Find a conformal map from the set $$\{z \in \mathbb{C}: |\operatorname{Im}z| < \pi \}\setminus \left[-\pi i; 0 \right]$$ to the upper half-plane.
I have used a composition of the following maps: $$z \mapsto e^{z},\space z \mapsto \frac{1}{z + 1}, \space z \mapsto z – \frac{1}{2}$$
Hence I have got the set $$\mathbb{C}\setminus\left(\left(-\infty; -\frac{1}{2} \right]\cup\left[\frac{1}{2};+\infty \right) \cup\left[0; \infty·i \right]\right)$$
But I'm wondering what the next steps are.
[Math] Finding a conformal map to the upper half-plane
complex-analysisconformal-geometry
Best Answer
Let \begin{align} \varphi &: z \mapsto \exp\left(\frac{\,z\,}{2}\right),\\ \psi &: z\mapsto i\frac{z-1}{z+1},\\ \phi &: z \mapsto \sqrt{z},\\ g &: z \mapsto \left(\frac{1+z}{1-z}\right)^2, \end{align} where $\sqrt{z}$ is defined in $\mathbb{C}\setminus [0,\infty)$ so that $\sqrt{re^{i\theta }}=\sqrt{r}e^{i\theta /2} $ for $0<\theta <2\pi$.
Also we define \begin{align} D&=\{z\in \mathbb{C} : |\operatorname{Im}\, z|<\pi\}\setminus [-\pi i; 0],\\ D_1&=\{z\in \mathbb{C} : \operatorname{Re}\, z>0\}\setminus \left\{z : z=e^{i\theta}, -\frac{\pi}{2}<\theta \le 0\right\},\\ D_2&=\{z\in \mathbb{C} : |z|<1\} \setminus [0; 1),\\ D_3&=\{z\in \mathbb{C} :\operatorname{Im}\,z>0, |z|<1\}. \end{align}
Note that \begin{align} &\varphi : D\to D_1,\quad \psi : D_1\to D_2, \quad \phi : D_2\to D_3,\\ &g : D_3\to \mathbb{H}, \end{align} where $\mathbb{H}$ is the upper half-plane. See the diagram below.
Thus we get $$ f(z)=g\circ\phi\circ\psi\circ\varphi (z)=\left(\frac{1+\sqrt{i\frac{e^\frac{z}{2}-1}{e^\frac{z}{2}+1}}}{1-\sqrt{i\frac{e^\frac{z}{2}-1}{e^\frac{z}{2}+1}}}\right)^2 $$ as a desired mapping function.