1550251515-Classical_Complex_Analysis__Gonzalez_

(jair2018) #1
Singularities/Residues/ Applications 659

singularity a, the values f(z) become arbitrarily close to any given complex
number b (including the point at infinity), or in another formulation, the
set of functional values w = f(z) is dense in the extended w-plane for z
restricted to any given deleted neighborhood of a. In symbols:

f(N'(a)) = C*


where N'(a) is a given deleted circular neighborhood of a contained in the

domain of definition of f.


Proof 1. Suppose that a -:/:-oo and that there is some complex number c
for which the stated property fails, i.e., such that


lf(z)-cl 2::: €


for all z in NHa) n DJ (DJ = domf). It follows that

1 < ~

lf(z)-cl €

for all z in NH a) n DJ. Hence the function

1
F(z) = f(z) -c

together with f(z), is analytic in 0 < lz - al < 81 for some 81 :::; 8, and

IF(z)I is bounded in 0 < lz - al < 81, that is, F(z) is locally bounded at


a. Thus a is a regular point for F. Since


1
f(z) = c + F(z)

this would imply that z =a is regular for f(z) if F(a) -:/:-O, or that z =a


is a pole of multiplicity m for f(z) if a is a zero for multiplicity m of F(z).
In either case we get a contradiction, since a is assumed to be an isolated


essential singularity of f..


2. If we suppose that lf(z)I :::; K for some K > 0 an.cl for all points

z E NJ(a)nDJ, then f(z) would be regular at a. Again, this is impossible.
This second case is in fact contained in the first. For we can take a b


such that lbl > K and then an t: > 0 such that N.(b) c {w: lwl > K}.


Then, by part 1 there is some z in N8(a) n DJ for which f(z) E Ne(b) and

hence lf(z)I > K.


  1. If a= oo, it suffices to consider g(z) = f(l/z) at z = 0. By p~rts 1
    and 2 given € > 0, 8 = 1/ R > O, b complex and K > O, we have


lg(z) - bl<€
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