746
From'(9.15-5) we hav~
4t '.'. - f(z)F(z) = f(z) + g(z)
so that
!le arg f(z) + b.e arg F(z) =!le arg[f(z) + g(z)]
or
!le arg f(z) =!la arg[f(z) + g(z)]
Chapter 9by taking (9.15-6) into account. Then, by the argument principle, we obtain
2rr(N - P) = 2rr(N' - P')and since P = P' it follows that N = N'.
Remark If in the statement of Theorem 9.21 we assume that g is mero-
morphic in R but has no poles on C, then obviously the conclusion of the
theorem is that N - P = N^1 - P'.
Examples 1. Consider the equation z^5 + 8z + 3 = 0. Let f(z) = 8z and
g(z) = z^5 +3. Here both f and g are holomorphic in C. On lzl = 1 we have
lf(z)I = Sizl = 8, lg(z)I:::; izl^5 + 3 = 4so that
lg(z)I < lf(z)I for lzl = 1 (9.15-7)Hence, by Rouche's theorem, the polynomial f ( z) + g( z) = z^5 + Sz + 3 has
inside lzl = 1 as many roots as f(z) = Sz, namely, one root.
Next, let fi(z) = z^5 and g 1 (z) = Sz + 3. On l:zl = 2 we have
lfi(z)I = lzl^5 = 32,so that
lg1(z)I < lfi(z)I for lzl = 2.
Then it follows that f1(z) + g1(z) = z^5 + 8z+ 3 has five roots inside izl = 2,
i.e., as many as Ji ( z) = z^5 • As a consequence, the given equation has four
roots in the ring 1 < lzl < 2. Note that there are no roots on lzl = 1 by
virtue of (9.15-7).- Still another proof of the fundamental theorem of algebra can be
obtained by a simple application of Rouche's theorem. Consider the general
equation of the nth degree (n ;::: 1) written in the form
zn + aizn-l + · · · + an-1Z + an = 0