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Complex Analysis and Differential Geometry

Notes 11.2 Rouche’s Theorem

Suppose f and g are analytic on and inside a simple closed contour C. Suppose that |f(z)| >
|g(z)| for all z  C. Then we shall see that f and f + g have the same number of zeros inside C.
This result is Rouche’s Theorem. To see why it is so, start by defining the function (t) on the
interval 0  t  1 :

1 f'(z) tg'(t)

 (t)   dz.
2 i C f(z) tg(z)


Observe that the denominator of the integrand is never zero:
|f(z) + tg(z)|  ||f(t) – t|g(t)||  ||f(t)| – |g(t)|| > 0.
Observe that ¸ is continuous on the interval [0,1] and is integer-valued—t) is the number of
zeros of f + tg inside C. Being continuous and integer-valued on the connected set [0,1], it must
be constant. In particular, (0) = (1). This does the job!
1 f'(z)
 (t)   dz
2 i C f(z)

is the number of zeros of f inside C, and

1 f'(z) g'(t)

 (t)   dz.
2 i C f(z) g(z)


is the number of zeros of f + g inside C.
Example:

How many solutions of the equation z – 5z + z – 2 = 0 are inside the circle |z| = 1? Rouche’s
5
3
6
Theorem makes it quite easy to answer this. Simply let f(z) = –5z and let g(z) = z + z – 2. Then
5
3
6
|f(z)| = 5 and |g(z)|  |z| + |z| + 2 = 4 for all |z| = 1. Hence |f(z)| > |g(z)| on the unit circle.
6
3
From Rouche’s Theorem we know then that f and f + g have the same number of zeros inside
|z| = 1. Thus, there are 5 such solutions.
The following nice result follows easily from Rouche’s Theorem. Suppose U is an open set
(i.e., every point of U is an interior point) and suppose that a sequence (f ) of functions analytic
n
on U converges uniformly to the function f. Suppose further that f is not zero on the circle
C ={z : |z : z | = R}  U. Then there is an integer N so that for all n  N, the functions f and f have
0
n
the same number of zeros inside C.
This result, called Hurwitz’s Theorem, is an easy consequence of Rouche’s Theorem. Simply
observe that for z  C, we have |f(z)| >  > 0 for some . Now let N be large enough to insure that
|f (z) – f(z)| <  on C. It follows from Rouche’s Theorem that f and f + (f – f) = f have the same
n
n
n
number of zeros inside C.
Example:
z 2 z
On any bounded set, the sequence (f ), where f (z) = 1 z  2  ...  n! n , converges uniformly to
n
n
f(z) = e , and f(z)  0 for all z. Thus for any R, there is an N so that for n > N, every zero of
z

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