Describe the function $f(z)=z^3$ for $z$ in the semidisk given by $ |z| \leq2, im(z)\geq 0 $.
SOLUTION(given by book):
We know that the points $z$ in the sector of the semidisk from $Arg(z)=0$ to $Arg(z)= \frac {2\pi} {3} $, when cubed, cover the entire disk $|w| \leq 8$. The cubes of the remaining z-points also fall in the disk, overlapping it in the upper half-plane.
This is the image the book has given, and is what I want to graph from scratch.
First of all, I understand that $|z|\leq2, im(z)\geq 0$ is the semi-circle in (a). What I don't understand, is how they are getting $Arg(z)=2\pi/3$ and $0$. I also don't understand what the horizontal, and vertical lines represent, and WHY they are there, and HOW they get there. I also understand how to map individual POINTS from z to w, but how do they map the ENTIRE function over like that?
If I were to not have these graphs as visual aids, how would I do this on my own from scratch?
I tried expanding $z^3$, which gives me $(x^3-3xy^2)+i(3x^2y-y^3)$, using $w=u(x,y)+iv(x,y)$, we can see $u(x,y) = x^3-3xy^2$ and $v(x,y)=3x^2y-y^3$.
From this point, I'm not quite sure where to go. I still have yet to figure out how to get those Arg values, what the lines mean, and how to map the WHOLE function out in both planes from scratch like this.
Best Answer
You want to write $z^3$ in polar coordinates, that is $z = re^{i\theta}$, then $z^3 = (re^{i\theta})^3 = r^3 (e^{i\theta})^3 = r^3 e^{i3\theta}$, so the argument always triples. So if $0 \leq \theta \leq \frac{2}{3}\pi$, then since $\arg z^3 = 3\theta$, then $0 \leq \arg z^3 \leq 2\pi$, so you are already getting all the $w$ with radius at most $2^3 = 8$, so the entire circle (and by circle I really mean the entire disc, that is including the inside), just by considering the horizontally lined region on the left. The reason why they lined the rest of the semicircle horizontally is that that is the part that just by itself covers half a circle on the right. Let's see: on the left we have $\frac{2}{3} \pi \leq \theta \leq \pi$, so if we multiply by 3 we get $2 \pi \leq \arg z^3 \leq 3\pi$, since angles are the same up to multiples of $2 \pi$ this is the same as saying that $0 \leq \arg z^3 \leq \pi$, that is the upper semicircle that they marked by a grid.