56.2. THE MITTAG-LEFFLER THEOREM 1777

The function, f is the sum of a rational function, R1, having poles at P1 with the specifiedsingular terms and an analytic function. Therefore, Q works on K1. Now consider Km form > 1. Then

Q(z) = R1 (z)+m+1

∑k=2

(Rk (z)−Qk (z))+∞

∑k=m+2

(Rk (z)−Qk (z)) .

As before, the infinite sum converges uniformly on Km+1 and hence on some open set, Ocontaining Km. Therefore, this infinite sum equals a function which is analytic on O. Also,

R1 (z)+m+1

∑k=2

(Rk (z)−Qk (z))

is a rational function having poles at ∪mk=1Pk with the specified singularities because the

poles of each Qk are not in Ω. It follows this function is meromorphic because it is ana-lytic except for the points in P. It also has the property of retaining the specified singularbehavior.

56.2.2 A Direct Proof Without Runge’s TheoremThere is a direct proof of this important theorem which is not dependent on Runge’s the-orem in the case where Ω = C. I think it is arguably easier to understand and the Mittag-Leffler theorem is very important so I will give this proof here.

Theorem 56.2.2 Let P ≡ {zk}∞

k=1 be a set of points in C which satisfies limn→∞ |zn| = ∞.For each zk, denote by Sk (z) a polynomial in 1

z−zkwhich is of the form

Sk (z) =mk

∑j=1

akj

(z− zk)j .

Then there exists a meromorphic function, Q defined on C such that the poles of Q are thepoints, {zk}∞

k=1 and the singular part of the Laurent expansion of Q at zk equals Sk (z) . Inother words, for z near zk,

Q(z) = gk (z)+Sk (z)

for some function, gk analytic in some open set containing zk.

Proof: First consider the case where none of the zk = 0. Letting

Kk ≡ {z : |z| ≤ |zk|/2} ,

there exists a power series for 1z−zk

which converges uniformly and absolutely on this set.Here is why:

1z− zk

=

(−1

1− zzk

)1zk

=−1zk

∞

∑l=0

(zzk

)l