the transformation of mapping the right half plane onto |w|<2

Question

I have a question about finding the transformation function.

Can you please find the transformation which maps the right half plane $R(z)>0$ onto the circle $|w|<2$?

Thank you.

Answer 1

The following is a proof I once wrote for a similar situation where the left half-plane was required instead of the right half-plane and where I dealt with a disk of radius $1$ instead of $2$. After the proof, I will quickly descbribe the changes for your situation.

First of all, let $\mathcal{H}$ be the left half-plane and $\mathbb{D}$ the disk of radius $1$. It is possible to define a map $\phi:\mathcal{H}\to \mathbb{D}$ as follows: $$w\overset{\phi}{\mapsto}\frac{w+1}{w-1}.$$

This map is indeed from $\mathcal{H}$ to $\mathbb{D}$, since for $w\in\mathcal{H}$ we have that $|\Re(w+1)|\leq|\Re(w-1)|$ and thus $|\phi(w)|=\frac{|w+1|}{|w-1|}\leq 1$. For $w\in \partial\mathcal{H}$ we have that $\Re(w)=0$, which implies that $w=\alpha i$ for some $\alpha\in\mathbb{R}$. Therefore $|\phi(w)|=\frac{\alpha i+1}{\alpha i -1}= \frac{\sqrt{\alpha^2+1^2}}{\sqrt{\alpha^2+(-1)^2}} = 1$.

The map $\phi$ is also an injective map. Suppose $\phi(\alpha)=\phi(\beta)$, then $\frac{\alpha+1}{\alpha-1}=\frac{\beta+1}{\beta-1}\iff \alpha\beta+\beta-\alpha-1 = (\alpha+1)(\beta -1) = (\beta+1)(\alpha -1) = \alpha\beta-\beta+\alpha-1 \iff 2\alpha=2\beta \iff \alpha=\beta.$ Moreover, since $1\not\in \mathcal{H}$, $\phi$ is holomorphic.

Required changes.

To scale the disk from radius $1$ to $2$, just multiply the image of the map $\phi$ by $2$. In order to flip the half-plane required, interchange numerator and denominator of the image of $\phi$. (The statement about $1\not\in\mathcal{H}$ then becomes one about $-1$.) The remarks about the boundary were necessary at the time, since I dealt with a disk and half-plane having boundary, where the boundaries were at some point required to be mapped to each other.