0
$\begingroup$

This question already has an answer here:

What is the explicit bijective conformal mapping $f(z):G_n\to\mathbb{D}$, $z\in\mathbb{C}$ for the following domain transformations:

$G_1=\{x+iy~|~x>1/2,y>0\}$ is the open region of the first quadrant.

$G_2=\{x+iy~|~x<0,y>0\}$ is the open second quadrant.

$G=_3\{x+iy~|~x<0,y<0\}$ is the open third quadrant.

$G_4=\{x+iy~|~x>0,y<0\}$ is the open fourth quadrant.

Where it is noted that $\mathbb{D}$ is the open unit disc defined by $\mathbb{D}=\{x+iy~|\sqrt{~x^2+y^2}<1\}$. I am aware that the aforementioned transformations have the specific form $f(z)=Ke^{cz}$, $\forall K\in\mathbb{C}$ or $f(z)=z^2$, where such mappings exist by the Riemann Mapping Theorem.

$\endgroup$

marked as duplicate by Namaste, tired, Adrian, Pragabhava, Adam Hughes Oct 18 '16 at 21:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Do you know the corresponding mapping for the open first quadrant, say? Your $G_\nu$ are readily transformed to that. $\endgroup$ – Hagen von Eitzen Aug 19 '15 at 20:36
  • $\begingroup$ The corresponding map it that case is $f(z)=z^2$. @HagenvonEitzen $\endgroup$ – Sergio Charles Aug 19 '15 at 20:47
  • $\begingroup$ Not really. $z\mapsto z^2$ maps the first quadrant to the upper half plane, not to the unit disk. $\endgroup$ – Hagen von Eitzen Aug 19 '15 at 20:53
  • $\begingroup$ It is therefore a composition of the Cayley transform $\kappa(w)$ and $f(z)$. In particular, it is $\kappa(z^2):\Pi_1 \to \mathbb{D}$ or $\kappa(z):z \to \frac{z^2+i}{z^2-i}$. @HagenvonEitzen $\endgroup$ – Sergio Charles Aug 19 '15 at 21:13
  • $\begingroup$ The problem itself seems different, but it is very similar. I was not aware of this post. Thank you. @mathbeing $\endgroup$ – Sergio Charles Oct 18 '16 at 12:23
1
$\begingroup$

Let $\mathbb H$ denote the upper half plane and let $\kappa: \mathbb H \to \mathbb D$ be the Cayley transform $$ z \mapsto {z - i \over z + i}$$

Let $f_2: \mathbb H \to \mathbb H , z \mapsto \sqrt{z}$. Note that $f_2$ maps the second quadrant conformally onto $\mathbb H$.

Hence $\kappa \circ f_2$ maps $G_2$ conformally onto $\mathbb D$.

Now you use similar maps $f_1, f_3, f_4$ for the other $G_i$.

$\endgroup$

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