Conformal mapping between $\mathbb{C}^{-}$ to $\mathbb{E}$ I need to find a conformal map between $\mathbb{C}^{-}:= \mathbb{C}\setminus ]-\infty;0]$ and $\mathbb{E} := \{z \in \mathbb{C} : |z| < 1\}$.
I only do know the Cayley map as well as the complex square root, complex exponential function and the complex logarithm.
How do I find such a map?
I am aware of the fact that the Cayley map does map every complex number in $\{z \in \mathbb{C}: \text{Im}(z) > 0\}$ to $\mathbb{E}$. I guess that might be a good starting point. Can anyone help?
 A: It is well known by elementary complex analysis that if $\alpha\in(0,2]$, then the map $z\mapsto z^{\alpha}$ maps the right-half plane to $\{z\in\mathbb{C}-\{0\}: |\arg(z)|<\alpha\cdot\pi/2\}$. Now for $\alpha=2$ the right half-plane is mapped through $z\mapsto z^2$ onto your domain, namely $\mathbb{C}\setminus(-\infty,0]$. Also, $z^2$ is a one-to-one map when restricted to the right-plane, since $z^2=w^2$ means that $z=w$ or $z=-w$, but if $z,w$ lie on the right half plane then $w$ lies on the left half plane.
Anyway, composing with the map $f(z)=\frac{1+z}{1-z}$ that maps the unit disk onto the right-half plane with a one-to-one manner, you get your conformal mapping.
A: Vercassivelaunos has basically answered it in the comments, but I'll post an answer here with a bit more detail.
As you know, the Cayley map gives a conformal bijection $\{z\in\mathbb{C}~|~\mathrm{Im}(z)>0\}\to\mathbb{E}$. So we need to find a conformal bijection $\mathbb{C}^-\to\{z\in\mathbb{C}~|~\mathrm{Im}(z)>0\}$.
To make $\mathbb{C}^-$ into a half plane, you need to use a square root. Which branch do you need to pick? As in, how exactly will it act on an element of $\mathbb{C}^-$?
Once you have your half-plane, it won't be oriented correctly, so you'll need to rotate it. By what angle will you need to rotate it, and how do you express this as a conformal map?
Composing these two things with the Cayley map will give you what you're looking for. I'll leave you to work out the details!
