1550251515-Classical_Complex_Analysis__Gonzalez_

Singularities/Residues/Applications 657

Theorem 9.3 ·Let z = a be an isolated singularity of f. Then a is a

pole of f iff

lira Z-+a lf(z)I = oo

Proof 1. Suppose that a f=. oo is a pole of multiplicity m ( m 2:: 1) of f.

By {9.2-4) we have

. f(z) = h(z)
(z -a)m

where h(z) is analytic in a neighborhood of a and h(a) f=. 0. It follows that

there is a 81 > 0 such that lz - al < 81 implies that

llh{z)l - lh(a)ll:::; lh(z) - h(a)I <^1 / 2 lh{a)I

or

so that

lh{z)I >^1 / 2 lh(a)I

if lz - al < 81 {Fig. 9.2). Hence we have

lf{z)I > l/2lh(a)I (9.3-1)

lz-alm

provided that 0 < lz - al < 81. Now, for a given K > 0 we have

l/2lh(a)I > K (9.3-2)

lz-alm

( )

1/m •

if 0 < lz - al < l~~I = 82 , say. Hence we obtain from {9.3-1)

and (9.3-2),

lf{z)I > K

Fig. 9.2