I've been reading about confromal maps and fractional linear maps, and have the following problem:
Find a conformal map taking the horizontal strip $\{z \in \mathbb{C}: \text{Re}(z) < 0 \space \text{ and } \space 0 < \text{Im}(z) < \pi \}$ to the first quadrant
I've read some solutions to problems of this nature, but seem to generally have trouble with the intuition behind them. I've basically just been doing guess work, and am seeking a more intuitive explanation.
If I could map the horizontal strip to a quadrant, then rotation solves the issue.
If I could map the horizontal strip to a half-plane, then some combination of rotation and $\sqrt{z}$ will work.
If I could map the horizontal strip to the full horizontal strip $0 < \text{Im}(z) < \pi$, then using $e^{z/2}$ should do the trick.
However, I can't seem to figure out how to map the horizontal half-strip to any of these three cases.
Use conformal map $w_1=e^z$ which is $$ u+iv=e^{x+iy} \Rightarrow \begin{cases} u=e^x\cos y,\\ v=e^x\sin y. \end{cases} \Rightarrow u^2+v^2=e^{2x}<1 $$ for $x<0$. Also when $0<y<\pi$ then $v>0$ thus $w_1$ send the horizontal strip $$\{z\in\mathbb{C}:{\bf Re}\,z<0 \space,\space 0<{\bf Im}\,z<\pi\}$$ to upper semi-circle of $|z|<1$. Now consider $w_2=\dfrac{1+z}{1-z}$ that maps the unit disk to right half plane and the upper part of $|z|<1$ to the first quadrant. The composition of $w_1$ and $w_2$ makes all you wanted.