Sufficient conditions for a mobius transformation to map the unit circle to itself.

Question

A question on Conway's Complex Functions of One Variable asks: Find necessary and sufficient conditions for a Mobius transformation $T(z)=\frac{az+b}{cz+d}$ to map the unit circle to itself. So if $\gamma$ is a circle, $T(\gamma)=\gamma$.

I've worked out the necessary conditions. Namely, if $T(\gamma)=\gamma$, then

1) $|a|^2+|b|^2=|c|^2+|d|^2$

2) $a\bar{b}=\bar{d}c$

3) $\bar{a}b=d\bar{c}$

How does one go about showing sufficiency? Should I simply assume conditions 1),2) and 3) and try to prove that $T(\gamma)=\gamma$? If so I can simply claim that all the implications I used to get these conditions also work backwards. Or just show that $|\frac{az+b}{cz+d}|=1$ by these conditions, which is rather simple. Is that all there is to this? I just wish the whole "necessary and sufficient" language was scrapped for some direct notation.

As an aside, I'm wondering if I'm using the words "necessary" and "sufficient" correctly in this context. Is what I showed in the first part the necessary conditions (that's what makes sense to me semantically, because they are necessary once I've assumed the map), or are they the sufficient conditions?

Any help is much appreciated.

Answer 1

The map $T(z) = \frac{az+b}{cz+d}$ sends the unit circle to itself if and only if for any $\zeta$ in the circle, $|T(\zeta)|=1$. Now you just have to translate this into conditions on the coefficients.

$|T(\zeta)|=1$ is equivalent to $$ |a|^2 + |b|^2 + 2Re(a\bar{b} \zeta) = |c|^2 +|d|^2+ 2Re(c\bar{d}\zeta) $$

Since $T$ is determined by the image of three points, just evaluate the equation above for three distinct values of $\zeta$ and you'll get the necessary and sufficient conditions. For $\zeta = 1, i, -1$, you'll get the conditions 1) and 2).

Answer 2

here we have a unit circle $\gamma$. we want to find the sufficient and necessary condition such that $T(\gamma)=\gamma$.

we know that for every point $z$ in the unit circle, $\bar{z}$ is also in it. so $|\bar{z}|=1$ so we have $|z^2|=1$ and $z\bar{z}=1$.

$|T(\gamma)|$ must be also equal to $1$. so for every $z$, $T(z)\bar{T(z)}=1$. after solving and simplify, we have:

$z\bar{z}(a\bar{a}-c\bar{c})+z(a\bar{b}-c\bar{d})+\bar{z}(\bar{a}b-\bar{c}d)+(b\bar{b}-\bar{d}d)=0$

it must be just like $z\bar{z}-1=0$.

so :

$a\bar{a}-c\bar{c}=1$

$a\bar{b}-c\bar{d}=0$

$\bar{a}b-\bar{c}d=0$

$b\bar{b}-\bar{d}d=-1$

this yields the sufficient condition $|a|^2+|b|^2=|c|^2+|d|^2 $ $(*)$

to finding the necessary condition we assume that $c=\lambda\bar{b}$ and this yields $d=\frac{\bar{a}}{\lambda}$. from inserting $d$ and $c$ in $(*)$, $|\lambda|$ must be equal to $1$. and if we take $\lambda=e^{i\alpha}$, $T(z)=e^{i\alpha}\frac{az+b}{\bar{b}z+\bar{c}}$ is the necessary condition.