Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find a conformal mapping between the sector $\{z\in\mathbb{C} : -\pi/4<\arg(z) <\pi/4\}$ and the open unit disc $D$.

I know that it should be a Möbius transformation, but other than that I am very stuck, any help would be much appreciated.

share|cite|improve this question
Remember that Mobius transformations take circles to circles and lines to lines. Since the boundary of the sector is neither a line nor a circle, Mobius transformations on their own can't possible get you there. – Brett Frankel Feb 8 '13 at 15:22
up vote 3 down vote accepted

Here is a plan: first, apply $z \to z^2$. It will conformally map your sector onto the half-plane $\mathrm{Re}(z) > 0$. Then find a Möbius transformation that will map this half-plane to the unit disk.

share|cite|improve this answer
Thanks, I didn't even consider composing it with another holomorphic map! – user61496 Feb 8 '13 at 15:08

You know that there is a conformal mapping from the unit disk to the upper half plane given by: $$z\mapsto -i\frac{z-i}{z+i}$$ Which sends

enter image description here


enter image description here

But then you know that the transformation $z\mapsto \sqrt z$ taking the principal value sends the upper half plane to the region you are desiring. This gives:

enter image description here

Reversing these mappings gives:

$$w \mapsto \frac{iw^2+1}{-w^2-i}$$

Which you will see is a conformal mapping sending the first quadrant to the unit disk.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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