I have a question about a conformal mapping. The map $f(z)=\frac{1+z}{1-z}$ takes the unit disk to the right half plane. Composing this map with $z^2$ gives $f(z)=(\frac{1+z}{1-z})^2$, which I think gives me the whole complex plane, except for the negative real axis. But how can I easily see (without trial and error computations) that the negative real axis will be excluded.
Thankful for any help with this!
If you include the boundary of the disk in your domain, then the boundary of the right half-plane will be in the image of $f$, and the negative real axis will be in the image of $z\mapsto f(z)^2$.
Your first function takes the upper half of the boundary to the upper half of the imaginary axis. If you include the upper half of the circle, including $-1$ but excluding $1$, that maps to the upper half of the imaginary axis including $0$ but excluding $\infty$, and then squaring takes that to the negative real axis including $0$ but excluding $\infty$.