742 Chapter^9
0 :::; t ~ 2rr, is given by
1 J f'(z) d - Res f'(z) =Res...!:_ !'(1/z)
N = 2rri f(z) z = z=oo f(z) z=O z^2 J(l/z)
c+1 nco + · · · + Cn-1Zn-l
=Res- =n
z=O z Co+···+ CnZn
i.e., the total number of zeros of a polynomial of degree n ~ 1 equals n.
This is an alternative proof of the fundamental theorem of algebra.
The following theorems are generalizations of Theorem 9.16.
Theorem 9.17 With the same assumptions and notations of Theo-
rem 9.16, suppose that the points ak (k = 1, ... , m) are c-points of the
function. f(z). Then we have
-^1 - f f' ( z) dz = N - P
2rri le+ f(z) - c
(9.14-4)Proof H suffices to let g( z) = f ( z )-c. Then the c-points off are the zeros
of,g while the poles are the same, and formula (9.14-1) applies.
Theorem 9.18 ' With the same assumptions and notations of Theo-
rem 9.16, suppose now that the contour C homotopic to a point in R, and
passing through none of the points ak, br, is not necessarily simple and thatit may or may not enclose all the points ak, br (k = 1, ... , m; r = 1, ... , n).
In addition, suppose that C is described a certain number of times in
either direction. Then we have1 J f'(z) m n
2 rri c f(z) dz= k=l L:aknc(ak)-L:,arnc(br) r=l
(9.14-5)where O:c( a) denotes, as before, the winding number of C with respect to a.Proof It follows from. the general form of the residue theorem (Theo-
rem 9.10).
Theorem 9.19 With the same assumptions and notations of the preced-
ing theorem, let h( z) be any holomorphic function in R. Then1 J J'(z) m n
2 rri c h(z) f(z) dz=~ akh(ak)nc(ak) - ~ ,8rh(br)!2c(br) (9.14-6)
Proof In a deleted neighborhood of the zero ak we havef'(z) = ~ + g'(z)
f(z) z - ak g(z)