Page 60 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 6: Cauchy’s Integral Formula

1 1 Notes
Now suppose p(z)  0 for all z. Then is also bounded on the disk |z|  R. Thus, is a
p(z) p(z)
bounded entire function, and hence, by Liouville’s Theorem, constant! Hence the polynomial is
constant if it has no zeros. In other words, if p(z) is of degree at least one, there must be at least
one z for which p(z ) = 0. This is, of course, the celebrated fundamental theorem of algebra.
0
0
6.4 Maximum Moduli

Suppose f is analytic on a closed domain D. Then, being continuous, |f(z)| must attain its
maximum value somewhere in this domain. Suppose this happens at an interior point. That is,
suppose |f(z)|  M for all z  D and suppose that |f(z )| = M for some z in the interior of D.
0
0
Now z is an interior point of D, so there is a number R such that the disk  centered at z having
0
0
radius R is included in D. Let C be a positively oriented circle of radius   R centered at z . From
0
Cauchy’s formula, we know
1 f(s)

f(z )  2 i s z 0 ds.
0


C
Hence,
1 2
it
f(z )  2 i  0 f(z   e )dt,
0
0

and so,
1 2
it
M  f(z )   f(z   e ) dt  M.
0
0
2 0
since |f(z + e )|  M. This means
it
0
1 2
it
M   f(z   e ) dt.
0
2 0
Thus,
This integrand is continuous and non-negative, and so must be zero. In other words, |f(z)| = M
for all z  C. There was nothing special about C except its radius   R, and so we have shown
that f must be constant on the disk .
It is easy to see that if D is a region (i.e. connected and open), then the only way in which the
modulus | f(z)| of the analytic function f can attain a maximum on D is for f to be constant.

6.5 Summary

Suppose f is analytic in a region containing a simple closed contour C with the usual

positive orientation and its inside, and suppose z is inside C. Then it turns out that
0
1 f(z)
f(z )  2 i z z 0 dz.

0


C
This is the famous Cauchy Integral Formula.

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