I'm working on a problem and part of it is to map the sector $\{z \in \mathbb{C}| \frac{\pi}{4} < \text{arg} z < \frac{3\pi}{4}\}$ to $\{z \in \mathbb{C}| \frac{-\pi}{2} < \text{arg} z < \frac{\pi}{2}\}$. I've tried the map $z \mapsto z^2$, which maps the original sector to $\{z \in \mathbb{C}| \frac{\pi}{2} < \text{arg} z < \frac{3\pi}{2}\}$, so now I only have to rotate it. I know that rotations preserve angles and are therefore conformal mappings, but I can't seem to get an explicit conformal mapping to rotate it. So, what am I missing here?
Thanks in advance!
EDIT: I know that $z \mapsto -z^2$ does the trick and I see it on my sketch, but I don't see it explicitly in the formula. More specifically: if $z = Re^{iy}$ with $ \frac{\pi}{2} < y < \frac{3\pi}{2}$, then $z \mapsto -z$ gives $-Re^{iy}$, with $\frac{\pi}{2} < y < \frac{3\pi}{2}$. I feel that it's terribly obvious, but I fail to see why this is equivalent with $\{z \in \mathbb{C}| \frac{-\pi}{2} < \text{arg} z < \frac{\pi}{2}\}$.
I figured out what went wrong. As mentioned in the question and in the answers, $z \mapsto -z$ does the trick. However, if $z = R e^{i \theta}$, then $z \mapsto - R e^{i\theta}$, which doesn't immediately show why this gives a rotation of $\pi$ degrees in counterclockwise direction. However, $R \geq0$, so we have to get rid of the minus sign (the radius of a complex number can't be negative). Use the relation $-1 = e^{\pi i}$. Then we have that $z \mapsto Re^{(\pi + i)\theta}$, which gives a rotation of $\pi$ degrees in counterclockwise direction.