Proving $-\frac{1}{2}(z+\frac{1}{z})$ maps upper half disk onto upper half plane

With $\;z=x+iy\;,\;\;x,y\in\Bbb R\;$:

$$-\frac12\left(z+\frac1z\right)=-\frac12\frac{z^2+1}z\cdot\frac{\overline z}{\overline z}=-\frac12\frac{z|z|^2+\overline z}{|z|^2}=$$

$$=-\frac1{2|z|^2}\left(x(|z|^2+1)+y(|z|^2-1)i\right)$$

The imaginary part of the above is

$$\frac{1-|z|^2}{2|z|^2}y>0\iff |z|<1\;,\;\text{whenever}\;\;y>0\;\ldots\;\text{and we're done.}$$

It turns out Ahlfors has solved this question in his Complex Analysis textbook. I repeat the argument here.

Define $\sqrt{z}$ to be the unique complex number $\zeta$ such that $\Re{\zeta} \gt 0$, defined on $\mathbb{C} - [0, -\infty)$

I will show $z(\omega) = -(\omega - \sqrt{\omega^2 - 1})$ lies inside the unit disk for all $\omega \in \mathbb{C} - [-1, 1]$ and hence provides a closed form expression for the inverse function I was looking for.

This would suffice because we see that if $z(\omega)$ lies in the unit disk like I claim, it must be the smaller of the two roots of the equation $z^2 + 2\omega z + 1 = 0$ (since the other root lies outside of disk by reciprocity), and I already showed that the smaller one would be inside the upper half disk when $\omega$ is in the upper half plane.

Consider the domain $\mathbb{C} - [-1, 1]$. Under the map $$z' = \frac{z+1}{z-1}$$ the domain gets mapped conformally to $\mathbb{C} - [0, -\infty)$ . Then we are able to take the square root with the branch defined previously $$z'' \rightarrow \sqrt{z'}$$ we obtain a map in the plane $\{ \Re z'' \gt 0 \}$ Then finally we map this region conformally to the disk by $$z''' = \frac{z''-1}{z''+1}$$.

In total we get a bijection between ${\sqrt{\frac{z+1}{z-1}} - 1}$ and the unit disk defined $$z'''= \frac{\sqrt{\frac{z+1}{z-1}} - 1}{\sqrt{\frac{z+1}{z-1}} + 1}$$ $$z''' = z - \sqrt{z^2 - 1}$$

And then negating a point in the disk leaves it in the disk so we get what we wanted.

Ignoring the $-\frac{1}{2}$ you have there, you basically want to find the inverse of the Joukowski map $\phi (z)= z+\frac{1}{z}$.

Look at this question (and its answer):

Inverse of the Joukowski map $\phi(z) = z + \frac{1}{z}$

You may also be interested in this paper for more information:

http://www.jpier.org/PIERL/pierl19/13.10091305.pdf

[2010. INVERSE JOUKOWSKI MAPPING. C.-H. Liang, X.-W. Wang, and X. Chen. Science and Technology on Antennas and Microwave Laboratory]