ch6


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 z0 is inside
C. Then it turns out that
fÅ»ðzÞð
1
fÅ»ðz0Þð =ð

z ?ð z0 dz.
2^ði
C
This is the famous Cauchy Integral Formula. Let s see why it s true.
Let Pð >ð 0 be any positive number. We know that f is continuous at z0 and so there is a
number Nð such that fÅ»ðzÞð ?ð fÅ»ðz0Þð <ð Pð whenever z ?ð z0 <ð Nð. Now let _ð >ð 0 be a number
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such that _ð <ð Nð and the circle C0 =ð áðz : z ?ð z0 =ð _ðâð is also inside C. Now, the function
| |
fÅ»ðzÞð
is analytic in the region between C and C0; thus
z?ðz0
fÅ»ðzÞð fÅ»ðzÞð
Xð Xð
z ?ð z0 dz =ð z ?ð z0 dz.
C C0
1
We know that Xð dz =ð 2^ði, so we can write
z?ðz0
C0
fÅ»ðzÞð fÅ»ðzÞð
1
Xð Xð Xð
z ?ð z0 dz ?ð 2^ðifÅ»ðz0Þð =ð z ?ð z0 dz ?ð fÅ»ðz0Þð z ?ð z0 dz
C0 C0 C0
fÅ»ðzÞð ?ð fÅ»ðz0Þð


z ?ð z0 dz.
C0
For zOðC0 we have
fÅ»ðzÞð ?ð fÅ»ðz0Þð
| |
fÅ»ðzÞð ?ð fÅ»ðz0Þð
z ?ð z0 =ð |z ?ð z0|

²ð .

Thus,
6.1
fÅ»ðzÞð fÅ»ðzÞð ?ð fÅ»ðz0Þð
Xð Xð
z ?ð z0 dz ?ð 2^ðifÅ»ðz0Þð =ð z ?ð z0 dz
C0 C0

²ð 2^ð_ð =ð 2^ðPð.

But Pð is any positive number, and so
fÅ»ðzÞð

z ?ð z0 dz ?ð 2^ðifÅ»ðz0Þð =ð 0,
C0
or,
fÅ»ðzÞð fÅ»ðzÞð
1 1
fÅ»ðz0Þð =ð
Xð Xð
z ?ð z0 dz =ð 2^ði z ?ð z0 dz,
2^ði
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 fÅ»ðzÞð 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 =ð 4 traversed once in the counterclockwise direction. Let s evaluate
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the integral
cos z
dz.

z2 ?ð 6z +ð 5
C
We simply write the integrand as
fÅ»ðzÞð
cos z cos z
=ð =ð ,
z ?ð 1
Å»ðz ?ð 5ÞðÅ»ðz ?ð 1Þð
z2 ?ð 6z +ð 5
where
cos z
fÅ»ðzÞð =ð .
z ?ð 5
Observe that f is analytic on and inside C, and so,
6.2
fÅ»ðzÞð
cos z
dz =ð dz =ð 2^ðifÅ»ð1Þð
Xð Xð
z ?ð 1
z2 ?ð 6z +ð 5
C C
cos 1 i^ð
=ð 2^ði =ð ?ð cos 1
1 ?ð 5 2
Exercises
1. Suppose f and g are analytic on and inside the simple closed curve C, and suppose
moreover that fÅ»ðzÞð =ð gÅ»ðzÞð for all z on C. Prove that fÅ»ðzÞð =ð gÅ»ðzÞð for all z inside C.
2. Let C be the ellipse 9x2 +ð 4y2 =ð 36 traversed once in the counterclockwise direction.
Define the function g by
s2 +ð s +ð 1
gÅ»ðzÞð =ð ds.

s ?ð z
C
Find a) gÅ»ðiÞð b) gÅ»ð4iÞð
3. Find
e2z dz,

z2 ?ð 4
C
where C is the closed curve in the picture:
e2z
4. Find Xð dz, where @ð is the contour in the picture:
z2?ð4

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
gÅ»ðsÞð
GÅ»ðzÞð =ð ds

s ?ð z
C
for all z 6ð C. We shall show that G is analytic. Here we go.
Consider,
GÅ»ðz +ðAðzÞð ?ð GÅ»ðzÞð
1 1 1
=ð ?ð gÅ»ðsÞðds

s ?ð z
Aðz Aðz s ?ð z ?ð Aðz
C
gÅ»ðsÞð
=ð ds.

