<< стр. 2(всего 4)СОДЕРЖАНИЕ >>

Exercises

12. Is the collection of all values of logпѓќi 1/2 пѓћ the same as the collection of all values of
1
log i ? Explain.
2

13. Is the collection of all values of logпѓќi 2 пѓћ the same as the collection of all values of
2 log i ? Explain.

3.8
14. Find all values of logпѓќz 1/2 пѓћ. (in rectangular form)

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

16. Find the principal value of
a) i i . b) пѓќ1 пЂї iпѓћ 4i

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

3.9
Chapter Four

Integration

4.1. Introduction. If пЃЊ : D п‚ё C is simply a function on a real interval D пЂЅ пѓџпЃЉ, пЃ‹пѓ  , then the
пЃ‹
integral пЃ˜ пЃЊпѓќtпѓћdt is, of course, simply an ordered pair of everyday 3 rd grade calculus
пЃЉ
integrals:

пЃ‹ пЃ‹ пЃ‹

пЃ˜ пЃЊпѓќtпѓћdt пЂЅ пЃ˜ xпѓќtпѓћdt пЂ« i пЃ˜ yпѓќtпѓћdt,
пЃЉ пЃЉ пЃЉ

where пЃЊпѓќtпѓћ пЂЅ xпѓќtпѓћ пЂ« iyпѓќtпѓћ. Thus, for example,

1

пЃ˜пѓџпѓќt 2 пЂ« 1пѓћ пЂ« it 3 пѓ dt пЂЅ 4 пЂ« 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 п‚ё C, where D is a subset of the complex
plane. LetвЂ™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. TurnerвЂ™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 п‚® b; but in case a пЂЅ b, we must specify an orientation
for the closed path C.) Next, let P be a partition of the curve; that is, P пЂЅ пѓЎz 0 , z 1 , z 2 , пЃµ , z n пѓў
is a finite subset of C, such that a пЂЅ z 0 , b пЂЅ z n , and such that z j comes immediately after
z jпЂї1 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

пЂѕ fпѓќz пЃ„ пѓћпЃЃz j ,
SпѓќPпѓћ пЂЅ j
jпЂЅ1

where z пЃ„ is a point on the arc between z jпЂї1 and z j , and пЃЃz j пЂЅ z j пЂї z jпЂї1 . (Note that for a
j
given partition P, there are many SпѓќPпѓћвЂ”depending on how the points z пЃ„ are chosen.) If
j
there is a number L so that given any пЃђ пЂѕ 0, there is a partition P пЃђ of C such that

|SпѓќPпѓћ пЂї L| пЂј пЃђ

whenever P пѓ‘ P пЃђ , 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 пЃ˜ fпѓќzпѓћdz.
C

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

пЃ˜ cfпѓќzпѓћdz пЂЅ c пЃ˜ fпѓќzпѓћdz
C C

for any complex constant c.

4.2
пЃ˜пѓќfпѓќzпѓћ пЂ« gпѓќzпѓћпѓћdz пЂЅ пЃ˜ fпѓќzпѓћdz пЂ« пЃ˜ gпѓќzпѓћdz
C C C

4.2 Evaluating integrals. Now, how on Earth do we ever find such an integral? Let
пЃЊ : пѓџпЃЉ, пЃ‹пѓ  п‚ё C be a complex description of the curve C. We partition C by partitioning the
пѓџпЃЉ, пЃ‹пѓ  пЃЉ пЂЅ t 0 пЂј t 1 пЂј t 2 пЂјпЃµ пЂј t n пЂЅ пЃ‹.
interval in the usual way: Then
пѓЎa пЂЅ пЃЊпѓќпЃЉпѓћ, пЃЊпѓќt 1 пѓћ, пЃЊпѓќt 2 пѓћ, пЃµ , пЃЊпѓќпЃ‹пѓћ пЂЅ bпѓў is partition of C. (Recall we assume that пЃЊ пЃ¶ пѓќtпѓћ п‚® 0
for a complex description of a curve C.) A corresponding Riemann sum looks like

n

пЂѕ fпѓќпЃЊпѓќt пЃ„ пѓћпѓћпѓќпЃЊпѓќt j пѓћ пЂї пЃЊпѓќt jпЂї1 пѓћпѓћ.
SпѓќPпѓћ пЂЅ j
jпЂЅ1

We have chosen the points z пЃ„ пЂЅ пЃЊпѓќt пЃ„ пѓћ, where t jпЂї1 п‚І t пЃ„ п‚І t j . Next, multiply each term in the
j j j
sum by 1 in disguise:

n

пЂѕ fпѓќпЃЊпѓќt пЃ„ пѓћпѓћпѓќ пЃЊпѓќt jtпѓћj пЂї tпЃЊпѓќt jпЂї1 пѓћ пѓћпѓќt j пЂї t jпЂї1 пѓћ.
пЂї
SпѓќPпѓћ пЂЅ j
jпЂї1
jпЂЅ1

I hope it is now reasonably convincing that вЂќin the limitвЂќ, we have

пЃ‹

пЃ˜ fпѓќzпѓћdz пЂЅ пЃ˜ fпѓќпЃЊпѓќtпѓћпѓћпЃЊ пЃ¶ пѓќtпѓћdt.
C пЃЉ

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

Example

We shall find the integral of fпѓќzпѓћ пЂЅ пѓќx 2 пЂ« yпѓћ пЂ« iпѓќxyпѓћ from a пЂЅ 0 to b пЂЅ 1 пЂ« i along three
different paths, or contours, as some call them.

