# Image of two homographic transformations (Möbius transformations)

1. Jan 2, 2014

### mahler1

The problem statement, all variables and given/known data.

1) Let $T$ be the transformation over the extended complex plane that sends the poins $0,i,-i$ to the points $0,1,\infty$. Prove that the image of the circle centered at the origin and of radius $1$ under this transformation is the line $\{re(z)=1\}$.

2) For $\alpha \in \mathbb C$ such that $|\alpha|\neq 1$, prove that the homographic transformation $T(z)=\dfrac{z-\alpha}{-\overline\alpha z+1}$ sends the circle $\{|z|=1\}$ to itself and sends $\alpha$ to $0$.

The attempt at a solution.

1) Let $T(z)=\dfrac{az+b}{cz+d}$. Then, $T(0)=0$, so $b=0$. $T(i)=1$ and from this condition we get that $ai=ci+d$, so $a=c-di$. From the last condition we get that $c(-i)+d=0$, so $d=ci$. Finally, $T(z)=\dfrac{2z}{z+i}$.

Sorry if this is obvious but how do I show that for any $z \in \{|z|=1\}$, $T(z)=w$ with $w$ of the form $w=1+bi$, $b \in \mathbb R$?.

2) I have the same problem, if $z \in \{|z|=1\}$, I want to show that $T(z)=w$ such that if $w=a+bi$, then $a^2+b^2=1$.

Edition:

I could solve 2), what I did was ${T(z)}^2=T(z) \overline {T(z)}$ and I could prove that the last expression equals to $1$.

I would appreciate some help for 1).

Last edited: Jan 2, 2014
2. Jan 3, 2014

### Dick

For 1) if you know Mobius transformations will map circles to either lines or circles then if you just compute T(1), T(i) and T(-i) and notice they are all on the line Re(z)=1 that will do it. If you want to be more explicit I'd multiply numerator and denominator of 2z/(z+i) by z* (remembering zz*=1 since |z|=1) and then use that z=a+bi where a^2+b^2=1 and convert it to rectangular form.

3. Jan 3, 2014

Thanks Dick!