5

Find a conformal map from the set $\{|z|<1, \Re{z} > 0\}\backslash [0,1/2]$ to the upper half plane.

The main problem I am encountering is that the boundary of the given domain is comprised of two segments which intersect at right angles. So I am unable to map it to the unit disk conformally. What is usually the best strategy to construct maps from/to domains with slits? I try to use the map $\{|z|<1\}\backslash (-1,0] -> \{|z|<1,\Re{z}>0\}$ via $\sqrt{z}$.

Sourav D
  • 1,274
  • 1
    This is what I tried.
    1. Use the map $z \rightarrow z^2$ to send ${|z|<1, \Re{z} > 0}\backslash [0,1/2]$ to ${|z|<1}\backslash [-1,1/4]$.
    2. Then use the map $z \rightarrow \frac{z-1/4}{1-z/4}$ to send it to ${|z|<1} \backslash [-1,0]$.
    3. Use $ z \rightarrow \sqrt{z}$ to send it to the right half disk.
    4. Then use $\frac{z-i}{z+1}$ followed by $z^2$ to send it to the upper half plane.

    I am not sure if the resulting map is conformal.

     – Sourav D Dec 13 '13 at 01:01
  • 1
    I think that since the map $z \rightarrow z^2$ is one-to-one in the right half disk and all the other subsequent maps are also injective in their domain of use, hence the resulting composition map is also injective and therefore conformal. Can anyone comment on this please. – Sourav D Dec 14 '13 at 01:19
  • This is very true ! – the8thone Dec 17 '13 at 16:48

2 Answers2

2

The first three steps from solution in a comment are fine:

  1. Use the map $z\mapsto z^2$ to send $\{|z|<1 : \operatorname{Re} z>0\}\setminus [0,1/2]$ to $\{|z|<1\}\setminus [−1,1/4]$.

  2. Then use the map $z\mapsto \dfrac{z−1/4}{1−z/4}$ to send it to $\{|z|<1\}\setminus[−1,0]$.

  3. Use $z\mapsto \sqrt{z}$ to send it to the right half disk.

  4. I think $\dfrac{z-i}{z+1}$ has a typo: should be $z+i$ in the denominator. Easily fixed. Anyway, my preference is to use the Joukowski map $z\mapsto z+z^{-1}$ for half-disks. It sends the upper half of the unit disk to the lower halfplane, and the lower half-disk to the upper halfplane. So, the remaining steps can be replaced with $z\mapsto -iz$ followed by $z\mapsto z+z^{-1}$.

since the map $z\mapsto z^2$ is one-to-one in the right half disk and all the other subsequent maps are also injective in their domain of use, hence the resulting composition map is also injective and therefore conformal.

This is correct.

What is usually the best strategy to construct maps from/to domains with slits?

Exactly what you used: a combination of square and square root maps that "push" the slit back into the boundary from which it sticks out.

Community
  • 1
Post No Bulls
  • 9,056
0

Consider the map $z \rightarrow \sqrt{z}.$ This will map your domain to a half-disk (and is conformal). Is it easier to find the map now?

Igor Rivin
  • 25,994
  • 1
  • 19
  • 40
  • I am sorry I do not quite understand your suggestion. The map $z \rightarrow \sqrt{z}$ will send the given domain to a sector of angle $\pi/2$ symmetric about the real axis and having a slit from $[0,1/\sqrt{2}]$. I don't see how to proceed from there. – Sourav D Dec 13 '13 at 00:48
  • No, the original angle was $2\pi.$ Half of $2\pi$ is $\pi,$ not $\pi/2.$ – Igor Rivin Dec 13 '13 at 01:40
  • My starting domain is the right half disk ($|z|<1, Re (z) >0 $)with a slit on the real axis - $[0,1/2]$. – Sourav D Dec 13 '13 at 01:58
  • Ah, ok, misread it, more soon. – Igor Rivin Dec 13 '13 at 02:16