I need to find the image $\Gamma_w$ of complex function $w=z^2+\frac{1}{z^2}$, with the domain $$\Gamma_z = \{ z \in \mathbb{C}\ |\ 0 \leq \arg z \leq \pi/2 \wedge |z| \geq 1\}.$$
Using $z=re^{i\phi}$, $r\geq1$, $0\leq\phi\leq\pi/2$, I end up with
$$w = (r^2+\frac{1}{r^2})\cos2\phi + i (r^2-\frac{1}{r^2})\sin2\phi = u + iv.$$ The only thing I can conclude here is that $u$ takes all values from $\mathbb{R}$, and that $v$ takes all values from $[0,+\infty)$. My intuition tells me that $\Gamma_w$ is the upper half plane, but I don't know how to show that. I would prefer the solution to be as rigorous as possible, because I have seen how similar problems are done but couldn't understand some steps.
Best Answer
Hint. Consider the Möbius transformation $M(z)=\frac{z-1}{z+1}$. Then $M^{-1}(z) = -\frac{z+1}{z-1}$ then $$z^2+\frac{1}{z^2}=2M^{-1}((M(z^2))^2).$$
Remark. Note that given $r\geq 1$ for $0\leq 2\phi\leq \pi$, $$w=z^2+\frac{1}{z^2}=\underbrace{\left(r^2+\frac{1}{r^2}\right)\cos(2\phi)}_x+i\underbrace{\left(r^2-\frac{1}{r^2}\right)\sin(2\phi)}_{y\geq 0}$$ draws, in the upper half plane, the semi-ellipse given by $$\frac{x^2}{\left(r^2+\frac{1}{r^2}\right)^2}+\frac{y^2}{\left(r^2-\frac{1}{r^2}\right)^2}=1\quad \text{with $y\geq0$}.$$ Pay attention to the degenerate case $r=1$.