First, let C 1 be the part of the parabola y пЂЅ x 2 connecting the two points. A complex
description of C 1 is пЃЊ 1 пѓќtпѓћ пЂЅ t пЂ« it 2 , 0 п‚І t п‚І 1:

4.3
1

0.8

0.6

0.4

0.2

0 0.2 0.4 0.6 0.8 1
x

Now, пЃЊ пЃ¶1 пѓќtпѓћ пЂЅ 1 пЂ« 2ti, and fпѓќ пЃЊ 1 пѓќtпѓћпѓћ пЂЅ пѓќt 2 пЂ« t 2 пѓћ пЂ« itt 2 пЂЅ 2t 2 пЂ« it 3 . Hence,

1

пЃ˜ fпѓќzпѓћdz пЂЅ пЃ˜ fпѓќ пЃЊ 1 пѓќtпѓћпѓћпЃЊ пЃ¶1 пѓќtпѓћdt
C1 0
1

пЂЅ пЃ˜пѓќ2t 2 пЂ« it 3 пѓћпѓќ1 пЂ« 2tiпѓћdt
0
1

пЂЅ пЃ˜пѓќ2t 2 пЂї 2t 4 пЂ« 5t 3 iпѓћdt
0

пЂЅ 4 пЂ« 5i
15 4

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

1

0.8

0.6

0.4

0.2

0 0.2 0.4 0.6 0.8 1
x

Here we have пЃЊ 2 пѓќtпѓћ пЂЅ t пЂ« it, 0 п‚І t п‚І 1. Thus, пЃЊ пЃ¶2 пѓќtпѓћ пЂЅ 1 пЂ« i, and our integral looks like

4.4
1

пЃ˜ fпѓќzпѓћdz пЂЅ пЃ˜ fпѓќ пЃЊ 2 пѓќtпѓћпѓћпЃЊ пЃ¶2 пѓќtпѓћdt
C2 0
1

пЂЅ пЃ˜пѓџпѓќt 2 пЂ« tпѓћ пЂ« it 2 пѓ пѓќ1 пЂ« iпѓћdt
0
1

пЃ˜ пѓџt пЂ« iпѓќt пЂ« 2t 2 пѓћпѓ dt
пЂЅ
0

пЂЅ 1 пЂ« 7i
2 6

Finally, letвЂ™s integrate along C 3 , the path consisting of the line segment from 0 to 1
together with the segment from 1 to 1 пЂ« 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 пЂ« i.
Then we have

пЃ˜ fпѓќzпѓћdz пЂЅ пЃ˜ fпѓќzпѓћdz пЂ« пЃ˜ fпѓќzпѓћdz.
C3 C 31 C 32

For C 31 we have пЃЊпѓќtпѓћ пЂЅ t, 0 п‚І t п‚І 1. Hence,

1

пЃ˜ fпѓќzпѓћdz пЂЅ пЃ˜ dt пЂЅ 1.
3
C 31 0

For C 32 we have пЃЊпѓќtпѓћ пЂЅ 1 пЂ« it, 0 п‚І t п‚І 1. Hence,

1

пЃ˜ fпѓќzпѓћdz пЂЅ пЃ˜пѓќ1 пЂ« t пЂ« itпѓћitdt пЂЅ пЂї 1 пЂ« 5 i.
3 6
C 32 0

4.5
Thus,

пЃ˜ fпѓќzпѓћdz пЂЅ пЃ˜ fпѓќzпѓћdz пЂ« пЃ˜ fпѓќzпѓћdz
C3 C 31 C 32

пЂЅ 5 i.
6

Suppose there is a number M so that |fпѓќzпѓћ| п‚І M for all zпЃЏC. Then

пЃ‹

пЃ˜ fпѓќzпѓћdz пЃ˜ fпѓќпЃЊпѓќtпѓћпѓћпЃЊ пЃ¶ пѓќtпѓћdt
пЂЅ
C пЃЉ
пЃ‹

п‚І пЃ˜|fпѓќпЃЊпѓќtпѓћпѓћпЃЊ пЃ¶ пѓќtпѓћ|dt
пЃЉ
пЃ‹

п‚І M пЃ˜|пЃЊ пЃ¶ пѓќtпѓћ|dt пЂЅ ML,
пЃЉ
пЃ‹
where L пЂЅ пЃ˜|пЃЊ пЃ¶ пѓќtпѓћ|dt is the length of C.
пЃЉ

Exercises

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

2. Evaluate пЃ˜ 1
dz, where C is the circle of radius 2 centered at 0 oriented
z
C
counterclockwise.

4. Evaluate пЃ˜ fпѓќzпѓћdz, where C is the curve y пЂЅ x 3 from пЂї1 пЂї i to 1 пЂ« i , and
C

for y пЂј 0
1
fпѓќzпѓћ пЂЅ .
4y for y п‚і 0

5. Let C be the part of the circle пЃЊпѓќtпѓћ пЂЅ e it in the first quadrant from a пЂЅ 1 to b пЂЅ i. Find as
small an upper bound as you can for пЃ˜ пѓќz 2 пЂї z 4 пЂ« 5пѓћdz .
C

4.6
6. Evaluate пЃ˜ fпѓќzпѓћdz where fпѓќzпѓћ пЂЅ z пЂ« 2 z and C is the path from z пЂЅ 0 to z пЂЅ 1 пЂ« 2i
C
consisting of the line segment from 0 to 1 together with the segment from 1 to 1 пЂ« 2i.

