 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 $\varphi :\Omega \to \mathbb{D}$, and transports the disk hyperbolic metric to $\Omega$.

Fix $w\in \Omega$ and choose $\varphi (w)=0$. Put $a_n=\varphi (z_n)$ and $F=f\circ {\varphi }^{-1}$. First suppose $F(0)\ne 0$. For radii avoiding its zeros, Jensen's formula givesThe monotone convergence theorem as $r\uparrow 1$ 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 satisfiesConsequently 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 $e^{\rho (w,w_0)}$.