I have the following problem:
Find an injective and holomorphic function that maps de unit disc to the set given by
$$\mathbb{C}\setminus\{z\in\mathbb{C}:\Re{z}=0,|\Im{z}|\geq1\}.$$
I believe that this set is like a strip $\{z\in\mathbb{C}:\Re(z)\neq 0, |\Im(z)|<1\}$.
Probably I'm wrong, but we can take the map $f:\mathbb{D}\to\mathbb{H}$ given by $f(z)=i\frac{1-z}{1+z}$, the map $g:\mathbb{H}\to\{z=x+iy:-\pi/2<x<\pi/2,y>0\}$ given by $g(z)=\sin{z}$, then rotate the previous half-strip by $z\mapsto iz$ and then extend and compress the half-strip $\{z=x+iy:-\pi/2<y<\pi/2, x>0\}$ to $\{z=x+iy:-1<y<1\}$. Then I don't know how to continue to remove $x=0$. Maybe this idea is super wrong or simply is not the way to get a right answer.
Can you help me? or give me a hint to take a way not to complicated?
Thanks in advance and sorry for the bad english.
I.
We consider a mapping $w=f(\zeta )$ $: \mathbb{H}\to T=\mathbb{H}\setminus \{w\in\mathbb{H}:\Re{w}=0,\Im{w}\geq1\},$ where $\mathbb{H}$ is the upper half plane. $T$ is a triangle, whose three vertices $w=i,-\infty,\infty$ with angles $2\pi, -\pi/2,-\pi/2$ respectively. We want to find $f$ such that $$ f(1)=i,\quad f(\infty)=-\infty, \quad f(0)=\infty. $$ By the Schwarz-Christoffel formula $w=f(\zeta )$ will be of the form \begin{align} f(\zeta )&=i+A\int_1^\zeta \eta^{-\frac{1}{2}-1}(\eta -1)^{1-1}d\eta=i+A \int_1^\zeta \eta^{-\frac{3}{2}}(\eta -1)d\eta\\ &=i+A \int_1^\zeta \left(\frac{1}{\sqrt{\eta}}-\frac{1}{\eta\sqrt{\eta}}\right)d\eta\\ &=i+2A\left( \sqrt{\zeta }+\frac{1}{\sqrt{\zeta} }-2\right). \end{align} Here $A$ is some constant which should be determined so that $f(\zeta )$ is real for all negative real $\zeta $, since $f((-\infty,0))=(-\infty,\infty)$. Easy calculation shows that $A=\frac{i}{2}$. Thus we have $$ f(\zeta )=\frac{i}{2}\left(\sqrt{\zeta }+\frac{1}{\sqrt{\zeta }}\right).$$ II.
Let $\mathbb{Q}$ be the first quadrant and consider $\zeta =\varphi (\xi )=\xi ^2$ which maps $\mathbb{Q}$ to $\mathbb{H}$. Then $$g(\xi )=(f\circ \varphi) (\xi )=\frac{i}{2}\left(\xi +\frac{1}{\xi }\right)$$ maps $\mathbb{Q}$ to $T$. By Schwalz reflection principle $g(\xi )$ maps $\mathbb{H}$ to $\mathbb{C}\setminus\{w\in\mathbb{C}:\Re{w}=0,|\Im{w}|\geq1\}$.
III.
Now we consider $$\xi =\phi(z) =i\frac{1+z}{1-z}$$ which maps $\mathbb{D}$ to $\mathbb{H}$. Then $$ w=(g\circ\phi)(z)=\frac{2z}{z^2-1}$$ maps $\mathbb{D}$ to $\mathbb{C}\setminus\{w\in\mathbb{C}:\Re{w}=0,|\Im{w}|\geq1\}$, which is a desired function.