4.3 Antiderivatives. Suppose D is a subset of the reals and пЃЊ : D п‚ё C is differentiable at t.
Suppose further that g is differentiable at пЃЊпѓќtпѓћ. Then letвЂ™s see about the derivative of the
composition gпѓќпЃЊпѓќtпѓћпѓћ. It is, in fact, exactly what one would guess. First,

gпѓќпЃЊпѓќtпѓћпѓћ пЂЅ uпѓќxпѓќtпѓћ, yпѓќtпѓћпѓћ пЂ« ivпѓќxпѓќtпѓћ, yпѓќtпѓћпѓћ,

where gпѓќzпѓћ пЂЅ uпѓќx, yпѓћ пЂ« ivпѓќx, yпѓћ and пЃЊпѓќtпѓћ пЂЅ xпѓќtпѓћ пЂ« iyпѓќtпѓћ. Then,

d gпѓќпЃЊпѓќtпѓћпѓћ пЂЅ пЂЇu dx пЂ« пЂЇu dy пЂ« i пЂЇv dx пЂ« пЂЇv dy .
пЂЇx dt пЂЇy dt пЂЇx dt пЂЇy 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 gпѓќпЃЊпѓќtпѓћпѓћ пЂЅ пЂЇu dx пЂї пЂЇv dy пЂ« i пЂЇv dx пЂ« пЂЇu dy
пЂЇx dt пЂЇx dt пЂЇx dt пЂЇx dt
dt
dx пЂ« i dy
пЂЅ пЂЇu пЂ« i пЂЇv
пЂЇx пЂЇx dt dt
пЂЅ g пЃ¶ пѓќпЃЊпѓќtпѓћпѓћпЃЊ пЃ¶ пѓќtпѓћ.

The nicest result in the world!

Now, back to integrals. Let F : D п‚ё C and suppose F пЃ¶ пѓќzпѓћ пЂЅ fпѓќzпѓћ in D. Suppose moreover
that a and b are in D and that C пѓђ D is a contour from a to b. Then

пЃ‹

пЃ˜ fпѓќzпѓћdz пЂЅ пЃ˜ fпѓќпЃЊпѓќtпѓћпѓћпЃЊ пЃ¶ пѓќtпѓћdt,
C пЃЉ

where пЃЊ : пѓџпЃЉ, пЃ‹пѓ  п‚ё C describes C. From our introductory discussion, we know that
d
FпѓќпЃЊпѓќtпѓћпѓћ пЂЅ F пЃ¶ пѓќпЃЊпѓќtпѓћпѓћпЃЊ пЃ¶ пѓќtпѓћ пЂЅ fпѓќпЃЊпѓќtпѓћпѓћпЃЊ пЃ¶ пѓќtпѓћ. Hence,
dt

4.7
пЃ‹

пЃ˜ fпѓќzпѓћdz пЂЅ пЃ˜ fпѓќпЃЊпѓќtпѓћпѓћпЃЊ пЃ¶ пѓќtпѓћdt
C пЃЉ
пЃ‹

пЃ˜ d FпѓќпЃЊпѓќtпѓћпѓћdt пЂЅ FпѓќпЃЊпѓќпЃ‹пѓћпѓћ пЂї FпѓќпЃЊпѓќпЃЉпѓћпѓћ
пЂЅ
dt
пЃЉ

пЂЅ Fпѓќbпѓћ пЂї Fпѓќaпѓћ.

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 пЃ˜ fпѓќzпѓћdz for C пѓђ D is path
C
independent.

Example

1 i
Let C be the curve y пЂЅ from the point z пЂЅ 1 пЂ« i to the point z пЂЅ 3 пЂ« . LetвЂ™s find
9
x2

пЃ˜ z 2 dz.
C

1
This is easyвЂ”we know that F пЃ¶ пѓќzпѓћ пЂЅ z 2 , where Fпѓќzпѓћ пЂЅ z 3 . Thus,
3

3
пЃ˜ z 2 dz пЂЅ 1 пѓќ1 пЂ« i пѓћ 3 пЂї 3 пЂ« 9
i
3
C

пЂЅ пЂї 260 пЂї 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

пЃ˜ fпѓќzпѓћdz,
Fпѓќzпѓћ пЂЅ
Cz

where C z is any path in D from z 0 to z. Here is important that the integral is path
independent, otherwise Fпѓќzпѓћ 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:

пЃ˜ fпѓќsпѓћds,
Fпѓќz пЂ« пЃЃzпѓћ пЂї Fпѓќzпѓћ пЂЅ
L пЃЃz

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

Next, observe that пЃ˜ ds пЂЅ пЃЃz. Thus, fпѓќzпѓћ пЂЅ пЃ˜ fпѓќzпѓћds, and we have
1
пЃЃz
L пЃЃz L пЃЃz

Fпѓќz пЂ« пЃЃzпѓћ пЂї Fпѓќzпѓћ
пЃ˜ пѓќfпѓќsпѓћ пЂї fпѓќzпѓћ пѓћds.
пЂї fпѓќzпѓћ пЂЅ 1
пЃЃz пЃЃz
L пЃЃz

Now then,

пЃ˜ пѓќfпѓќsпѓћ пЂї fпѓќzпѓћ пѓћds
1 1 |пЃЃz| maxпѓЎ|fпѓќsпѓћ пЂї fпѓќzпѓћ| : sпЃЏL пЃЃz пѓў
п‚І
пЃЃz пЃЃz
L пЃЃz

п‚І maxпѓЎ|fпѓќsпѓћ пЂї fпѓќzпѓћ| : sпЃЏL пЃЃz пѓў.

