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.

Use the principal logarithm $\mathrm{Log}z=\log |z|+i\mathrm{Arg}z$ with $-\pi <\mathrm{Arg}z<\pi$, whose branch cut is the negative real axis. For this choice,has three simple poles away from the cut: $z=1$, $z=ia$, and $z=-ia$. The pole at $1$, caused by $\mathrm{Log}1=0$, must be included.

Choose $0<\epsilon <\min \{1,a\}$ and $R>\max \{1,a\}$, and use a positively oriented keyhole contour around the negative real axis: the large circle is traversed counterclockwise, the upper bank from $-R$ to $-\epsilon$, the small circle clockwise, and the lower bank from $-\epsilon$ to $-R$. Formally, use radial segments at arguments $\pm (\pi -\eta )$, then let $\eta \downarrow 0$ with $R,\epsilon$ fixed. The function has continuous limiting values on each compact bank, which justifies this limit.

Put $L=\log a$. The residues areThe first follows from $\mathrm{Log}z=(z-1)+O((z-1{)}^2)$; the other two follow from the simple zeros of $z^2+a^2$. Their sum isThese expressions are valid also for $a=1$, since $L\pm i\pi /2$ never vanish for real $L$.

On the outer circle, $|\mathrm{Log}z|\ge \log R$ for $R>1$ and $|z^2+a^2|\ge R^2-a^2$. On the inner circle, $|\mathrm{Log}z|\ge |\log \epsilon |$ and $|z^2+a^2|\ge a^2-{\epsilon }^2$. Thus the estimation lemma bounds the two arc integrals byrespectively, as $R\to \infty$ and $\epsilon \downarrow 0$.

On the upper bank, $z=-x+i0$ gives $\mathrm{Log}z=\log x+i\pi$; on the lower bank it gives $\log x-i\pi$. Accounting for the opposite orientations, their combined integral isThe real improper integral converges: its integrand is bounded near zero by $1/(a^2{\pi }^2)$ and at infinity by $1/({\pi }^2{x}^2)$. Taking the arc limits in the residue theorem gives $-2\pi iI=2\pi i\Sigma$, henceThe final subtraction is exactly the contribution of the pole at $1$.