# Determining conformal mapping to unit disk with initial conditions

Find a conformal map $f: D\to B$, where $$D = \{z\in\mathbb{C} : \frac{\pi}{4}<\mbox{arg}z<\frac{3\pi}{4}\}$$ and $B$ is the unit disk with conditions: $$f(0)=i\ \ \mbox{and}\ \ f(i)=0$$ from what I managed to gather from reading around about similar problems is that we start by attempting to map region $D$ into something that's more comfortable to work with.
1. How do we decide what to map $D$ to, initially and then how do we do it?
2. How should one use the conditions given for $f$ to determine such $f$?

How should I start tackling such problems?

Some thoughts:

My idea is, if we can somehow rotate $D$ into $D' :=\{z\in\mathbb{C} : 0<\arg z<\pi/2\}$, which should be of the form: $$z\mapsto Az$$ for some $A\in\mathbb{C}$ (to be verified, if it works), the mapping is clearly injective and it's differentiable with $(Az)'\neq 0$. Proceed by $w\mapsto w^2$ which gives us a conformal mapping from $D$ to the upper half plane.

Since composition of injective functions is injective and of differentiable functions is differentiable, then, in principle, the composition of conformal mappings should also be conformal.

A problem: Conformity requires $f'(z)\neq 0$ for all $z\in D$. Can we exclude $0$ from D?

• Looks to me like $0\notin D$ – zhw. May 6 '16 at 15:28

## 1 Answer

Your ideas are good. Here are some additional hints: You can rotate your domain $D$ onto $D^\prime$ by multiplying with a constant of modulus $1$ and argument $-\pi/4$, i.e. $$z\mapsto e^{-\frac{\pi}{4}i} z.$$ After doing this, you already mentioned that squaring could be a good idea to map $D^\prime$ onto the upper half-plane $\mathbb{H}$. Finally, the conformal mapping $$z\mapsto \frac{z-i}{z+i}$$ maps $\mathbb{H}$ onto the unit disk (verify that!). I leave it to you to compose these maps.

Note that, in order to satisfy your additional conditions on $f$, you can multiply the last map with any unimodular constant ($|c|=1$). This will just rotate your result, which obviously leaves the disk invariant.