# Mapping a complex function

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.

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.

• So this question is just asking me HOW it maps into the bigger circle, and not asking me to draw them from scratch? – SignalProcessed Sep 2 '16 at 1:57
• I am not sure exactly how it's laid out in your book, but if I would ask a "describe" kind of question, I would be looking to see a paragraph or two describing what the function does, perhaps accompanying a picture. If the picture is already given for you, then I assume you just talk about it, or perhaps explain why it is the right picture. You should think not in terms of "Is this the right answer?" or "Was the author/teacher looking for this?" but think rather "Would this answer this question if I had asked this question myself?" – Jiri Lebl Sep 2 '16 at 20:51

With these type of problems, you basically see if the image of the function provides a surjection into a nice region. In this case, we want to show that $f(z) = z^3$ "hits" every point of the disk centered at the origin with radius $8$ in the image space.

Indeed, this is the case, take

$$w \in D(0, 8) \implies w = r e^{i\theta} = f(z) \quad 0 \leq r < 8$$ $$\implies z = r^{1/3}e^{i\theta/3}, \; r^{1/3} e^{i\theta/3 + 2\pi/3}, \; r^{1/3}e^{i\theta/3 + 4\pi/3}$$

where I have included three potential points of the preimage, because a given complex value has three inverses under the map $w = z^3$. Note that at least one of those three preimages will lie in the upper semicircle.

This shows that for our given domain of the upper semicircle of $D(0,2)$, the function $f(z) = z^3$ is "onto" (another word for having a surjection) $D(0, 8)$ in the image plane. Moreover, we can't get outside of this disk, because

$$w \not \in D(0, 8) \implies w = re^{i\theta} \quad r \geq 8 \implies |w^{1/3}| = r^{1/3} \geq 2$$

which is not in the upper semicircle of radius $2$.

The point of the $Arg(z) = 2\pi / 3$ and $Arg(z) = 0$ lines is that you only need this region of the upper semicircle to cover the entire disk $D(0,8)$. This is because, given any point $w \in D(0,8)$, one of those third roots of unity listed above will satisfy $Arg(z) \in [0, 2\pi/3)$, so in fact, the points in $Arg(z) \in [2\pi/3, \pi)$ contained in $D(0,2)$ are redundant in that they contribute nothing new to the image of the function under the given domain of $D(0,2)$.

• I apologize for reclarification, but I'm pretty new to analysis in general, but what is $D(0,8)$ is that just domain, or some set w belongs to? – SignalProcessed Sep 1 '16 at 23:57
• Sorry $D(0, 8)$ is the (open) disk centered at $z = 0$ and with radius $8$. In general, $D(z_0, r)$ has a center at $z = z_0$ and radius $r$. – J. Marx-Kuo Sep 2 '16 at 4:28