It is well known (by Schwarz-Christoffel) that if $k \in (0,1)$, then the Jacobi elliptic function $\mathrm{sn}(\cdot,k)$ provides a biholomorphic map from the rectangle $(-K(k),K(k)) \times (0,K'(k))$ to the upper half plane, $\mathbb{R} \times (0,\infty)$. Here, $K(k)$ is the complete elliptic integral of the first kind and $K'(k)=K(\sqrt{1-k^2})$. Using this, I have been able to show that $z \mapsto \frac{\mathbb{dc}(z,k)-\sqrt{k}}{\mathrm{dc}(z,k)+\sqrt{k}} = \frac{\mathrm{dn}(z,k)-\sqrt{k}\mathrm{cn}(z,k)}{\mathrm{dn}(z,k)+\sqrt{k}cn(z,k)}$ provides a biholomorphic map from the rectangle to the doubly slit disk $\mathbb{D}^1 \setminus \left(\left(-1,-\frac{1-\sqrt{k}}{1+\sqrt{k}}\right] \cup \left[\frac{1-\sqrt{k}}{1+\sqrt{k}},1\right)\right)$ that maps the point $iK'(k)/2$ (the center of the rectangle) to the origin, the edges $[-K(k),K(k)] \times \{0\}$ and $[-K(k),K(k)] \times \{K'(k)\}$ to the slits, and finally the edges $\{-K(k)\} \times [0,K'(k)]$ and $\{K(k)\} \times [0,K'(k)]$ to $\mathbb{S}^1$.
I require something more, namely (at least nearly) the same thing, but mapping a point $z_0$ (all of the form $z_0=x_0+ iK'(k)/2$ suffices) in the rectangle to the origin instead. I do not necessarily require the slits in the disk to be of equal length or lying on the same line, but I still need the left and right edges sent to slits (orthogonal to the unit circle) and the bottom and top edges to the unit circle.
All my attempts at this have failed, and I'm starting to think that such a map may not exist (or be accessible). However, complex analysis is not my area, so that's why I'm posting here. Does anyone know if this is possible?
From a map as you are looking for, one could obtain a biholomorphic map of $\mathbb D^1$ to itself that extends continuously to four points of the boundary (corresponding to the vertices of the rectangle) and leaves theses fixed. As we know all automorphisms of $\mathbb D^1$, this is not possible.