Find the image of the half plane $\operatorname{Re}(z) > 0$ under the linear fractional transformation

Question

I am trying to find the image of the half plane $\operatorname{Re}(z) > 0$ under the linear fractional transformation that maps $0, i, -i$ to $1, -1, 0$ respectively.

I figure that this transformation is $\frac{(z-1)}{(z+1)}$ ? Or is that off too?

Answer 1

Your mapping should be $$w=\frac{z+i}{-3z+i}$$

You can get this by starting with $w=\frac{az+b}{z+c}$ and plugging in the given values to get $a,b,c$

Answer 2

Here’s my strategy: you notice that each triple consists of three points on a straight line. So it should be possible to make a “similar” transformation of points on the real lime, and then go back to get the transformation that you want.

So, if you can get a transformation $f$ that sends $(0,1,-1)$ to $(1,-1,0)$, then your desired $g$ will be $g(z)=f(z/i)$.

Answer 3

Set $f=\frac {az+b}{cz+d}$. Plug in $0$, get $b=d$. Plug in $i$. Get $\frac {ai+b}{ci+b}=-1$. So $c=-a+2bi$. Plug in $-i$. Get $ai=b$. So $c=3bi$. So $f(z)=\frac {-biz+b}{3biz+b}=\frac{1-iz}{1+3iz}=\frac{z+i}{-3z+i}$.

$f$ maps the $y$-axis to the $x$-axis. Plug in a test point, say $1+\frac13 i$. Get $f(1+\frac13 i)=\frac{1+\frac43 i}{-3}$, which is below the $x$-axis. So $Re(z)\gt 0$ is mapped by $f$ to $Im(z)\lt 0$.