Find the image of $f(z)=\tan^2(z)$, $D=\{z: 0<\operatorname{Re}(z)<\pi/2\}$. I don't know if what I did was correct. I have no way to confirm it. I got that the result is $w=f(z)$, so $\operatorname{Im}(w)=0$, $-1 \leq \operatorname{Re}(w) \leq 0$.
First I used the fact that $\tan^2(z)=(\sin(z)/\cos(z))^2$ and I opened up the brackets, where $\sin(z)=(e^{iz}-e^{-iz})/2i$, and $\cos(z)=(e^{iz}+e^{-iz})/2$.
I graphed the complex plane, and saw the region that was given, I called a few segments, let's say the segment where $(y=0, x:0 \to \pi/2)$, I called this segment I, then segment II is $(x=\pi/2, y=0 \to \infty)$, and so on, I got a few segments, I wanted to see how $\tan^2(z)$ copies each segment, then kind of puzzle in the mappings and see what the total mapping is.
What I got is that the mapping is literally the interval $[-1,0]$. But I am not sure, and I have no way to confirm..