Å»ðs ?ð z ?ð AðzÞðÅ»ðs ?ð zÞð
C
Next,
GÅ»ðz +ðAðzÞð ?ð GÅ»ðzÞð gÅ»ðsÞð
1 1
?ð ds =ð ?ð gÅ»ðsÞðds
Xð Xð
Aðz
Å»ðs ?ð z ?ð AðzÞðÅ»ðs ?ð zÞð
Å»ðs ?ð zÞð2 Å»ðs ?ð zÞð2
C C
Å»ðs ?ð zÞð ?ð Å»ðs ?ð z ?ð AðzÞð
=ð gÅ»ðsÞðds

Å»ðs ?ð z ?ð AðzÞðÅ»ðs ?ð zÞð2
C
gÅ»ðsÞð
=ð Aðz ds.

Å»ðs ?ð z ?ð AðzÞðÅ»ðs ?ð zÞð2
C
Now we want to show that
6.4
gÅ»ðsÞð
lim Aðz ds =ð 0.

Å»ðs ?ð z ?ð AðzÞðÅ»ðs ?ð zÞð2
Aðz¸ð0
C
To that end, let M =ð maxáð gÅ»ðsÞð : s 5ð Câð, and let d be the shortest distance from z to C.
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Thus, for s 5ð C, we have s ?ð z Å‚ð d >ð 0 and also
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s ?ð z ?ð Aðz Å‚ð s ?ð z ?ð Aðz Å‚ð d ?ð Aðz .
| | | | | | | |
Putting this all together, we can estimate the integrand above:
gÅ»ðsÞð
M
²ð
Å»ðs ?ð z ?ð AðzÞðÅ»ðs ?ð zÞð2 Å»ðd ?ð Aðz Þðd2
| |
for all s 5ð C. Finally,
gÅ»ðsÞð
M
Aðz ds ²ð Aðz lengthÅ»ðCÞð,
Xð | |
Å»ðs ?ð z ?ð AðzÞðÅ»ðs ?ð zÞð2 Å»ðd ?ð Aðz Þðd2
| |
C
and it is clear that
gÅ»ðsÞð
lim Aðz ds =ð 0,

Å»ðs ?ð z ?ð AðzÞðÅ»ðs ?ð zÞð2
Aðz¸ð0
C
just as we set out to show. Hence G has a derivative at z, and
gÅ»ðsÞð
GvðÅ»ðzÞð =ð ds.

Å»ðs ?ð zÞð2
C
Truly a miracle!
Next we see that Gvð has a derivative and it is just what you think it should be. Consider
6.5
GvðÅ»ðz +ðAðzÞð ?ð GvðÅ»ðzÞð
1 1 1
=ð ?ð gÅ»ðsÞðds

Aðz Aðz
Å»ðs ?ð z ?ð AðzÞð2 Å»ðs ?ð zÞð2
C
Å»ðs ?ð zÞð2 ?ð Å»ðs ?ð z ?ð AðzÞð2 gÅ»ðsÞðds
1


Aðz
Å»ðs ?ð z ?ð AðzÞð2Å»ðs ?ð zÞð2
C
2Å»ðs ?ð zÞðAðz ?ð Å»ðAðzÞð2 gÅ»ðsÞðds
1


Aðz
Å»ðs ?ð z ?ð AðzÞð2Å»ðs ?ð zÞð2
C
2Å»ðs ?ð zÞð ?ð Aðz
=ð gÅ»ðsÞðds

Å»ðs ?ð z ?ð AðzÞð2Å»ðs ?ð zÞð2
C
Next,
GvðÅ»ðz +ðAðzÞð ?ð GvðÅ»ðzÞð gÅ»ðsÞð
?ð 2 ds

Aðz
Å»ðs ?ð zÞð3
C
2Å»ðs ?ð zÞð ?ð Aðz
2
=ð ?ð gÅ»ðsÞðds

Å»ðs ?ð z ?ð AðzÞð2Å»ðs ?ð zÞð2 Å»ðs ?ð zÞð3
C
2Å»ðs ?ð zÞð2 ?ð AðzÅ»ðs ?ð zÞð ?ð 2Å»ðs ?ð z ?ð AðzÞð2 gÅ»ðsÞðds


Å»ðs ?ð z ?ð AðzÞð2Å»ðs ?ð zÞð3
C
2Å»ðs ?ð zÞð2 ?ð AðzÅ»ðs ?ð zÞð ?ð 2Å»ðs ?ð zÞð2 +ð 4AðzÅ»ðs ?ð zÞð ?ð 2Å»ðAðzÞð2 gÅ»ðsÞðds


