To prove $-\frac{1}{2}(z+\frac{1}{z})$ maps upper half disk onto upper half plane, I have been trying to find a formula for the inverse. To this end, I chose $\omega$ in the upper half plane.
$$\omega = -\frac{1}{2}(z+\frac{1}{z})$$
$$0 = z^2+2 \omega z +1$$
At this point I would prefer to use the quadratic formula because I want a formula for the inverse, but I was unable to prove that one of the two solutions given lie inside the unit disk. I have come up with an unsatisfactory existence proof below.
There are two solutions for $z$, call them $z_1$ and $z_2$. By the form of the polynomial $z_1 z_2 = 1$ and $z_1 + z_2 = -2 \omega$. wlog let $|z_1| \lt 1$ and $|z_2| \gt 1$. Since $z_1$ and $z_2$ lie on opposite sides of the real line (because they are inverses), and their sum is in the lower half plane, the smaller one $z_1$ must lie in the upper half plane and hence the upper half disk. So there exists an inverse, but I don't have a closed form solution.
(This is exercise 8.5 in Stein and Shakarchi Complex Analysis )
Best Answer
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
$$\frac{1-|z|^2}{2|z|^2}y>0\iff |z|<1\;,\;\text{whenever}\;\;y>0\;\ldots\;\text{and we're done.}$$