1550251515-Classical_Complex_Analysis__Gonzalez_

Singularities/Residues/ Applications 741

in some deleted neighborhood of a. Hence a is a simple pole of f' / f with

residue -a.

Theorem 9.16 Suppose that f(z) is a meromorphic function in a region

R having in this region the zeros ak with multiplicities ak ( k = 1, ... , m ),

and the poles br with multiplicities f3r (r = 1, ... , n ), respectively (and no

other zero or pole). Let

and

and let C be a simple closed contour homotopic to a point in R and

enclosing the points ak and br (k = 1, ... , m; r = 1, ... , n). Then

1 J f'(z)

27ri f(z) dz=N-P (9.14-1)

c+

Proof By Theorem 9.15 at each zero ak the function f' / f has a simple
pole with residue ak, and at each pole br the function f' / f has a simple

pole with residue -fir· Hence by the residue theorem we have

1 J f'(z) m n

-. - dz= L ak - L f3r = N - P

27ri c+ f(z) k--1 r-_ 1

Remark Formula (9.14-1) can be written in the form

-.^1 J -d" d logf(z)dz = N - P

27ri z

(9.14-2)

c+

The integral on the left of (9.14-2) is called the logarithmic residue of f(z)

with respect to the contour c+. We note that N represents the total

number of zeros of f inside C and P the total number of poles of f inside

C, counting multiplicities.

Corollary 9.6 If f is analytic in R, then

__!_ j f'(z) dz= N

27ri f(z)

(9.14-3)

c+

In particular, suppose that f is a polynomial of degree n > 1, and let

f(z) = c 0 zn + c 1 zn-l + · · · + Cn· We know that for r large enough f(z) h~
no zeros for lzl 2:: r. Hence the number of zeros of f(z) inside C: z = re•t,