Page 59 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes with z = 0 and f(s) = e . Thus,
s
e z
0
i
 ie    3  dz.
C z
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 | f(z)|  M for all z. Then it must be true that f’(z) = 0 identically. To see this, suppose
M
that f’(w)  0 for some w. Choose R large enough to insure that  f'(w) . Now let C be a circle
R
centered at 0 and with radius  > max{R, |w|}. Then we have :

M 1 f(s)

R  f'(w)  2 i (s w) 2 ds


C
1 M M
 2  ,
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 p(z) = a z + a z + ... + a z + a be a polynomial.
n
n–1
0
1
n
n–1
Then
 a n 1 a n 2 a 0  n
p(z)   a     2  ...  n  z .
n
 z z z 
a a
Now choose R large enough to insure that for each j = 1, 2,...,n, we have n j  n whenever

z j 2n
|z| > R. (We are assuming that a  0. ) Hence, for |z| > R, we know that
n
a a a n
p(z)  a  n 1  n 2  ...  0 z


n z z 2 z n
a a a n
 a  n 1 – z n 2 – ...– z n 0 z


n
z
2
a a a n
n
> a  2n n – 2n – ...– 2n n z
n
a
n
> n z .
2
Hence, for |z| > R,
1 2 2 .
p(z)  a z n  a R n
n
n

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