Singularities/Residues/ Applications 659singularity 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 thedomain 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) -ctogether 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.
- 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