Singularities/Residues/ Applications 745tion, and not passing through any of the zeros or poles off, by applying
formula (9.14-5) we obtainm n 1 I dw
LakDc(ak)-Lf3rDc(br) =
2
7ri UJ" = Da1(0)
k=l r=l C'
Henceb,.01 argw = 27r0ci(O) = 27r [~ akDc(ak) - ~ /3rDc(br)] (9.15-4)
Theorem 9.21 (Rouche's Theorem). Suppose that:1. f is meromorphic in a region R.
- g is holomorphic in R.
- lg(z)I < lf(z)I for all z E C, where C is a simple closed contour
homotopic to a point in R, and such that no poles of f lie on C.
Then f and f + g have the same number of zeros inside C (counting
multiplicities).
Proof First we note that because of (3), lf(z)I > 0 for z E C, and alsolf(z) + g(z)l 2'.: lf(z)l-lg(z)I > 0
for z E C, so that neither f nor f + g have any zeros on C. Let
N = number of zeros of f inside C
P = number of poles of f inside C
N' =number of zeros off+ g inside C
P' = number of poles of f + g inside CSince g is holomorphic in R we have P = P'. We must prove that N = N'.
Consider the function
w = F(z) = f(z) + g(z) = 1 + g(z)
f(z) f(z)
(9.15-5)For z E C we have
lg(z)Ilw -11 = lf(z)I < 1
Since w remains in an open unit disk with center at 1 as z describes C once
in the positive direction, we have, letting C' = F( C),
b,.01 argw = b,.c argF(z) = 0 (9.15-6)