2
$\begingroup$

Given the open vertical strip $G=\{x+iy~|~0<x<1,~-\infty<y<\infty\}$, what is the explicit conformal injective map characterizing $w=f(z):G\to\mathbb{D}$?

It is noted that if there exists a mapping $w=M(z):G\to\mathbb{H}$, where $\mathbb{H}=G=\{x+iy~|y>0, (x,y)\in\mathbb{R}\}$. Therefore, the composition of $w=M(z)$ with the Cayley transform given by $\kappa(z):z\to \frac{z+i}{z-i}$, namely $\kappa(w(z)):w\to \frac{w+i}{w-i}$ is such a conformal mapping. However, what is the particular $w=M(z):G\to\mathbb{H}$?

$\endgroup$
  • 1
    $\begingroup$ $z \mapsto e^{i\pi z}$. $\endgroup$ – user98602 Aug 16 '15 at 1:44
  • $\begingroup$ How would you prove this or show that $w=M(z):G\to\mathbb{H}$, where $w=M(z)=e^{i\pi z}$? @MikeMiller $\endgroup$ – Sergio Charles Aug 16 '15 at 22:38
2
$\begingroup$

I'll fill in the details of the comment and write out the full solution:

I'm going to leave it as a composition of maps. First, let $f_1(z)=\pi z,$ which brings us from $G$ to the strip $\Sigma=\{z:\ \Re z\in (0,\pi)\}.$ As the previous comment suggested, let's consider $f_2(z)=e^{iz}.$ By Euler's formula, if $z=x+iy,$ then $f_2(z)=e^{-y}\left(\cos x+i\sin x\right).$ Here, $x\in (0,\pi),$ so $\cos x\in (-1,1)$ and $\sin x\in (0,1).$ Also, $y\in \mathbb{R},$ and so $e^{-y}\in \mathbb{R},$ as well. Hence, $\Re f_2(z)\in \mathbb{R},$ and $\Im f_2(z)\in (0,\infty)$ (all bijectively, clearly). Thus, $f_2$ sends $\Sigma$ to the upper half plane $H$. The standard map $f_3(z)=\frac{z-i}{z+i}$ sends $H$ to the unit disk $D_1(0).$ In total, the composition $F=f_3\circ f_2\circ f_1$ sends $G$ to the unit disk.

I know that this is an old question, but it's a fairly standard problem, so hopefully this write-up helps anyone else who might come across the same exercise with the same confusions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.