= Solution
The map $h(z)=e^{\pi z/2}$ is a <conformal bijection> from the width-two strip to the <upper half-plane>, and $h(i)=i$. Its inverse there is $h^{-1}(w)=2L(w)/\pi$.
Write $e^{\pi\beta/2}=A+iB$, where $B>0$. Conjugate the preceding affine half-plane map by $h$:
$$
\boxed{g_\beta(z)=\frac2\pi L\left(A+B e^{\pi z/2}\right).}
$$
Every argument of $L$ lies in the <upper half-plane>. The formula is therefore a conformal <bijection> of the strip onto itself, and
$$
g_\beta(i)=\frac2\pi L(A+iB)=\frac2\pi L(e^{\pi\beta/2})=\beta.
$$
The final equality uses $0<\operatorname{Im}(\pi\beta/2)<\pi$, so no multiple of $2\pi i$ is lost. This is a <strip automorphism with a prescribed image of its midpoint>.
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