# Mapping the complex right half plane to the interior of a circle

I don't know how to approach this problem: Let $P$ be the complex plane; the function $$z'(z)=\frac{az+b}{cz+d}$$ maps $P\rightarrow P$, where $z'$, $z$ are complex variables and ($a$, $b$, $c$, $d$) are complex constants. What is the function $z'(z)$ that maps the superior semi-plane defined by $\Re(z)>0$, to the interior of a circle of radius 1 defined by $|z'|<1$? What is the inverse of this function $z(z')$ (maps the interior of a circle of radius 1 to the superior semi-plane)?

I would appreciate any relevant hints to find the ($a$, $b$, $c$, $d$) needed for such a transformation (for example, I find that the condition $|z'|<1$ yields $|z|>|\frac{b-d}{c-a}|$).

• What is superior semi plane? The complex upper half plane? Commented Apr 20, 2018 at 2:42
• @See-WooLee my mistake I was referring to the complex right plane. Commented Apr 20, 2018 at 19:51

A complex number $z$ is in the upper half space iff its distance to $i$ is strictly smaller than its distance to $-i$, iff $\vert {z-i\over z+i} \vert <1$. Therefore
$h(z)={z-i\over z+i}$ does the job, as $z\in \cal H \iff \vert h(z) \vert <1 \iff h(z)\in D$
You can replace $i$ by any complex number $z_0$ with positive imaginary part, and $h$ by $z-z_0\over z-{\bar z_0}$.
• I'm sorry, the title of the question was incorrect, instead of $\Im(z)>0$, I was referring to $\Re(z)>0$. In any case your answer ($\Im(z)>0$) can be made more general by $z'(z)=\alpha\frac{z-z_0}{z-z^*_0}$ where $|\alpha|=1$ ($\Im(z_0)>0$ as before). If instead we consider $\Re(z)>0$, the solution is now (by rotation) $z'(z)=\alpha\frac{z+i z_0}{z+i z^*_0}$, with $|\alpha|=1$, and $\Im(z_0)>0$ as before. Commented Apr 20, 2018 at 20:08
I find it easier to compute the inverse map first. The map will map boundary to boundary, so it has to map some point on the unit circle to $\infty$ If we choose this point to be $1$ we need $c=-d$. Then it has to map $-1$ to the real line so $\Im\left(\frac{-a+b}{-c+d}\right)= 0.$ Also $0$ is mapped to the upper half plane so $\Im\left(\frac{b}{d}\right)> 0$ It's easy to find $a,b,c,d$ that satisfy these conditions, and then to find the inverse map.
• A conformal mapping will always map the boundary to the boundary. If you are looking for the most general mapping, just map an arbitrary point to $\infty.$ Commented Apr 20, 2018 at 21:40