Homework Help: 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!