# Conformal Mapping onto the Unit Disc

Given the open vertical strip $G=\{x+iy~|~0<x<1,~-\infty<y<\infty\}$, what is the explicit conformal injective map characterizing $w=f(z):G\to\mathbb{D}$?

It is noted that if there exists a mapping $w=M(z):G\to\mathbb{H}$, where $\mathbb{H}=G=\{x+iy~|y>0, (x,y)\in\mathbb{R}\}$. Therefore, the composition of $w=M(z)$ with the Cayley transform given by $\kappa(z):z\to \frac{z+i}{z-i}$, namely $\kappa(w(z)):w\to \frac{w+i}{w-i}$ is such a conformal mapping. However, what is the particular $w=M(z):G\to\mathbb{H}$?

• $z \mapsto e^{i\pi z}$. – user98602 Aug 16 '15 at 1:44
• How would you prove this or show that $w=M(z):G\to\mathbb{H}$, where $w=M(z)=e^{i\pi z}$? @MikeMiller – Sergio Charles Aug 16 '15 at 22:38

I'm going to leave it as a composition of maps. First, let $$f_1(z)=\pi z,$$ which brings us from $$G$$ to the strip $$\Sigma=\{z:\ \Re z\in (0,\pi)\}.$$ As the previous comment suggested, let's consider $$f_2(z)=e^{iz}.$$ By Euler's formula, if $$z=x+iy,$$ then $$f_2(z)=e^{-y}\left(\cos x+i\sin x\right).$$ Here, $$x\in (0,\pi),$$ so $$\cos x\in (-1,1)$$ and $$\sin x\in (0,1).$$ Also, $$y\in \mathbb{R},$$ and so $$e^{-y}\in \mathbb{R},$$ as well. Hence, $$\Re f_2(z)\in \mathbb{R},$$ and $$\Im f_2(z)\in (0,\infty)$$ (all bijectively, clearly). Thus, $$f_2$$ sends $$\Sigma$$ to the upper half plane $$H$$. The standard map $$f_3(z)=\frac{z-i}{z+i}$$ sends $$H$$ to the unit disk $$D_1(0).$$ In total, the composition $$F=f_3\circ f_2\circ f_1$$ sends $$G$$ to the unit disk.