0

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 $< 0$

syphracos
  • 486
  • 2
  • 12
  • 1
    Are you aware that the function $z \mapsto z^2$ doubles the angle of every point? – Martin R Jan 10 '23 at 20:45
  • @MartinR yes. We have $z = |z| e^{i arg(z)}$. Then it follows that $z^2 = |z|^2 e^{i2arg(z)}$. But this is then $z^2 = |z|^2 \cdot ( \cos(2arg(z)) + i\sin(2arg(z)) )$. $Re(z^2)$ would then be $|z|^2 \cos(2arg(z))$, but this can be $\lt 0$. What Am I missing? – syphracos Jan 10 '23 at 20:52
  • $z \mapsto z^2$ maps $S_{\pi/4}$ to the right half-plane, but not every sector $S_\alpha$. – Martin R Jan 10 '23 at 20:53
  • 1
    @MartinR okay, now I see it.. would then $z^{\frac{\pi}{2\alpha}}$ map $S_\alpha$ to the right half plane? Because $\arg(z) \lt \alpha$, and, If we would allow $\arg(z) = \alpha$, the maximum that we can have would be $ z^{\frac{\pi}{2\alpha}} = |z|^{\frac{\pi}{2\alpha}} e^{i\frac{\pi arg(z)}{2\alpha}} = |z|^{\frac{\pi}{2\alpha}} ( \cos(\frac{\pi \alpha}{2\alpha})) + i\sin(\frac{\pi \alpha}{2\alpha}) = |z|^{\frac{\pi}{2\alpha}} ( \cos(\frac{\pi }{2}) + i\sin(\frac{\pi}{2}$ – syphracos Jan 10 '23 at 21:09
  • You should say you have posted a similar question yesterday – Jean Marie Jan 10 '23 at 23:12

0 Answers0