I want to show that the mapping given by $$w = f(z) = - \frac{1}{2} \left( z + \frac{1}{z} \right)$$ is a bijective mapping from the upper half disc to the upper half plane.
One-to-one case is straightforward. For onto, for $w \in \mathbb{C}$ such that $\text{Im}(w) > 0$, how to show that the value $$z = \sqrt{w^2 - 1} - w $$ is in the upper-half of the unit disk, i.e., $|z| < 1$ with $\text{Im} (z) > 0$.
I tried substituting $w = u + iv $ with $v > 0$ but wasn't successful. Any help is much appreciated.
The equation $\,z^2 + 2 w\,z + 1 = 0\,$ is a quadratic in $z$, which has two complex roots with product equaling $\,1\,$ by Vieta's relations, so one of them must be inside the unit circle. The question then reduces to showing that $\,z\,$ is in the upper-half of the unit disk when $w$ is in the upper half plane.
Using that $\displaystyle\,\frac{1}{z}=\frac{\bar z}{|z|^2}\,$ and $\,a - \bar a = 2i\,\text{Im}(a)\,$:
$$ -2 \cdot 2i\, \text{Im}(w)=z + \frac{\bar z}{|z|^2}-\left(\bar z + \frac{z}{|z|^2}\right)=\left(z-\bar z\right)\left(1 - \frac{1}{|z|^2}\right) = 2i\,\text{Im}(z)\,\left(1-\frac{1}{|z|^2}\right) \\ \implies\;\;\;\; \underbrace{2\, \text{Im}(w)}_{\text{w in the upper} \\ \text{half-plane}} = -\,\underbrace{\underbrace{\text{Im}(z)}_{\gt 0} \bigg/ \underbrace{\left(1-\frac{1}{|z|^2}\right)}_{\lt\,0}}_{\text{z in the upper-half} \\ \text{ of the unit disk}} \;\gt\; 0 $$