all z. Then it must be true

that f z 0 identically. To see this, suppose that f w ‚® 0 for some w. Choose R large

enough to insure that M |f w|. Now let C be a circle centered at 0 and with radius

R

6.8

maxR, |w|. Then we have :

fs

˜

M |f w| ‚ 1 dz

2i s w 2

C

‚ 1 M 2 M ,

2 2

a contradiction. It must therefore be true that there is no w for which f w ‚® 0; or, in other

words, f z 0 for all z. This, of course, means that f is a constant function. What we

have shown has a name, Liouville™s Theorem:

The only bounded entire functions are the constant functions.

Let™s put this theorem to some good use. Let pz a n z n « a n1 z n1 «µ «a 1 z « a 0 be a

polynomial. Then

pz a n « a n1 « a n2 «µ « an0

zn.

z z

2

z