57.1. ANALYTIC FUNCTION WITH PRESCRIBED ZEROS 1793

of Ω1. It is necessary to show f is analytic at this point also and right now, f is not evendefined at w.

The {wn} are bounded because {h(zn)} is bounded and limn→∞ |wn−h(zn)|= 0 whichimplies |wn−h(zn)| ≤ C for some constant, C. Therefore, there exists δ > 0 such that ifz ∈ B′ (w,δ ) , then for all n,∣∣∣∣∣ h(zn)−w( 1

z−w

)−wn

∣∣∣∣∣=∣∣∣∣h(zn)−wn

h(z)−wn

∣∣∣∣< 12.

Thus 57.1.9 holds for all z ∈ B′ (w,δ ) and n so by Theorem 57.0.2, the infinite product in57.1.8 converges uniformly on B′ (w,δ ) . This implies f is bounded in B′ (w,δ ) and so w isa removable singularity and f can be extended to w such that the result is analytic. It onlyremains to verify f (w) ̸= 0. After all, this would not do because it would be another zeroother than those in the given list. By 57.1.10, a partial product is of the form

N

∏n=1

(h(z)−h(zn)

h(z)−wn

)egn(z) (57.1.11)

where

gn (z)≡

(h(zn)−wn

h(z)−wn+

12

(h(zn)−wn

h(z)−wn

)2

+ · · ·+ 1n

(h(zn)−wn

h(z)−wn

)n)

Each of the quotients in the definition of gn (z) converges to 0 as z→ w and so the partialproduct of 57.1.11 converges to 1 as z→ w because

(h(z)−h(zn)h(z)−wn

)→ 1 as z→ w.

If f (w) = 0, then if z is close enough to w, it follows | f (z)|< 12 . Also, by the uniform

convergence on B′ (w,δ ) , it follows that for some N, the partial product up to N must alsobe less than 1/2 in absolute value for all z close enough to w and as noted above, this doesnot occur because such partial products converge to 1 as z→ w. Hence f (w) ̸= 0. Thisproves the corollary.

Recall the definition of a meromorphic function on Page 1638. It was a function whichis analytic everywhere except at a countable set of isolated points at which the function hasa pole. It is clear that the quotient of two analytic functions yields a meromorphic functionbut is this the only way it can happen?

Theorem 57.1.8 Suppose Q is a meromorphic function on an open set, Ω. Then there existanalytic functions on Ω, f (z) and g(z) such that Q(z) = f (z)/g(z) for all z not in the setof poles of Q.

Proof: Let Q have a pole of order m(z) at z. Then by Corollary 57.1.7 there exists ananalytic function, g which has a zero of order m(z) at every z ∈ Ω. It follows gQ has aremovable singularity at the poles of Q. Therefore, there is an analytic function, f such thatf (z) = g(z)Q(z) . This proves the theorem.

Corollary 57.1.9 Suppose Ω is a region and Q is a meromorphic function defined on Ω

such that the set, {z ∈Ω : Q(z) = c} has a limit point in Ω. Then Q(z) = c for all z ∈Ω.