$z^{3}$ is the conjugate of $z\in\mathbb{C}$

Question

We have this problem in math class at computer science department and we students can hardly agree on that. The problem is the following.

We were asked to solve: $$z^3 = \overline{z}$$ where $z=a+b\text{i}$ ($a,b\in\mathbb{R}$), i.e. $$(a+bi)^3 = a - bi.$$

Now some students used the De Moivre formula and got $3$ solutions while others used algebra and found $5$ solutions: $\{1, -1, i, -i, 0\}$

More details:

Well, some students used the De Moivre:

$$r^{3}\text{cis}(3\theta) = r\text{cis}(-\theta)$$

and then split the equation to the forms of: $$r^3 = r$$ and so $r$ can be $1$ or - because $-1$ falls ($r\geq 0$), and

$$3\theta = \theta + k \cdot 2\pi$$

if I remember right, and then for $k = 0, 1, 2$, they got some answers. But again, every student got different answers and we are all confused. Is it even possible to get $5$ answers to "$z^3$-form" equations?

Answer 1

A quick solution: $$z^3 = \bar{z} \implies |z|^3 = |z| \implies |z| =0 \text{ or }1$$

If $z \neq 0$, since $|z| = 1$, put $z = e^{i \theta}$. With this: $$z^3 = \bar{z} \implies e^{3i\theta} = e^{-i\theta} \implies 3i\theta = -i\theta + 2k\pi i \implies \theta = \frac{k\pi}{2},$$with $k \in \Bbb Z$. So the solution set is $\{ 0,1,i,-1,-i \}$.

Answer 2

Let $z=re^{i\theta}$. Then $r^3e^{3i\theta} = re^{-i\theta}$. If $r\neq 0$, we get $r^2e^{3i\theta} = e^{-i\theta}$, implying that $r=1$ and $4i\theta = n\cdot 2i\pi$ for some $n\in\mathbb{Z}$.

This shows that for $r=1$ all multiples of $\frac{\pi}{2}$ work for $\theta$, i.e. $\theta\in \{0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$, corresponding to the values of $z\in\{1,i,-1,-i\}$.

Also, since $r=0$ works, $z=0$ is also a solution.

Answer 3

Of course $0$ is a solution, so let's assume $z\ne0$. Taking modules, we have $$ |z|^3=|z| $$ so $|z|^2=1$ that entails $|z|=1$. Therefore $\bar{z}=z^{-1}$ and the equation becomes $z^4=1$.

If you are with the “three solutions” side, you're wrong.

Of course $0$ is a solution, so let's assume $z\ne0$. Taking modules, we have $$ |z|^3=|z| $$ so $|z|^2=1$ that entails $|z|=1$. Therefore $\bar{z}=z^{-1}$ and the equation becomes $z^4=1$.

If you are with the “three solutions” side, you're wrong.

Using the trigonometric form, $z=r\operatorname{cis}\theta$, the equation becomes $$ r^3\operatorname{cis}3\theta=r\operatorname{cis}(-\theta) $$ and so $r^3=r$, so $r=1$ (again assuming $z\ne0$), which gives $$ 3\theta=-\theta+2k\pi $$ that means $$ \theta=k\frac{\pi}{2} $$ Note that $k$ can assume four distinct values and still determine different solutions.

Answer 4

Here is a not so quick solution, but quite detailed. We can rewrite $z=\rho e^{\mathrm{i}\theta}$. We have

$$\begin{align} z^{3}&=\overline{z}\\ \rho^{3}e^{3\mathrm{i}\theta} &=\rho e^{-i\theta}\\ \rho\left(\rho^{2}e^{3\mathrm{i}\theta}-e^{-i\theta}\right) &=0 \end{align}$$

Hence a first solution is $\rho=z_{1}=0$. The other possible solutions will be given by:

$$\begin{align} \rho^{2}e^{3\mathrm{i}\theta}-e^{-\mathrm{i}\theta} &=0\\ \rho^{2}e^{4\mathrm{i}\theta}-1 &=0\\ \rho^{2}e^{4\mathrm{i}\theta} &= 1\\ \rho^{2}e^{4\mathrm{i}\theta} &= e^{\mathrm{i}\cdot (0+2k\pi)} \tag{$k\in\mathbb{Z}$} \end{align}$$

So that $\rho=1$ because two complex numbers can't be equal if they do not have the same modulus. It becomes (assuming $k\in\mathbb{Z}$):

$$\begin{align} e^{4\mathrm{i}\theta} &= e^{\mathrm{i}\cdot (0+2k\pi)}\\ 4\theta&=2k\pi\\ \theta &=k\frac{\pi}{2} \end{align}$$

And it gives you $4$ other solutions since $\theta = 0$, $\theta = \pi/2$, $\theta =\pi$, $\theta=3\pi/2$ are the only possible values for $\theta$ leading to distinct solutions (the other possibles values are these $\theta+2k\pi$).

These $4$ solutions are $$\{-1,1,-\mathrm{i},\mathrm{i}\}$$ And all the solutions are: $$\{-1,0,1,-\mathrm{i},\mathrm{i}\}$$