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?

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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.