We know f is continuous at z, and so lim maxпѓЎ|fпѓќsпѓћ пЂї fпѓќzпѓћ| : sпЃЏL пЃЃz пѓў пЂЅ 0. Hence,
пЃЃzп‚ё0

Fпѓќz пЂ« пЃЃzпѓћ пЂї Fпѓќzпѓћ
пЃ˜ пѓќfпѓќsпѓћ пЂї fпѓќzпѓћ пѓћds
1
пЂї fпѓќzпѓћ пЂЅ lim
lim
пЃЃz пЃЃz
пЃЃzп‚ё0 пЃЃzп‚ё0
L пЃЃz

пЂЅ 0.

4.9
In other words, F пЃ¶ пѓќzпѓћ пЂЅ fпѓќzпѓћ, and so, just as promised, f has an antiderivative! LetвЂ™s
summarize what we have shown in this section:

Suppose f : D п‚ё 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 пЃћ пЂ« 2i. Evaluate the integral

пЃ˜ cos z dz.
2
C

8. a)Let Fпѓќzпѓћ пЂЅ log z, 0 пЂј arg z пЂј 2пЃћ. Show that the derivative F пЃ¶ пѓќzпѓћ пЂЅ 1 .
z
пЃћ 7пЃћ
b)Let Gпѓќzпѓћ пЂЅ log z, пЂї 4 пЂј arg z пЂј 4 . Show that the derivative G пѓќzпѓћ пЂЅ 1 .
пЃ¶
z
c)Let C 1 be a curve in the right-half plane D 1 пЂЅ пѓЎz : Re z п‚і 0пѓў from пЂїi to i that does not
pass through the origin. Find the integral

пЃ˜ 1 dz.
z
C1

d)Let C 2 be a curve in the left-half plane D 2 пЂЅ пѓЎz : Re z п‚І 0пѓў from пЂїi to i that does not
pass through the origin. Find the integral.

пЃ˜ 1 dz.
z
C2

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

пЃ˜ 1 dz.
z
C

10. a)Let Hпѓќzпѓћ пЂЅ z c , пЂїпЃћ пЂј arg z пЂј пЃћ. Find the derivative H пЃ¶ пѓќzпѓћ.
b)Let Kпѓќzпѓћ пЂЅ z c , пЂї пЃћ пЂј arg z пЂј 7пЃћ . What is the largest subset of the plane on which
4 4
Hпѓќzпѓћ пЂЅ Kпѓќzпѓћ?
c)Let C be any path from пЂї1 to 1 that lies completely in the upper half-plane. (Upper

4.10
half-plane пЂЅ пѓЎz : Im z п‚і 0пѓў.) Find

пЃ˜ Fпѓќzпѓћdz,
C

where Fпѓќzпѓћ пЂЅ z i , пЂїпЃћ пЂј arg z п‚І пЃћ.

11. Suppose P is a polynomial and C is a closed curve. Explain how you know that
пЃ˜ Pпѓќzпѓћdz пЂЅ 0.
C

4.11
Chapter Five

CauchyвЂ™s Theorem

5.1. Homotopy. Suppose D is a connected subset of the plane such that every point of D is
an interior pointвЂ”we call such a set a regionвЂ”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 п‚ё D,
where S is the square S пЂЅ пѓЎпѓќt, sпѓћ : 0 п‚І s, t п‚І 1пѓў, such that Hпѓќt, 0пѓћ describes C 1 and Hпѓќt, 1пѓћ
describes C 2 , and for each fixed s, the function Hпѓќt, sпѓћ 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
Kпѓќt, sпѓћ пЂЅ Hпѓќt, 1 пЂї sпѓћ 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 пЃЊпѓќtпѓћ пЂЅ c. We thus speak of a closed curve C being homotopic to a
constantвЂ”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 пЂЅ пѓЎz : 1 пЂј |z| пЂј 5пѓў. Suppose C 1 is the circle described by
пЃЊ 1 пѓќtпѓћ пЂЅ 2e i2пЃћt , 0 п‚І t п‚І 1; and C 2 is the circle described by пЃЊ 2 пѓќtпѓћ пЂЅ 4e i2пЃћt , 0 п‚І t п‚І 1. Then
Hпѓќt, sпѓћ пЂЅ пѓќ2 пЂ« 2sпѓћe i2пЃћt 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
пЃЊ 3 пѓќtпѓћ пЂЅ 4e пЂїi2пЃћt , 0 п‚І t п‚І 1. A homotopy between C 1 and C 3 is not too easy to constructвЂ”in
fact, it is not possible! The moral: orientation counts. From now on, the term вЂќclosed
curveвЂќ will mean an oriented closed curve.

