How can i transform $\{z \in \mathbb{C}: 0 <Im(z) <1\}/[0,i/2]$ to $\{ z \in \mathbb{C} Im(z)>0\}$?

Question

it would be direct by $e^{\pi z}$ if we didn't have that finite slit in the imaginary axe, so how can i deal with that slit?

Answer 1

This is slightly complicated.
I explain the solution step-by-step.

First step:
The function $e^{\pi z}$ maps $$D=\{z\in \mathbb{C}: 0<\operatorname{Im}\, z<1\}\setminus [0,i/2]$$ to $$E=\{z\in \mathbb{C}: 0<\operatorname{Im}\, z\}\setminus \{z\in\mathbb{C}: z=e^{i\theta },0\le \theta \le \pi/2\}.$$ $E$ is the upper half plane with a circular slit in the first quadrant.

Second step: This is a renewal.
The old one is not so easy. It is to be erased.
Let $\varphi (z)=\frac{z-1}{z+1}.$
Note that $\varphi $ maps the upper half plane to the upper half plane and the image of the circular slit $\{z\in\mathbb{C}: z=e^{i\theta },0\le \theta \le \pi/2\}$ by $\varphi $ is the line segment $[0,i]$.
Hence we see that $\varphi $ maps $E$ to $F=\{z\in\mathbb{C}: \operatorname{Im}z>0\}\setminus [0,i]$.
($\varphi$ and $F$ are changed.)

Third step:
Let $\phi(z)=\sqrt{z},$ where $\sqrt{z}$ is defined in $\mathbb{C}\setminus [0,\infty)$ so that $$\sqrt{re^{i\theta }}=\sqrt{r}e^{i\frac{\theta }{2}},\quad z=re^{i\theta }, 0\le \theta <2\pi.$$ Then $\phi(z^2+1)=\sqrt{z^2+1}$ maps $F$ to the upper half plane.

Answer 2

The Möbius transformation $\varphi(z)=\frac{z-1}{z+1}$ doesn't map the upper half plane to the upper half plane because determinant = $(ad-bc)\neq1$.