 Solution

The two boundary circles meet at $z=\pm 1$: subtracting their equations gives $\mathrm{Im}z=0$, and either equation then gives $(\mathrm{Re}z{)}^2=1$. Send these points to zero and infinity by the Möbius transformationThe point $-1$ is on the boundary, so $T$ is holomorphic throughout $B$. Write $w=u+iv$. Substituting the inverse gives the exact equivalencesThusThere is no missing point from the inverse here: $w=1$ lies outside this sector. The image is therefore exactly the stated sector, rather than merely a subset of it.

Rotate by $e^{-i\pi /4}$ to obtain the first quadrant, and take the power $2/3$ using its holomorphic logarithm with $0<\arg \zeta <\pi /2$. The required map isHere the principal logarithm works because the expression inside it lies in the first quadrant. It multiplies its argument by $2/3$, giving $0<\arg \Phi <\pi /3$, and maps its modulus $r>0$ to $r^{2/3}$, which ranges over all positive real numbers.

For explicit surjectivity and injectivity, given $\xi$ with $0<\arg \xi <\pi /3$, definewhere ${\xi }^{3/2}$ uses $0<\arg \xi <\pi /3$. Then $\pi /4<\arg w<3\pi /4$, so $w\ne 1$ and $z\in B$, and these are inverse operations. Each stage is holomorphic with nonzero derivative, proving that $\Phi$ is the requested conformal bijection.