5.1
Another Example

Let D be the set obtained by removing the point z пЂЅ 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 пЃЂ and пЃѓ
are homotopic in D, while K and пЃЂ 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 CauchyвЂ™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 Hпѓќt, sпѓћ be a homotopy
between C 1 and C 2 . For each s, the function пЃЊ s пѓќtпѓћ describes a closed curve C s in D. Let
Iпѓќsпѓћ be given by

пЃ˜ fпѓќzпѓћdz.
Iпѓќsпѓћ пЂЅ
Cs

Then,

5.2
1

пЃ˜ fпѓќHпѓќt, sпѓћпѓћ пЂЇHпѓќt, sпѓћ dt.
Iпѓќsпѓћ пЂЅ
пЂЇt
0

Now letвЂ™s look at the derivative of Iпѓќsпѓћ. We assume everything is nice enough to allow us
to differentiate under the integral:

1

пЃ˜ fпѓќHпѓќt, sпѓћпѓћ пЂЇHпѓќt, sпѓћ dt
I пЃ¶ пѓќsпѓћ пЂЅ d
пЂЇt
ds
0
1
пЂЇ 2 Hпѓќt, sпѓћ
пЂЇHпѓќt, sпѓћ пЂЇHпѓќt, sпѓћ
пЃ˜ пЃ¶
пЂЅ f пѓќHпѓќt, sпѓћпѓћ пЂ« fпѓќHпѓќt, sпѓћпѓћ dt
пЂЇs пЂЇt пЂЇsпЂЇt
0
1
пЂЇ 2 Hпѓќt, sпѓћ
пЂЇHпѓќt, sпѓћ пЂЇHпѓќt, sпѓћ
пЃ˜ пЃ¶
пЂЅ f пѓќHпѓќt, sпѓћпѓћ пЂ« fпѓќHпѓќt, sпѓћпѓћ dt
пЂЇs пЂЇt пЂЇtпЂЇs
0
1
пЂЇ fпѓќHпѓќt, sпѓћпѓћ пЂЇHпѓќt, sпѓћ dt
пЃ˜
пЂЅ
пЂЇt пЂЇs
0
пЂЇHпѓќ1, sпѓћ пЂЇHпѓќ0, sпѓћ
пЂЅ fпѓќHпѓќ1, sпѓћпѓћ пЂї fпѓќHпѓќ0, sпѓћпѓћ .
пЂЇs пЂЇs
But we know each Hпѓќt, sпѓћ describes a closed curve, and so Hпѓќ0, sпѓћ пЂЅ Hпѓќ1, sпѓћ for all s. Thus,

пЂЇHпѓќ1, sпѓћ пЂЇHпѓќ0, sпѓћ
I пЃ¶ пѓќsпѓћ пЂЅ fпѓќHпѓќ1, sпѓћпѓћ пЂї fпѓќHпѓќ0, sпѓћпѓћ пЂЅ 0,
пЂЇs пЂЇs

which means Iпѓќsпѓћ is constant! In particular, Iпѓќ0пѓћ пЂЅ Iпѓќ1пѓћ, or

пЃ˜ fпѓќzпѓћdz пЂЅ пЃ˜ fпѓќzпѓћdz.
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 пЃ˜ fпѓќzпѓћdz пЂЅ пЃ˜ fпѓќzпѓћdz.
C1 C2

An easy corollary of this result is the celebrated CauchyвЂ™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
пЃ˜ fпѓќzпѓћdz пЂЅ 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 necessaryвЂ”specifically, the results can be
proved true without all our smoothness assumptionsвЂ”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

пЃ˜ fпѓќzпѓћdz пЂЅ пЃ˜ fпѓќzпѓћdz пЂ« пЃ˜ fпѓќzпѓћdz.
C C1 C2

Exercises

4. Prove CauchyвЂ™s Theorem.

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

5.4
пЃ˜ 1 dz.
z
S

6. a)Find пЃ˜ 1
dz, where C is any circle centered at z пЂЅ 1 with the usual counterclockwise
zпЂї1
C
orientation: пЃЊпѓќtпѓћ пЂЅ 1 пЂ« Ae 2пЃћit , 0 п‚І t п‚І 1.
b)Find пЃ˜ zпЂ«1 dz, where C is any circle centered at z пЂЅ пЂї1 with the usual counterclockwise
1

C
orientation.

c)Find пЃ˜ 1
dz, where C is the ellipse 4x 2 пЂ« y 2 пЂЅ 100 with the counterclockwise
z 2 пЂї1
C
orientation. [Hint: partial fractions]

d)Find пЃ˜ 1
dz, where C is the circle x 2 пЂї 10x пЂ« y 2 пЂЅ 0 with the counterclockwise
z 2 пЂї1
C
orientation.

8. Evaluate пЃ˜Log пѓќz пЂ« 3пѓћdz, where C is the circle |z| пЂЅ 2 oriented counterclockwise.
C

9. Evaluate пЃ˜ 1
dz where C is the circle described by пЃЊпѓќtпѓћ пЂЅ e 2пЃћit , 0 п‚І t п‚І 1, and n is an
zn
C
integer п‚® 1.

10. a)Does the function fпѓќzпѓћ пЂЅ 1 have an antiderivative on the set of all z п‚® 0? Explain.
z
1
b)How about fпѓќzпѓћ пЂЅ z n , n an integer п‚® 1 ?

2
11. Find as large a set D as you can so that the function fпѓќzпѓћ пЂЅ 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. CauchyвЂ™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

fпѓќzпѓћ
пЃ˜
1
fпѓќz 0 пѓћ пЂЅ z пЂї z 0 dz.
2пЃћi
C

This is the famous Cauchy Integral Formula. LetвЂ™s see why itвЂ™s true.

