14. SINGULARITIES AND WEIERSTRASS THEOREM
If
is analytic only in a punctured disk
, then
 is said to have an isolated
singularity
at
Let A denote the set of all isolated
singularities. We say that the isolated singularity
is an artificial (or removable)
singularity if f can be extended to
so that the resulting function
is holomorphic on
 .
An isolated singularity of
at
is called a pole
of
if
where
 ,
is analytic in
and
 .
The integer
is said to be the order of the pole
and if
 ,
then we say that
is a simple pole of
.
We say that the
isolated singularity
is an essential singularity if
is neither an artificial singularity
nor a pole.
A function
is called a meromorphic
function if
is analytic
for all values in
C except at points where poles
occur.
Theorem.
An isolated singularity of
at
is a pole if and only if
Proof ( )
This follows immediately from the definition of a pole.
( )
Let
Then
has an isolated singularity at
is not identically
and
has a removable singularity at
Then it is easy to show that there
exists a positive integer
such that
in a neighborhood of
where
is analytic there with
Thus
is analytic in a neighborhood of
Now
with
 and the proof is complete.
We can now write down the following immediate-to-prove
but practical:
Corollary.
has a pole of order
at
if and only if
 has
a zero of order
at
Examples.
1)
is an artificial singularity of
 since
2)
For
each integer n,
 is
a simple pole of
3)
is an essential singularity of
since
Weierstrass Theorem. Let
A be a singular subset of an open subset D
of C
and let f be holomorphic in D-A. If
 is
an essential singularity of f, then for every
r>0 with
D-A we have
 X.
Proof. The proof is by
contradiction. Assume that
 X.
Then there exists
 X
such that
  .
Then there exists
 such
that
 .
So
  .
Thus the function
is holomorphic in
and
 . That is, g is bounded in
. By a previous proposition
is an artificial (removable)
singularity of g. So we may regard g as a
holomorphic function on
 We now seek a contradiction in each
of the following two exclusive cases:
Case 1.
 : There exists
 such
that
and therefore
such that
 .
So
 .
Therefore
by the triangle inequality. That is, f is
bounded on
 .
Moreover, if
 then
f is bounded being continuous there. As a
result, f is bounded on
 .
By a previous proposition,
is a removable singularity of f
which is a contradiction.
Case 2.
 : That is
is a zero of g. Let m
denote the order of this zero. Thus there exists a
holomorphic function
 on
 such that
 and
 .
We have
 
That is,
 or
Notice that h is holomorphic on
and
 .
This means that
is a pole of f of order m
which is again a contradiction.
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