I'm asked to show that a function takes the upper half disk to the upper half plane.
$$ f(z) = \left(\frac{1+z}{1-z}\right)^2 $$
I have many ways to go about this, such as, I could define the modulus $|r|<\ 1$ and $0< \theta < \pi$, then say $z=e^{i\theta}$. But im unsure how to translate that onto the upper half plane where $1>\Im(z)$ . I could define $z=x+iy$ where $x,y \in \mathbb{R}$, but it would appear to be hellish to work out the whole algebra.
What is the most optimal way to go about this?
If $z=x+iy$, then $$\frac{1+z}{1-z}=\frac{1-x^2-y^2+2iy}{(1-x)^2+y^2} $$
Since $z$ is in the upper half disk, we know that $y>0$ and $x^2+y^2<1$. Therefore $\Im(\frac{1+z}{1-z})>0$ and $\Re(\frac{1+z}{1-z})>0$, so $\frac{1+z}{1-z}$ is in the first quadrant. If we square it, then the result is in the upper half plane.
So this at least shows that $f$ maps the upper half disk into the upper half plane. If you want to show that maps the upper half disk onto the upper half plane, it might be best to try to find the inverse of $f$. This will involve square roots, but that's fine because you can just choose the principal branch.