Let пЃђ пЂѕ 0 be any positive number. We know that f is continuous at z 0 and so there is a
number пЃЋ such that |fпѓќzпѓћ пЂї fпѓќz 0 пѓћ| пЂј пЃђ whenever |z пЂї z 0 | пЂј пЃЋ. Now let пЃџ пЂѕ 0 be a number
such that пЃџ пЂј пЃЋ and the circle C 0 пЂЅ пѓЎz : |z пЂї z 0 | пЂЅ пЃџпѓў is also inside C. Now, the function
fпѓќzпѓћ
zпЂїz 0 is analytic in the region between C and C 0 ; thus

fпѓќzпѓћ fпѓќzпѓћ
пЃ˜ пЃ˜
z пЂї z 0 dz пЂЅ z пЂї z 0 dz.
C C0

We know that пЃ˜ 1
dz пЂЅ 2пЃћi, so we can write
zпЂїz 0
C0

fпѓќzпѓћ fпѓќzпѓћ
пЃ˜ пЃ˜ пЃ˜ 1
z пЂї z 0 dz пЂї 2пЃћifпѓќz 0 пѓћ пЂЅ z пЂї z 0 dz пЂї fпѓќz 0 пѓћ z пЂї z 0 dz
C0 C0 C0

fпѓќzпѓћ пЂї fпѓќz 0 пѓћ
пЃ˜
пЂЅ z пЂї z 0 dz.
C0

For zпЃЏC 0 we have
|fпѓќzпѓћ пЂї fпѓќz 0 пѓћ|
fпѓќzпѓћ пЂї fпѓќz 0 пѓћ
пЂЅ
z пЂї z0 |z пЂї z 0 |
пЃђ
п‚І пЃџ.

Thus,

6.1
fпѓќzпѓћ пЂї fпѓќz 0 пѓћ
fпѓќzпѓћ
пЃ˜ пЃ˜
z пЂї z 0 dz пЂї 2пЃћifпѓќz 0 пѓћ пЂЅ z пЂї z 0 dz
C0 C0
пЃђ
п‚І пЃџ 2пЃћпЃџ пЂЅ 2пЃћпЃђ.

But пЃђ is any positive number, and so

fпѓќzпѓћ
пЃ˜ z пЂї z 0 dz пЂї 2пЃћifпѓќz 0 пѓћ пЂЅ 0,
C0

or,
fпѓќzпѓћ fпѓќzпѓћ
пЃ˜ пЃ˜
1 dz пЂЅ 1
fпѓќz 0 пѓћ пЂЅ z пЂї z 0 dz,
z пЂї z0
2пЃћi 2пЃћi
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 fпѓќzпѓћ for every z on the simple closed curve, then we know the value for
the function at every point inside the curveвЂ”quite remarkable indeed.

Example

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

пЃ˜ cos z dz.
2
z пЂї 6z пЂ« 5
C

We simply write the integrand as

fпѓќzпѓћ
cos z cos z
пЂЅ пЂЅ ,
zпЂї1
пѓќz пЂї 5пѓћпѓќz пЂї 1пѓћ
z 2 пЂї 6z пЂ« 5
where
fпѓќzпѓћ пЂЅ cos z .
zпЂї5
Observe that f is analytic on and inside C, and so,

6.2
fпѓќzпѓћ
пЃ˜ пЃ˜
cos z dz пЂЅ dz пЂЅ 2пЃћifпѓќ1пѓћ
zпЂї1
2
z пЂї 6z пЂ« 5
C C

пЂЅ 2пЃћi cos 1 пЂЅ пЂї iпЃћ cos 1
1пЂї5 2

Exercises

1. Suppose f and g are analytic on and inside the simple closed curve C, and suppose
moreover that fпѓќzпѓћ пЂЅ gпѓќzпѓћ for all z on C. Prove that fпѓќzпѓћ пЂЅ gпѓќzпѓћ for all z inside C.

2. Let C be the ellipse 9x 2 пЂ« 4y 2 пЂЅ 36 traversed once in the counterclockwise direction.
Define the function g by

пЃ˜ s 2 пЂ« s пЂ« 1 ds.
gпѓќzпѓћ пЂЅ sпЂїz
C

Find a) gпѓќiпѓћ b) gпѓќ4iпѓћ

3. Find

пЃ˜ e 2z dz,
z2 пЂї 4
C

where C is the closed curve in the picture:

4. Find пЃ˜ e 2z
dz, where пЃЂ is the contour in the picture:
z 2 пЂї4
пЃЂ

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

gпѓќsпѓћ
пЃ˜
Gпѓќzпѓћ пЂЅ s пЂї z ds
C

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

Consider,
Gпѓќz пЂ« пЃЃzпѓћ пЂї Gпѓќzпѓћ
пЃ˜
пЂЅ1 1 1
пЂї s пЂї z gпѓќsпѓћds
пЃЃz пЃЃz s пЂї z пЂї пЃЃz
C
gпѓќsпѓћ
пЃ˜
пЂЅ ds.
пѓќs пЂї z пЂї пЃЃzпѓћпѓќs пЂї zпѓћ
C

Next,

Gпѓќz пЂ« пЃЃzпѓћ пЂї Gпѓќzпѓћ gпѓќsпѓћ
пЂїпЃ˜ пЃ˜ 1 1
ds пЂЅ пЂї gпѓќsпѓћds
пЃЃz пѓќs пЂї z пЂї пЃЃzпѓћпѓќs пЂї zпѓћ
пѓќs пЂї zпѓћ 2 пѓќs пЂї zпѓћ 2
C C

