1550251515-Classical_Complex_Analysis__Gonzalez_

(jair2018) #1
Singularities/Residues/Applications

where the last term is analytic at ak. It follows that

h(z) f'(z) = akh(z) + h(z) g'(z)
f(z) z - ak g(z)
and
f'(z)
!~~ h(z) f(z) = akh(ak)

Similarly, at a pole br we obtain
f'(z)
z=br Res h(z)-f( Z ) = -fJrh(br)

743

I
Hence on applying the general form of the residue theorem, formula (9.14-6)
results.

Corollary 9.7 If in (9.14-6) we let h(z) = z, we obtain


1 J J'(z) m n


2 7ri c z f(z) dz= Lakakf!c(ak)-k=l Lf3rbrr!c(br) r=l


If, in addition, we assume that.C is a simple closed contour described once
in the positive direction, and that C encloses all the points ak and bn we get

1 J J'(z) m n
27ri z f(z) dz= L akak - L f3rbr
c+ k=l r=l

(9.14-7)

The right-hand side of (9.14-7) represents the difference between the
sum of the values of the zeros and the sum of the values of the poles of

f inside C (counting multiplicities). In particular, if C encloses just the

simple zero a 1 and no poles, we have

1 J f'(z)


2 7ri z f(z) dz= a^1

c+

9.15 The Argument Principle and Its Consequences


(9.14-8)

Theorem 9.20 (The Argument Principle). Suppose that f(z) is a mero-
morphic function in a region R, and let C be a simple closed contour

. homotopic to a point in R and passing through none of the zeros or poles


of f. Then as z describes C once in the positive direction the argument

of w = f ( z) increases (or decreases) by a multiple of 271" according to the


formula

Acargf(z) = 27r(N-P)
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