0

I'm looking to show that the transformation $z=Z^3$ maps the region $0<arg(z)<pi/3$ to the half plane $Im(z)>0$

I'm trying to define the boundaries in the form $Z=s$, $Z=se^{i(pi/n)}$ where, in my case $n$ is $3$ and $0<s<infinity$

I'm struggling to figure out what $s$ is.

Thanks.

juper
  • 267
  • 1
  • 10

1 Answers1

1

Hint: I think that a graphical representation is very useful. Have a look at the picture below showing the example of $z=1.1 e^{i\pi/4}$ (which belongs to the first sector ($0 \leq arg(z) \leq \pi/3$)), mapped onto $z^3=1.1^3e^{i 3 \pi/4}$.

This figure displays the (half) unit circle for the sake of representing where are situated points $e^{k i \pi/3},$ but consider that these points, when multiplied by any $s>0$ become separating rays between sectors, and particular that any point $z$ belonging to the first sector has an image belonging to one of three sectors, i.e., the superior half plane.

enter image description here

Jean Marie
  • 81,803