# Find conformal mapping from sector to unit disc

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.

• 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

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.

• Thanks, I didn't even consider composing it with another holomorphic map! – user61496 Feb 8 '13 at 15:08
• Why is that conformal? Is not it that f'(0)=0? – Ido4848 Nov 15 '17 at 15:52
• @Ido4848 Sure, $f'(0)=0$, but $0$ does not belong to the given sector, so it's irrelevant. Within the sector $f'(0) \neq 0$. – Dan Shved Nov 16 '17 at 17:41
• Why 0 is not within the sector? Is $Arg(0)=0$? – Ido4848 Nov 19 '17 at 6:58
• The argument of $0$ is undefined. – Dan Shved Nov 19 '17 at 13:30

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

to

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:

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.