My question is that to show the linear fractional transformation $ f(z)=\frac{2z-1}{2-z} $ maps the cicrle $ C:|z|=1$ into itself. Also to prove that $f(z)$ is conformal in $D=\{z:|z|\leq1\}$.
First part is easy to prove.As we can see that if we take $w=f(z)$ then $ w=\frac{2z-1}{2-z}\Rightarrow 2w-wz=2z-1\Rightarrow 2z+wz=1+2w\Rightarrow z(2+w)=1+2w\Rightarrow z=\frac{1+2w}{2+w}$
Now since $|z|=1$, then $|\frac{1+2w}{2+w}|=1$
Let, $w=u+iv$, then \begin{align*} |1+2u+i2v|=|2+u+iv|&\Rightarrow (1+2u)^2+4v^2=(2+u)^2+v^2\\ &\Rightarrow 1+4u+4u^2+4v^2=4+4u+u^2+v^2\\ &\Rightarrow3u^2+3v^2=3\\ &\Rightarrow u^2+v^2=1\\ &\Rightarrow |w|=1 \end{align*}
Hence the first part. Now how to show the second part? any help will be really appreciated. thank you very much.