= Solution
We use the complete <hyperbolic metric> of curvature $-1$; on the <unit disc> its length element is $2|dz|/(1-|z|^2)$. This normalization matters in the exponential summability criterion. A nonzero bounded <holomorphic function> on the entire plane is constant by <Liouville's theorem>, so the prescribed infinite zero set forces $\Omega\ne\mathbb C$. The <Riemann mapping theorem> supplies a <biholomorphism> $\phi:\Omega\to\mathbb D$, and transports the disk <hyperbolic metric> to $\Omega$.
Fix $w\in\Omega$ and choose $\phi(w)=0$. Put $a_n=\phi(z_n)$ and $F=f\circ\phi^{-1}$. First suppose $F(0)\ne0$. For radii avoiding its zeros, <Jensen's formula> gives
$$
\sum_{|a_n|<r}\log\frac r{|a_n|}\le\log\|F\|_\infty-\log|F(0)|.
$$
The <monotone convergence theorem> as $r\uparrow1$ gives $\sum_n-\log|a_n|<\infty$, hence $\sum_n(1-|a_n|)<\infty$. If $F$ has a zero at zero, factor off its finite order first; the resulting <holomorphic function> is still bounded, since division by that power is bounded away from zero and extends near zero. Thus the <Blaschke condition> holds in either case, with the possible origin zero treated as one additional finite term.
The <Hyperbolic distance in the Poincare disc> satisfies
$$
\rho(w,z_n)=\log\frac{1+|a_n|}{1-|a_n|},\qquad e^{-\rho(w,z_n)}=\frac{1-|a_n|}{1+|a_n|}\le1-|a_n|.
$$
Consequently \b[the exponential distances are summable]. This holds for every basepoint: the <triangle inequality> bounds the ratio of $e^{-\rho(w,z_n)}$ and $e^{-\rho(w_0,z_n)}$ between $e^{-\rho(w,w_0)}$ and $e^{\rho(w,w_0)}$.
Back to article page
