In the picture below, the symmetry principle was used to show that the rectangle of Figure $36$a is mapped onto Figure $36$b. I did that. Shortly thereafter the function $\mathrm{sn}( \alpha \cdot z)$ is considered, where $\alpha$ is such that $\mathrm{sn}(\alpha \cdot K) = 1$. Then, it claims that with this condition on $\alpha$, the function $\mathrm{sn}(\alpha \cdot z)$ maps Figure $36$a onto the upper half-plane. I can't see why this statement is true. I had some ideas but none with success, if someone could give me a light, I would be grateful.
Idea $1$: Since the function $\mathrm{sn}(z)$ is analytical, in particular, continuous, then it maps boundary to boundary, and connected set to connected set. Thus, it is enough to show that the boundary of the interior of the rectangle is mapped to the real axis, but I couldn't do it.
The other idea was to remember that, by definition, the function $\mathrm{sn}(z)$ maps the interior of the rectangle with vertices $-K, K, K+iK' \hspace{1mm} \mbox{and} \hspace{1mm} -K+iK'$ onto the upper half-plane. Thus, if the function $\mathrm{sn}(\alpha \cdot z)$ is considered, it maps the inside of the rectangle with vertices $- \dfrac{K}{\alpha}, \dfrac{K}{\alpha}, \dfrac{K+iK'}{\alpha} \hspace{1mm} \mbox{and} \hspace{1mm} -\dfrac{K+iK'}{\alpha}$ onto the upper half-plane. My idea was to show that this last rectangle and the rectangle of Figure 36a are the same. I think I should use the condition $\mathrm{sn}(\alpha \cdot K) = 1$ to conclude this fact but I couldn't do it.
Applying the symmetry principle to an inversion of the rectangle in Fig $34$a with respect to its upper side, we find that the function $w=\mathrm{sn}(z;q)$ maps the rectangle of Fig. $36$a onto the full $w$-plane which is furnished with a slit as indicated in Fig. $36$b. Consider now the function $\mathrm{sn}(\alpha z)$ (where $\alpha$ is such that $\mathrm{sn}(\alpha K=1)$) which maps the rectangle in Fig. $36$a onto the upper half-plane.
