56.2. THE MITTAG-LEFFLER THEOREM 1779

Definition 56.2.3 Suppose f is analytic on {z ∈ C : |z|> r} . Then f is said to have a re-movable singularity at ∞ if the function, g(z) ≡ f

( 1z

)has a removable singularity at 0. f

is said to have a pole at ∞ if the function, g(z) = f( 1

z

)has a pole at 0. Then f is said to

be meromorphic on Ĉ if all its singularities are isolated and either poles or removable.

So what is f like for these cases? First suppose f has a removable singularity at ∞

( f( 1

z

)= g(z) has a removable singularity at 0). Then zg(z) converges to 0 as z→ 0. It

follows g(z) must be analytic near 0 and so can be given as a power series. Thus f (z) is ofthe form f (z)= g

( 1z

)=∑

∞n=0 an

( 1z

)n. Next suppose f has a pole at ∞. This means g(z) has

a pole at 0 so g(z) is of the form g(z) = ∑mk=1

bkzk +h(z) where h(z) is analytic near 0. Thus

in the case of a pole at ∞, f (z) is of the form f (z) = g( 1

z

)= ∑

mk=1 bkzk +∑

∞n=0 an

( 1z

)n.

It turns out that the functions which are meromorphic on Ĉ are all rational functions.To see this, suppose f is meromorphic on Ĉ and note that there exists r > 0 such that f (z)is analytic for |z| > r. This is required if ∞ is to be isolated. Therefore, there are onlyfinitely many poles of f for |z| ≤ r,{a1, · · · ,am} , because by assumption, these poles areisolated and this is a compact set. Let the singular part of f at ak be denoted by Sk (z) .Then f (z)−∑

mk=1 Sk (z) is analytic on all of C. Therefore, it is bounded on |z| ≤ r. In

one case, f has a removable singularity at ∞. In this case, f is bounded as z→ ∞ and∑k Sk also converges to 0 as z→ ∞. Therefore, by Liouville’s theorem, f (z)−∑

mk=1 Sk (z)

equals a constant and so f −∑k Sk is a constant. Thus f is a rational function. In the othercase that f has a pole at ∞, f (z)−∑

mk=1 Sk (z)−∑

mk=1 bkzk = ∑

∞n=0 an

( 1z

)n−∑mk=1 Sk (z) .

Now f (z)−∑mk=1 Sk (z)−∑

mk=1 bkzk is analytic on C and so is bounded on |z| ≤ r. But now

∑∞n=0 an

( 1z

)n−∑mk=1 Sk (z) converges to 0 as z→ ∞ and so by Liouville’s theorem, f (z)−

∑mk=1 Sk (z)−∑

mk=1 bkzk must equal a constant and again, f (z) equals a rational function.

56.2.4 Great And Glorious Theorem, Simply Connected Regions

Here is given a laundry list of properties which are equivalent to an open set being simplyconnected. Recall Definition 51.7.21 on Page 1645 which said that an open set, Ω is simplyconnected means Ĉ \Ω is connected. Recall also that this is not the same thing at all assaying C \Ω is connected. Consider the outside of a disk for example. I will continue touse this definition for simply connected because it is the most convenient one for complexanalysis. However, there are many other equivalent conditions. First here is an interestinglemma which is interesting for its own sake. Recall n(p,γ) means the winding number of γ

about p. Now recall Theorem 51.7.25 implies the following lemma in which BC is playingthe role of Ω in Theorem 51.7.25.

Lemma 56.2.4 Let K be a compact subset of BC, the complement of a closed set. Thenthere exist continuous, closed, bounded variation oriented curves

{Γ j}m

j=1 for which Γ∗j ∩K = /0 for each j, Γ∗j ⊆Ω, and for all p ∈ K,

m

∑k=1

n(Γk, p) = 1.