 Solution

There is no such map for the domain printed in the paper. The planar domain $A=\mathbb{C}\setminus [a,b]$ is not simply connected. For example, a sufficiently large positively oriented circle $\gamma$ surrounds $[a,b]$ and lies in $A$, whileThe integrand is holomorphic on $A$. If $A$ were simply connected, the Cauchy integral theorem would force this integral to be zero. A conformal bijection is a homeomorphism and preserves simple connectivity, whereas $H$ is simply connected. Thus a conformal bijection $A\to H$ is impossible.

The explicit construction also pinpoints the missing point. DefineThis Möbius transformation sends the removed line segment to the nonpositive real axis, together with infinity. More precisely, its image on the stated planar domain issince $T(z)=1$ has no finite solution when $a\ne b$. Conversely, its inverse $z=(bw-a)/(w-1)$ is defined for every $w$ in this set and cannot lie on $[a,b]$, proving this image description exactly.

Take the principal square root of $T(z)$, characterized by positive real part, and then multiply by $i$. The result is the conformal mapEach stage is injective and onto the indicated image; the square root maps the slit plane to the right half-plane, with the point $1$ omitted here. Hence this is a conformal bijection onto the punctured upper half-plane. Its inverse isIf the source is instead $(\mathbb{C}\cup \{\infty \})\setminus [a,b]$ on the Riemann sphere, the same map extends with $F(\infty )=i$ and becomes a conformal bijection onto all of $H$. Indeed, in the local coordinate $u=1/z$ it has expansion $F=i+\frac{i}{2}(b-a)u+O(u^2)$, whose derivative at $u=0$ is nonzero. This gives the sphere version of the requested construction without silently changing the printed planar domain.