пѓќs пЂї zпѓћ пЂї пѓќs пЂї z пЂї пЃЃzпѓћ
пЃ˜
пЂЅ gпѓќsпѓћds
пѓќs пЂї z пЂї пЃЃzпѓћпѓќs пЂї zпѓћ 2
C
gпѓќsпѓћ
пЂЅ пЃЃz пЃ˜ ds.
пѓќs пЂї z пЂї пЃЃzпѓћпѓќs пЂї zпѓћ 2
C

Now we want to show that

6.4
gпѓќsпѓћ
пЃЃz пЃ˜ пЂЅ 0.
lim ds
пѓќs пЂї z пЂї пЃЃzпѓћпѓќs пЂї zпѓћ 2
пЃЃzп‚ё0
C

To that end, let M пЂЅ maxпѓЎ|gпѓќsпѓћ| : s пЂµ Cпѓў, and let d be the shortest distance from z to C.
Thus, for s пЂµ C, we have |s пЂї z| п‚і d пЂѕ 0 and also
|s пЂї z пЂї пЃЃz| п‚і |s пЂї z| пЂї |пЃЃz| п‚і d пЂї |пЃЃz|.

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

gпѓќsпѓћ M
п‚І
пѓќs пЂї z пЂї пЃЃzпѓћпѓќs пЂї zпѓћ 2 пѓќd пЂї |пЃЃz|пѓћd 2
for all s пЂµ C. Finally,

gпѓќsпѓћ
пЃЃz пЃ˜ M
ds п‚І |пЃЃz| lengthпѓќCпѓћ,
2
пѓќd пЂї |пЃЃz|пѓћd 2
пѓќs пЂї z пЂї пЃЃzпѓћпѓќs пЂї zпѓћ
C

and it is clear that

gпѓќsпѓћ
пЃЃz пЃ˜ пЂЅ 0,
lim ds
пѓќs пЂї z пЂї пЃЃzпѓћпѓќs пЂї zпѓћ 2
пЃЃzп‚ё0
C

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

gпѓќsпѓћ
пЃ˜
G пЃ¶ пѓќzпѓћ пЂЅ ds.
пѓќs пЂї zпѓћ 2
C

Truly a miracle!

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

6.5
G пЃ¶ пѓќz пЂ« пЃЃzпѓћ пЂї G пЃ¶ пѓќzпѓћ
пЃ˜
пЂЅ1 1 1
пЂї gпѓќsпѓћds
пЃЃz пЃЃz пѓќs пЂї z пЂї пЃЃzпѓћ 2 пѓќs пЂї zпѓћ 2
C

пѓќs пЂї zпѓћ 2 пЂї пѓќs пЂї z пЂї пЃЃzпѓћ 2
пЃ˜
пЂЅ1 gпѓќsпѓћds
пЃЃz пѓќs пЂї z пЂї пЃЃzпѓћ 2 пѓќs пЂї zпѓћ 2
C

2пѓќs пЂї zпѓћпЃЃz пЂї пѓќпЃЃzпѓћ 2
пЃ˜
пЂЅ1 gпѓќsпѓћds
пЃЃz пѓќs пЂї z пЂї пЃЃzпѓћ 2 пѓќs пЂї zпѓћ 2
C

2пѓќs пЂї zпѓћ пЂї пЃЃz
пЃ˜
пЂЅ gпѓќsпѓћds
пѓќs пЂї z пЂї пЃЃzпѓћ 2 пѓќs пЂї zпѓћ 2
C

Next,

G пЃ¶ пѓќz пЂ« пЃЃzпѓћ пЂї G пЃ¶ пѓќzпѓћ gпѓќsпѓћ
пЂї 2пЃ˜ ds
пЃЃz пѓќs пЂї zпѓћ 3
C

2пѓќs пЂї zпѓћ пЂї пЃЃz
пЃ˜ 2
пЂЅ пЂї gпѓќsпѓћds
2 2
пѓќs пЂї zпѓћ 3
пѓќs пЂї z пЂї пЃЃzпѓћ пѓќs пЂї zпѓћ
C

2пѓќs пЂї zпѓћ 2 пЂї пЃЃzпѓќs пЂї zпѓћ пЂї 2пѓќs пЂї z пЂї пЃЃzпѓћ 2
пЃ˜
пЂЅ gпѓќsпѓћds
пѓќs пЂї z пЂї пЃЃzпѓћ 2 пѓќs пЂї zпѓћ 3
C

2пѓќs пЂї zпѓћ 2 пЂї пЃЃzпѓќs пЂї zпѓћ пЂї 2пѓќs пЂї zпѓћ 2 пЂ« 4пЃЃzпѓќs пЂї zпѓћ пЂї 2пѓќпЃЃzпѓћ 2
пЃ˜
пЂЅ gпѓќsпѓћds
пѓќs пЂї z пЂї пЃЃzпѓћ 2 пѓќs пЂї zпѓћ 3
C

3пЃЃzпѓќs пЂї zпѓћ пЂї 2пѓќпЃЃzпѓћ 2
пЃ˜
пЂЅ gпѓќsпѓћds
пѓќs пЂї z пЂї пЃЃzпѓћ 2 пѓќs пЂї zпѓћ 3
C

