First is the definition of a meromorphic function. It is one which is analytic except for poles. We assume
here that all functions have complex values.
Definition 17.1.1Let f : Ω → ℂ where Ω is an open subset of ℂ. Then f : Ω → C is called meromorphicif all singularities are either poles or removable. Recall that α is a pole for f if for all z nearα,
∞∑ k ∑n bk
f (z) = ak(x− α) + ------k
k=0 k=1 (z − α)
Thus f is analytic in some deleted ball B^{′}
(α,δ)
meaning for all z such that 0 <
|z − α|
< δ. The complexvalued meromorphic functions defined on Ω will be denoted as ℳ
(Ω )
. When f is analytic in some deletedball around α and
∞∑ ∑∞
f (z) = ak(x− α)k + --bk--k
k=0 k=1 (z − α)
where infinitely many of the b_{k}are nonzero, we say that α is an essential singularity.
It is shown that the poles cannot have a limit point in Ω just as a nonzero analytic function cannot have
a limit point in Ω. Here is a lemma.
Lemma 17.1.2If f has a pole at a, then lim_{z→a}
|f (z)|
= ∞. Also if f ∈ℳ
(Ω )
then the polescannot have a limit point in Ω and there are at most countably many poles. For f ∈ℳ
Consider the second claim. Suppose α_{m} is a pole and lim_{m→∞}α_{m} = α ∈ Ω and f is analytic at α. Then
from the first part, there exists β_{m} such that
|αm − βm |
< 1∕m but
|f (βm)|
> m. Then β_{m}→ α but
lim_{m→∞}
|f (βm)|
does not exist. In particular,
|f (βm )|
fails to converge to f
(α )
showing that f cannot be
analytic at α. Thus it must be the case that α is a pole. But now this is not allowed either
because at a pole the function is analytic on a deleted ball centered at the pole and it is assume
here that lim_{m→∞}α_{m} = α. Thus there are finitely many poles in every compact subset of Ω.
However,
{ ( ) } -------
Ω = ∪ ∞k=1 z : dist z,ΩC ≤ 1 ∩B (0,k) ≡ ∪∞k=1Kk
k
where K_{k} is compact. If Ω = ℂ, let K_{k} =B
(0,k)
. Then by what was just shown, there are finitely many
poles in K_{k} and so the number of poles is at most countable.
Finally, consider the last claim. It is obvious that α is a zero if and only if lim_{z→α}f
(z)
= 0. It was
shown above that at poles lim_{z→α}
|f (z)|
= ∞. Then suppose the limit condition holds. Why is α a pole?
This happens because of the Casorati Weierstrass theorem, Theorem 16.6.4. Every singularity is isolated for
a meromorphic function by definition. Thus there is a Laurent expansion for f near α. If the principal part
is an infinite series, then by this theorem, the values of f near α are dense in ℂ and so lim_{z→α}
|f (z)|
does not even exist. Therefore, this principal part must be a finite sum and so α is a pole.
■
Here is an alternate definition of meromorphic.
Definition 17.1.3f ∈ℳ
(Ω )
for Ω an open set means f is analytic in
B′(α,r) ≡ {z : 0 < |z − α| < r}
for some r > 0 for every α ∈ Ω. At α ∈ Ω either lim_{z→α}
(z − α )
f
(z)
= 0 so α is removable or
lim_{z→α}
|f (z)|
= ∞ which corresponds to α a pole.
The fact that the poles can have no limit point in Ω is fairly significant. It shows that if P is the set of
poles and if Ω is connected, then Ω ∖ P is an open connected set if Ω is.
Lemma 17.1.4Suppose Ω is an open connected set and suppose P is a set of isolated pointscontained in Ω such that P has no limit point in Ω. Then Ω ∖ P is a connected open set.
Proof:Suppose Ω ∖ P = A ∪ B where A,B have empty intersection and A has no limit points of B
while B has no limit points of A and neither is empty. For p ∈ P, B
(p,r)
contains points of A∪B because
this cannot consist entirely of points of P due to what was just shown, that p is not a limit point of P. For
r sufficiently small, all points in B^{′}
(p,r)
are in A ∪ B because p is not a limit point of P but, since Ω is
open, B
(p,r)
⊆ Ω for r small enough. Also, each A,B is an open set since Ω ∖ P is open. Now it is
obvious that B^{′}
(p,r)
is a connected open set contained in A ∪ B and so B^{′}
(p,r)
must be
contained in either A or in B. Let P_{A} be those points of P such that B^{′}
(p,r)
⊆ A for all r
small enough and let P_{B} be defined similarly. Then Ω =
(A∪ PA )
∪
(B ∪ PB )
. If p ∈ P_{A},
then B^{′}
(p,r)
⊆ A and so p is not a limit point of B. From what was shown above, p is not a
limit point of P_{B} either. If x ∈ A, then x is not a limit point of B. Neither is it a limit point
of P_{B} because P_{B} has no limit points in Ω as shown in Lemma 17.1.2. Similarly, B ∪ P_{B}
has no limit points of A ∪ P_{A}. This separates Ω and is a contradiction to Ω being connected.
■
The following is a useful lemma.
Lemma 17.1.5Let f ∈ℳ
(Ω)
and suppose P_{f}the set of poles consist of
{α1,⋅⋅⋅,αn}
. Then there existsa function g analytic on Ω such that for all z
∕∈
P_{f},
∑n
f (z)− Si(z) = g (z)
i=1
where S_{i}
(z)
is the singular part corresponding to α_{i}. That is, for z near α_{i},
mi
f (z) = h (z) + ∑ ---bk---, h analytic near α (17.1)
i k=1 (z − αi)k i i
(17.1)
Proof: Note that f
(z)
−∑_{i=1}^{n}S_{i}
(z)
is meromorphic on Ω. However, it has no poles. Indeed, if α is a
pole, then it must be one of the α_{i} since all the S_{i} would be analytic at α if this were not the case. But
which is not ∞. Thus the function can be redefined at each of the finitely many poles so that the result,
called g
(z)
is analytic on Ω. ■
Because of this lemma, it is all right to be a little sloppy and simply write f
(z)
−∑_{i=1}^{n}S_{i}
(z)
equals
an analytic function.
Proposition 17.1.6Let Ω be a connected open set. Then ℳ
(Ω)
is a field with the usualconventions about summation and multiplication of functions.
Proof:It is tedious but routine to verify that ℳ
(Ω)
is a ring. The part of this which is not entirely
obvious is whether the product of two meromorphic functions is meromorphic. It is clear that fg is analytic
on B^{′}
(α,r)
for small enough r. The only difficulty arises when α is a zero for f but a pole for g.
Thus f
(z)
= ∑_{k=r}^{∞}a_{k}
(z − α)
^{k},g
(z)
= h
(z)
+ ∑_{k=1}^{m}
--bk-
(z−α)k
for h analytic. But this means
f
(z)
g
(z)
=
∑m
f (z)h(z)+ f (z) ---bk-k
k=1(z − α)
If r is as large as m, then fg has a removable singularity. Otherwise, it will have a pole. Thus the product
of meromorphic functions is indeed meromorphic. There are various cases to consider for the sum of two of
these meromorphic functions also but it is routine to verify that the sum is meromorphic. So suppose
f ∈ℳ
(Ω)
and f≠0. Then it is analytic on Ω ∖P where P is the set of poles. Thus the set of zeros has no
limit point. Consider 1∕f. It is analytic in B^{′}
(α, r)
for r small enough. If α is a zero, then
lim_{z→α}
|f1(z)|
= ∞ and 1∕f is analytic near α. If α is not a zero of f, then lim_{z→α}