I need to show that if |$z_0$|< $R$, the mapping $T(z)=\frac{R(z−z_0)}{R^2-\bar{z_0}z}$ takes open disk of radius $R$ one to one onto the disk of radius 1 and takes $z_0$ to the origin.
I tried to use Maximum module theorem. Since |$z_0$|< $R$ , then $T(z)$ is continuous on the closed bounded set ( ($R^2-z_0z$) can't be zero ) as $z_0$ approaches zero $|T(z)|=|\frac{Rz}{R^2}|=|\frac{z}{R}|=1$
or I'm thinking of plug in $R e^{i\theta}$ for z and see what's happening
But I'm not happy and satisfied of my work, I'm not convinced . Any mistakes please ?
I think that this way is OK. The circle goes to the circle and the interior to the interio, for example because of Maximum module theorem.