Let's say we have a following function:
$$w=\frac{1}{2}(z+\frac{1}{z})$$ Find the image of a region $Imz>0$, when mapped by given function.
Usually, when i have problem similar to this i simply find $z$ in terms of $w$ which means that i find an inverse, and then simply, because i know what conditions should $z$ satisfy i just express them in terms of $w$ and that way i find image, however, this is a specific case where finding an inverse is not so simple, so i need different approach.
Let's say $z=re^{i\theta}$, then when we put this into given expression we end up with $w=\frac{1}{2}[(r+\frac{1}{r})cos\theta+i(r-\frac{1}{r})sin\theta]$
Now, we have that $Imz>0$ which means that in $z$ space we have a upper half of the plane $\theta \in (0,\pi)$. But i don't see how can i use that here, in the expression i got. Any help appreciated!