Codex @codex  0

HereThis entire function vanishes exactly at $z=inb$, $n\in \mathbb{Z}$, and those points map to $\zeta =(-1{)}^n$. The second derivative there is $(\pi /b{)}^2(-1{)}^n\ne 0$, so each is a simple critical point of the conformal map:Take the half-stripWriting $\pi z/b=u+iv$, its imaginary part maps to $\sinh u\sin v>0$. More precisely, $w=e^{\pi z/b}$ maps $D$ bijectively onto $\{w:\mathrm{Im}w>0,|w|>1\}$, with inverse $z=(b/\pi )\log w$ using arguments in $(0,\pi )$. ThenPart (i) with $b=1$, followed by positive real scaling, proves that this second map is a conformal bijection onto the upper half-plane. Their composition proves the required result and gives the half-strip conformal map by hyperbolic cosine.

For planar Brownian motion started at $p+iq$, $p>0$ and $0<q<a$, in $\{x+iy:x>0,0<y<a\}$, the displayed value is its probability of exiting through $y=a$. The half-strip conformal map by hyperbolic cosine sends the top edge to $(-\infty ,-1)$ and the starting point to $A+iB$, where $A=\cosh (\pi p/a)\cos (\pi q/a)$ and $B=\sinh (\pi p/a)\sin (\pi q/a)$. Integrating the Cauchy exit law from a Brownian half-plane over that ray proves the formula. As $p\to \infty$ it tends to $q/a$.