= Solution
Here
$$
F'(z)=\frac\pi b\sinh\frac{\pi z}{b}.
$$
This entire function vanishes exactly at $z=inb$, $n\in\mathbb Z$, and those points map to $\zeta=(-1)^n$. The second <derivative> there is $(\pi/b)^2(-1)^n\ne0$, so each is a simple critical point of the <conformal map>:
$$
\boxed{z=inb\ (n\in\mathbb Z),\qquad\text{critical values }\zeta=\pm1.}
$$
Take the half-strip
$$
\boxed{D=\{z:\operatorname{Re}z>0,\ 0<\operatorname{Im}z<b\}.}
$$
Writing $\pi z/b=u+iv$, its imaginary part maps to $\sinh u\sin v>0$. More precisely, $w=e^{\pi z/b}$ maps $D$ bijectively onto $\{w:\operatorname{Im}w>0,\ |w|>1\}$, with inverse $z=(b/\pi)\log w$ using arguments in $(0,\pi)$. Then
$$
F(z)=\frac12\left(w+\frac1w\right).
$$
Part (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>.
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