Define a map on the unit disc $\mathbb{D}=\{z\in \mathbb{C} \mid |z| \text{ <1 }\}$
$$w(z)=f(z)=\frac{1}{1+z^2}$$ Question Prove that $w(z)=\frac{1}{1+z^2}$ maps the unit disc $D$ onto the plane $\Re(w)>1/2.$
Attempt $$w=\frac{1}{1+z^2}$$ then, $$z^2=\frac{1-w}{w}$$ $|z|<1$ is mapped to $\Re(w)>1/2$
Define, $S=\{w\in \mathbb{C} \mid \Re(w) \text{ >1/2 }\}$
$f:\mathbb{D}\to S$
Let, $w_0\in S$ then $\Re(w_0)>1/2$ then there exists
$z_0=\sqrt{\frac{1-w_0}{w_0}}\in \mathbb{D}$ such that $$f(z_0)=w_0$$
It is plain from the first equation $w \neq 0$. You already have obtained, $$ z^2 = \frac{1-w}{w} $$ and $|z^2| < 1$ if and only if $|z|<1$, thus $|z|<1$ if and only if, $$ \left\lvert \frac{1-w}{w} \right\rvert < 1. $$ That is true if and only if we also have, $$ \left\lvert \frac{1-\overline w}{\overline w} \right\rvert < 1. $$ Taking the product, the last equation is true if and only if, $$ \left\lvert \frac{1-w}{w} \right\rvert^2 = \frac{1-w}{w} \cdot \frac{1-\overline w}{\overline w} < 1, $$ which holds if and only if, $$ 1 - 2 \Re w +|w|^2 < |w|^2 $$ so that $\Re w > \tfrac{1}{2}$. This shows $w \in S$ as required. But each step step in the above argument is reversible so that if $w \in S$ then $|z| < 1$.