Identifying and sketching $\operatorname{Re}(z^3)=1$

Question

Identify and sketch the locus of $\operatorname{Re}(z^3)=1$.

I tried solving the exercise like this \begin{align*} z &= x+iy \\ z^3 &= (x+iy)^3 = x^3-3xy^2+i3x^2 y-iy^3 \\ \operatorname{Re}(z^3) &= x^3-3xy^2=1 \end{align*} But I can't figure out what this equation represents or how to sketch it.

Why is this post receiving downvotes? OP's question is clear, and he has shown his effort. — cansomeonehelpmeout, Sep 03 '18 at 17:00
Well you have the equation in cartesian form. Find out the $x$ & $y$ intercepts. Then check where the graph is tending to at $-\infty$ and $+\infty$. Check for any points of local maximum or minimum and for points of inflection. Then sketch the graph. — Harshit Joshi, Sep 03 '18 at 18:03

score 4 · Accepted Answer · answered Sep 03 '18 at 21:38

Writing $z$ in polar coordinates gives us $\operatorname{Re}(z^3)=r^3\cos(3t)$. There are couple of interesting things to note about equation $$r^3\cos(3t) = 1.$$

First of all, if $\cos(3t) = 0$, $r$ explodes to infinity. Since roots of $\cos$ are $\pi/2 + k\pi,\ k\in\mathbb Z$, we get that $r$ explodes to infinity for angles $\pi/6 + k \pi/3,\ k\in\mathbb Z$. Visually:

The red lines will be asymptotes for our graph. They also happen to be places where $\cos(3t)$ changes sign. This is important since $r$ must be positive. Our graph will, thus, lie in the areas with $+$ signs:

Let us now exploit symmetries of the equation $\operatorname{Re}(z^3)=1$ (they should be visible from the previous picture already).

Rotational symmetry: If $\omega = e^{i\frac{2\pi}3}$, then $\omega^3 = 1$ and multiplication by $\omega$ is rotation by $120^\circ$. If $z_0$ is a solution to the equation, then $$\operatorname{Re}((\omega z_0)^3) = \operatorname{Re}(\omega^3z_0^3) = \operatorname{Re}(z_0^3) =1,$$ so $\omega z_0$ is a solution as well. We conclude that our graph has rotational symmetry. (Alternatively, use $\cos(3(t + 2\pi/3)) = \cos(3t + 2\pi) = \cos(3t)$)
Reflection symmetry: If $z_0$ is a solution, then so is $\overline{z_0}$: $$\operatorname{Re}(\overline{z_0}^3) = \operatorname{Re}(\,\overline{z_0^3}\,) = \operatorname{Re}(z_0^3) = 1.$$ Thus, our graph is symmetric with respect to the real axis.

(Side note: if you know some group theory, these symmetries generate dihedral group $D_3$)

Finally, note that $0\leq\cos(3t)\leq 1$ implies that $r^3\geq 1$. Also, $z_0 = 1$ is an obvious solution.

Using all of this information, we get that our graph looks like this:

score 0 · Answer 2 · answered Sep 03 '18 at 18:51

I'm unsure if I'm understanding your question correctly ,but you can use polar coordinates, so that

$$ z^3 = r^3 e^{i3\theta}$$

implies that $Re(z^3) = r^3\cos(3\theta)$. So, the locus of all points that satisfy of $r^3\cos(3\theta)$ is a "triple pinwheel" construction that has maximum distance from the origin is $(x^2 + y^2)^{\frac{3}{2}}$. You should be able to show this using high school polar plotting skills, or by getting a hint and putting it into mathematica.

Identifying and sketching $\operatorname{Re}(z^3)=1$

2 Answers2