= Solution
{c}
The two boundary <circles> meet at $z=\pm1$: subtracting their equations gives $\operatorname{Im}z=0$, and either equation then gives $(\operatorname{Re}z)^2=1$. Send these points to zero and infinity by the <Möbius transformation>
$$
w=T(z)=\frac{z-1}{z+1},\qquad z=\frac{1+w}{1-w}.
$$
The point $-1$ is on the boundary, so $T$ is <holomorphic> throughout $B$. Write $w=u+iv$. Substituting the inverse gives the exact equivalences
$$
\begin{aligned}
|z+i|>\sqrt2
&\iff |1-iw|>|1-w|\iff u+v>0,\\
|z-i|<\sqrt2
&\iff |1+iw|<|1-w|\iff v>u.
\end{aligned}
$$
Thus
$$
T(B)=\{u+iv:v>|u|\}=\{w:\pi/4<\arg w<3\pi/4\}.
$$
There 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.
\Image[/past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2021-q6-crescent.png]
{title=The circular crescent becomes a right-angle sector, then a sector of angle pi over three}
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 is
$$
\boxed{\Phi(z)=\exp\left(\frac23\operatorname{Log}\left(e^{-i\pi/4}\frac{z-1}{z+1}\right)\right).}
$$
Here 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 $0<\arg\xi<\pi/3$, define
$$
w=e^{i\pi/4}\xi^{3/2},\qquad z=\frac{1+w}{1-w},
$$
where $\xi^{3/2}$ uses $0<\arg\xi<\pi/3$. Then $\pi/4<\arg w<3\pi/4$, so $w\ne1$ and $z\in B$, and these are inverse operations. Each stage is <holomorphic> with nonzero <derivative>, proving that $\Phi$ is the requested <conformal bijection>.
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