On the Solution of the Dirichlet Problem with Rational Holomorphic ...

5 (2005), No. 2 The Dirichlet Problem with Rational Holomorphic Data 449
Since K(z, a 2 ) and f ′ (z) are rational, this implies that f(z) is algebraic. Thus,
by Lemma 1, f is rational, as desired.
For ease of reading, we shall write the implication (iii) ⇒ (ii) as a theorem.
Theorem 2. Let Ω be a domain as in Theorem 1, and suppose that a Riemann
map f : Ω → D is rational. Then the solution to the Dirichlet problem
{
∆u = 0 in Ω
u(z, z) = R(z)
on ∂Ω
is rational for every rational function R(z) without poles on ∂Ω.
Proof. Fix a rational function R(z), and let a 1 , a 2 , . . . , a n be the poles of R in Ω.
Expanding R in terms of its singular parts, we have
(5) R(z) =
n∑
s k (z) + ˜R(z) =
k=1
n∑
p k
∑
k=1 j=1
c k,j
(z − a k ) j + ˜R(z),
where the p k and c k,j are fixed constants, and the function ˜R(z) is holomorphic
in Ω and rational. We now propose the following lemma.
Lemma 2. For any fixed k, 1 ≤ k ≤ n, there exist constants b k,j
rational function Q k (z) ∈ O(Ω) such that
(6)
p k
∑
j=1
p
b k,j
[f(z) − f(a k )] = ∑ k
c k,j
j (z − a k ) + Q k(z).
j
j=1
∈ C and a
Proof of Lemma 2. Expanding f(z) as a power series about the point a k yields
i.e.
Thus, we have
(7)
f(z) = f(a k ) + f ′ (a k )(z − a k ) + O(z − a k ) 2 ,
f(z) − f(a k ) = (z − a k )[f ′ (a k ) + O(z − a k )].
b k,j
[f(z) − f(a k )] = b k,j
1
j (z − a k ) j [f ′ (a k ) + O(z − a k )] j
b k,j
1
=
f ′ (a k ) j (z − a k ) j [1 + O(z − a k )] . j
Recall that f is one-to-one in Ω, so that f ′ (a k ) ≠ 0 (since a k ∈ Ω). Hence,
it is permissible to have this term in the denominator of the right-hand side of