I have a sector $S_{\alpha} = \{ z \in \mathbb C \ | \ |arg(z)| \lt \alpha \} $ with $0 \lt \alpha \leq \pi$.
My question is: Does the function $\phi: z \to z^2$ map the sector $S_\alpha$ to the right half plane? And if it does, how exactly? I can't find a solution because if I square $z$ I get $z^2 = z \cdot z = (x+iy)(x+iy) = x^2 - y^2 + i2xy$, where $Re(z) = x^2 - y^2$. And this can be $< 0$