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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.

azimut
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user61496
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    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 08 '13 at 15:22

2 Answers2

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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

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to

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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.

Community
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guest196883
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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.

azimut
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Dan Shved
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