Å»ðs ?ð z ?ð AðzÞð2Å»ðs ?ð zÞð3
C
3AðzÅ»ðs ?ð zÞð ?ð 2Å»ðAðzÞð2 gÅ»ðsÞðds


Å»ðs ?ð z ?ð AðzÞð2Å»ðs ?ð zÞð3
C
Hence,
GvðÅ»ðz +ðAðzÞð ?ð GvðÅ»ðzÞð gÅ»ðsÞð 3AðzÅ»ðs ?ð zÞð ?ð 2Å»ðAðzÞð2 gÅ»ðsÞðds
?ð 2 ds =ð
Xð Xð
Aðz
Å»ðs ?ð zÞð3 Å»ðs ?ð z ?ð AðzÞð2Å»ðs ?ð zÞð3
C C
Å»ð 3m +ð 2 Aðz ÞðM
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²ð Aðz ,
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Å»ðd ?ð AðzÞð2d3
where m =ð maxáð s ?ð z : s 5ð Câð. It should be clear then that
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GvðÅ»ðz +ðAðzÞð ?ð GvðÅ»ðzÞð gÅ»ðsÞð
lim ?ð 2 ds =ð 0,

Aðz
Å»ðs ?ð zÞð3
Aðz¸ð0
C
or in other words,
6.6
gÅ»ðsÞð
GvðvðÅ»ðzÞð =ð 2 ds.

Å»ðs ?ð zÞð3
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
fÅ»ðsÞð
2^ðifÅ»ðzÞð =ð ds

s ?ð z
C
and so with g =ð f in the formulas just derived, we have
fÅ»ðsÞð fÅ»ðsÞð
1 2
fvðÅ»ðzÞð =ð ds, and fvðvðÅ»ðzÞð =ð ds
Xð Xð
2^ði 2^ði
Å»ðs ?ð zÞð2 Å»ðs ?ð zÞð3
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 Xð fÅ»ðzÞðdz =ð 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 ¸ð C is continuous and such that Xð fÅ»ðzÞðdz =ð 0 for every closed curve in D, then f is
C
analytic in D.
Example
Let s evaluate the integral
ez dz,

z3
C
where C is any positively oriented closed curve around the origin. We simply use the
equation
fÅ»ðsÞð
2
fvðvðÅ»ðzÞð =ð ds

2^ði
Å»ðs ?ð zÞð3
C
6.7
with z =ð 0 and fÅ»ðsÞð =ð es. Thus,
ez
^ðie0 =ð ^ði =ð dz.

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 ?ð i =ð 2 with the positive orientation. Evaluate
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1 1
a) Xð dz b) Xð dz
z2+ð4 Å»ðz2+ð4Þð2
C C
7. Suppose f is analytic inside and on the simple closed curve C. Show that
fvðÅ»ðzÞð fÅ»ðzÞð
dz =ð dz
Xð Xð
z ?ð w
Å»ðz ?ð wÞð2
C C
for every w 6ð C.
8. a) Let Jð be a real constant, and let C be the circle LðÅ»ðtÞð =ð eit, ?ð^ð ²ð t ²ð ^ð. Evaluate
eJðz dz.

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

eJð cos t cosÅ»ðJð sin tÞðdt =ð ^ð.

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 fÅ»ðzÞð ²ð M for all z. Then it must be true
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that fvðÅ»ðzÞð =ð 0 identically. To see this, suppose that fvðÅ»ðwÞð ®ð 0 for some w. Choose R large
M
enough to insure that <ð fvðÅ»ðwÞð . Now let C be a circle centered at 0 and with radius
| |
R
6.8
_ð >ð maxáðR, w âð. Then we have :
| |
fÅ»ðsÞð
M 1
<ð fvðÅ»ðwÞð ²ð dz
| | Xð

