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Let $S_{\alpha} = \{ z \in \mathbb C \ | \ |arg(z)| \lt \alpha \} $ with $0 \lt \alpha \leq \pi$. I have to find a conformal mapping $\phi: S_{\alpha} \to B(0,1)$.

What would be a good approach to this? Could I map the sector to, for example, the upper half plane and then to the unit ball?

syphracos
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    “Could I map the sector to ... the upper half plane and then to the unit ball?” – Yes, that is a very good approach. – Martin R Jan 09 '23 at 20:35
  • @MartinR thank You! Is $B(0,1)$ actually the unit disk, i.e. the unit ball but in 2 dimensions? – syphracos Jan 09 '23 at 20:43
  • @MartinR https://math.stackexchange.com/questions/298075/find-conformal-mapping-from-sector-to-unit-disc here I found that $f(z) = z^2$ would map $z$ from the sector ${z \in \mathbb C \ | \ \frac{-\pi}{4} \lt arg(z) \lt \frac{\pi}{4} }$ to the upper half plane. Would this also be true in my case? If so, how can I get this result? That $z^2$ sends my sector to the upper half plane. – syphracos Jan 09 '23 at 21:17
  • No, read attentively the reference : not the upper half plane but the right half plane. You will have to multiply by $i$ afterward... – Jean Marie Jan 09 '23 at 22:10
  • @JeanMarie oh I see what You mean, thanks! So with $z^2$ I map it onto the right half plane, multiply by $i$ to rotate it and get the upper half plane, and then use $\frac{z-i}{z+i}$ to map it to the unit disc? – syphracos Jan 10 '23 at 15:12
  • You have well understood... – Jean Marie Jan 10 '23 at 16:17
  • @JeanMarie but can You maybe explain why does $z^2$ map to the right half plane? How can I "derive" this for my case with $0 \leq |arg(z)| \lt \pi$ ? – syphracos Jan 10 '23 at 19:22

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