Hence,

G пЃ¶ пѓќz пЂ« пЃЃzпѓћ пЂї G пЃ¶ пѓќzпѓћ 3пЃЃzпѓќs пЂї zпѓћ пЂї 2пѓќпЃЃzпѓћ 2
gпѓќsпѓћ
пЂї 2пЃ˜ пЃ˜
ds пЂЅ gпѓќsпѓћds
пЃЃz пѓќs пЂї zпѓћ 3 пѓќs пЂї z пЂї пЃЃzпѓћ 2 пѓќs пЂї zпѓћ 3
C C
пѓќ|3m| пЂ« 2|пЃЃz|пѓћM
п‚І |пЃЃz| ,
пѓќd пЂї пЃЃzпѓћ 2 d 3

where m пЂЅ maxпѓЎ|s пЂї z| : s пЂµ Cпѓў. It should be clear then that

G пЃ¶ пѓќz пЂ« пЃЃzпѓћ пЂї G пЃ¶ пѓќzпѓћ gпѓќsпѓћ
пЂї 2пЃ˜ ds пЂЅ 0,
lim
пЃЃz пѓќs пЂї zпѓћ 3
пЃЃzп‚ё0
C

or in other words,

6.6
gпѓќsпѓћ
G пЃ¶пЃ¶ пѓќzпѓћ пЂЅ 2 пЃ˜ ds.
пѓќs пЂї zпѓћ 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

fпѓќsпѓћ
пЃ˜
2пЃћifпѓќzпѓћ пЂЅ s пЂї z ds
C

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

fпѓќsпѓћ fпѓќsпѓћ
пЃ˜ пЃ˜
1 ds, and f пЃ¶пЃ¶ пѓќzпѓћ пЂЅ 2
f пЃ¶ пѓќzпѓћ пЂЅ ds
2пЃћi 2пЃћi
2
пѓќs пЂї zпѓћ 3
пѓќs пЂї zпѓћ
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 пЃ˜ fпѓќzпѓћdz пЂЅ 0 for every closed
C
curve in D. Then we know that f has an antiderivative in DвЂ”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 MoreraвЂ™s Theorem:

If f:D п‚ё C is continuous and such that пЃ˜ fпѓќzпѓћdz пЂЅ 0 for every closed curve in D, then f is
C
analytic in D.

Example

LetвЂ™s evaluate the integral

пЃ˜ e z dz,
z3
C

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

fпѓќsпѓћ
пЃ˜
2
f пЃ¶пЃ¶ пѓќzпѓћ пЂЅ ds
2пЃћi пѓќs пЂї zпѓћ 3
C

6.7
with z пЂЅ 0 and fпѓќsпѓћ пЂЅ e s . Thus,

пЃ˜ e z dz.
пЃћie 0 пЂЅ пЃћi пЂЅ
z3
C

Exercises

5. Evaluate

пЃ˜ sin z dz
z2
C

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

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

a) пЃ˜ b) пЃ˜
1 1
dz dz
z 2 пЂ«4 пѓќz 2 пЂ«4пѓћ 2
C C

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

f пЃ¶ пѓќzпѓћ fпѓќzпѓћ
пЃ˜ пЃ˜
z пЂї w dz пЂЅ dz
пѓќz пЂї wпѓћ 2
C C

for every w пЂ¶ C.

8. a) Let пЃЉ be a real constant, and let C be the circle пЃЊпѓќtпѓћ пЂЅ e it , пЂїпЃћ п‚І t п‚І пЃћ. Evaluate

e пЃЉz dz.
пЃ˜ z
C

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

пЃћ

пЃ˜ e пЃЉ cos t cosпѓќпЃЉ sin tпѓћdt пЂЅ пЃћ.
0

6.3. LiouvilleвЂ™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 |fпѓќzпѓћ| п‚І M for all z. Then it must be true
that f пЃ¶ пѓќzпѓћ пЂЅ 0 identically. To see this, suppose that f пЃ¶ пѓќwпѓћ п‚® 0 for some w. Choose R large
enough to insure that M пЂј |f пЃ¶ пѓќwпѓћ|. Now let C be a circle centered at 0 and with radius
R

6.8
пЃџ пЂѕ maxпѓЎR, |w|пѓў. Then we have :

fпѓќsпѓћ
пЃ˜
M пЂј |f пЃ¶ пѓќwпѓћ| п‚І 1 dz
пЃџ 2пЃћi пѓќs пЂї wпѓћ 2
C

п‚І 1 M 2пЃћпЃџ пЂЅ M ,
пЃџ
2пЃћ пЃџ 2

a contradiction. It must therefore be true that there is no w for which f пЃ¶ пѓќwпѓћ п‚® 0; or, in other
words, f пЃ¶ пѓќzпѓћ пЂЅ 0 for all z. This, of course, means that f is a constant function. What we
have shown has a name, LiouvilleвЂ™s Theorem:

The only bounded entire functions are the constant functions.

LetвЂ™s put this theorem to some good use. Let pпѓќzпѓћ пЂЅ a n z n пЂ« a nпЂї1 z nпЂї1 пЂ«пЃµ пЂ«a 1 z пЂ« a 0 be a
polynomial. Then

pпѓќzпѓћ пЂЅ a n пЂ« a nпЂї1 пЂ« a nпЂї2 пЂ«пЃµ пЂ« an0
zn.
z z
2
z
 << стр. 2(всего 4)СОДЕРЖАНИЕ >>