2^ði
Å»ðs ?ð wÞð2
C
1 M M
²ð 2^ð_ð =ð ,

2^ð
_ð2
a contradiction. It must therefore be true that there is no w for which fvðÅ»ðwÞð ®ð 0; or, in other
words, fvðÅ»ðzÞð =ð 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 pÅ»ðzÞð =ð anzn +ð an?ð1zn?ð1 +ðuð +ða1z +ð a0 be a
polynomial. Then
an?ð1 an?ð2 a0
pÅ»ðzÞð =ð an +ð +ð +ðuð +ð zn.
z
zn
z2
an?ðj |an|
Now choose R large enough to insure that for each j =ð 1, 2, uð , n, we have <ð
zj 2n
whenever z >ð R. (We are assuming that an ®ð 0. ) Hence, for z >ð R, we know that
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an?ð1 +ð an?ð2 +ðuð +ð a0 |z|n
pÅ»ðzÞð Å‚ð an ?ð
| | | |
z
zn
z2
an?ð1 ?ð an?ð2 ?ðuð ?ð a0 |z|n
Å‚ð an ?ð
| |
z
zn
z2
an an an | |n
| | | | | |
>ð an ?ð ?ð ?ðuð ?ð z
| |
2n 2n 2n
an | |n
| |
>ð z .
2
Hence, for z >ð R,
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1 2 2
<ð ²ð .
an z an Rn
pÅ»ðzÞð | || |n | |
1 1
Now suppose pÅ»ðzÞð ®ð 0 for all z. Then is also bounded on the disk z ²ð R. Thus,
| |
pÅ»ðzÞð pÅ»ðzÞð
is a bounded entire function, and hence, by Liouville s Theorem, constant! Hence the
polynomial is constant if it has no zeros. In other words, if pÅ»ðzÞð is of degree at least one,
there must be at least one z0 for which pÅ»ðz0Þð =ð 0. This is, of course, the celebrated
6.9
Fundamental Theorem of Algebra.
Exercises
9. Suppose f is an entire function, and suppose there is an M such that Re fÅ»ðzÞð ²ð M for all
z. Prove that f is a constant function.
10. Suppose w is a solution of 5z4 +ð z3 +ð z2 ?ð 7z +ð 14 =ð 0. Prove that w ²ð 3.
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11. Prove that if p is a polynomial of degree n, and if pÅ»ðaÞð =ð 0, then pÅ»ðzÞð =ð Å»ðz ?ð aÞðqÅ»ðzÞð,
where q is a polynomial of degree n ?ð 1.
12. Prove that if p is a polynomial of degree n Å‚ð 1, then
pÅ»ðzÞð =ð cÅ»ðz ?ð z1Þðk1Å»ðz ?ð z2Þðk2uð Å»ðz ?ð zjÞðkj,
where k1, k2, uð , kj are positive integers such that n =ð k1 +ð k2 +ðuð +ðkj.
13. Suppose p is a polynomial with real coefficients. Prove that p can be expressed as a
product of linear and quadratic factors, each with real coefficients.
6.4. Maximum moduli. Suppose f is analytic on a closed domain D. Then, being
continuous, fÅ»ðzÞð must attain its maximum value somewhere in this domain. Suppose this
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happens at an interior point. That is, suppose fÅ»ðzÞð ²ð M for all z 5ð D and suppose that
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fÅ»ðz0Þð =ð M for some z0 in the interior of D. Now z0 is an interior point of D, so there is a
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number R such that the disk Cð centered at z0 having radius R is included in D. Let C be a
positively oriented circle of radius _ð ²ð R centered at z0. From Cauchy s formula, we
know
fÅ»ðsÞð
1
fÅ»ðz0Þð =ð

s ?ð z0 ds.
2^ði
C
Hence,
2^ð
1
fÅ»ðz0Þð =ð fÅ»ðz0 +ð _ðeitÞðdt,

2^ð
0
and so,
6.10
2^ð
1
M =ð fÅ»ðz0Þð ²ð | |
fÅ»ðz0 +ð _ðeitÞð dt ²ð M.
| | Xð
2^ð
0
since fÅ»ðz0 +ð _ðeitÞð ²ð M. This means
| |
2^ð
1
M =ð | |
fÅ»ðz0 +ð _ðeitÞð dt.

2^ð
0
Thus,
2^ð 2^ð
1 1
M ?ð | | Xðßð | |Ä…ð
fÅ»ðz0 +ð _ðeitÞð dt =ð M ?ð fÅ»ðz0 +ð _ðeitÞð dt =ð 0.

2^ð 2^ð
0 0
This integrand is continuous and non-negative, and so must be zero. In other words,
fÅ»ðzÞð =ð M for all z 5ð C. There was nothing special about C except its radius _ð ²ð R, and so
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we have shown that f must be constant on the disk Cð.
I hope it is easy to see that if D is a region (=ðconnected and open), then the only way in
which the modulus fÅ»ðzÞð of the analytic function f can attain a maximum on D is for f to be
| |
constant.
Exercises
14. Suppose f is analytic and not constant on a region D and suppose fÅ»ðzÞð ®ð 0 for all z 5ð D.
Explain why fÅ»ðzÞð does not have a minimum in D.
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15. Suppose fÅ»ðzÞð =ð uÅ»ðx, yÞð +ð ivÅ»ðx, yÞð is analytic on a region D. Prove that if uÅ»ðx, yÞð attains a
maximum value in D, then u must be constant.
6.11


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