Conformal mapping to upper half plane I need to find conformal mapping of $U = \{z \in \mathbb{C}:Im(z) >0\}\setminus\{ it:t\in [1;\infty)\}$ to upper half plane. I tried to square $U$ and then use inverse of $w = \frac{1}{2}(z + \frac{1}{z})$ and on paper it seems like i am on the right way, but cannot understand what's wrong. Any hints?
 A: By this answer the map
$$z\mapsto\sqrt{\frac{z^2}{z^2+1}}$$
achieves the desired objective. It is the successive application of

*

*$z^2$: maps $U$ to $U_1=\mathbb C\setminus((-\infty,-1]\cup[0,\infty))$

*$\frac z{z+1}$: maps $U_1$ to $U_2=\mathbb C\setminus[0,\infty)$

*$\sqrt z$: maps $U_2$ to upper half-plane

A: It would be a good idea to embed $U$ into the Riemann sphere $\overline{\mathbb C}$, since it lends itself better to geometrical intuition. See this picture:

The blue line is the real axis, green is the imaginary axis, and red is the unit circle. On this picture, $U$ is the backside of the sphere (with $\mathrm i$ at the center), but with the upper part of the dashed green line removed. Squaring yields $U'=\mathbb C\backslash{(-\infty,-1]\cup[0,\infty)}$. That's the entire sphere, but with 3/4 of the blue circle missing (the arc starting at $0$ and going counterclockwise to $-1$. Which is not far removed from the slashed complex plane ($\mathbb C^-:=\mathbb C\backslash(-\infty,0]$), which is the sphere with a different part of the blue circle missing (the clockwise arc from $\infty$ to $1$). If you can find a conformal mapping from $U'$ to $\mathbb C^-$, and a conformal mapping from $\mathbb C^-$ to the upper half plane, you're done (just compose all of the conformal mappings).
