Find all solutions for $z^3=\overline{z}$

Question

Im having a problem. How do you solve

$$z^3=\overline{z}$$ $$z=a+bi$$ and $$\overline{z}=a-b$$ Thanks in advance

Answer 1

Notice that $|z^3| = |z|^3$, but $|\overline{z}| = |z|$. So you know $|z|^3 = |z|$, and therefore $|z| = 0$ or $|z| = 1$.

You also know that $z\overline{z}$ is real, so $z^4$ is real (in fact, $z^4 = |z|^2$). Together with the fact that $|z| = 0$ or $|z| = 1$, we have that $z$ is either $0$ or a $4$th root of unity.

That means $z$ can be $0$, $1$, $-1$, $i$, or $-i$.

Answer 2

$(a + bi)^3 = a - bi$

$a^3 +3a^2bi -3ab^2 -b^3i = a-bi$

Equating real:

$a^3 -3ab = a$

$a(a^2 - 3b^2 -1) = 0$

Equating Imaginary:

$3a^2b - b^3 = -b$

$b(b^2 - 3a^2 - 1) = 0$

Then we get $a = 0$ or $a = \pm \sqrt{1 + 3b^2}$

Similarily, $b = 0$ or $b = \pm \sqrt{1 + 3a^2}$

Now we plug in the solutions to get:

$a = 0, b = 0, \pm 1$

$b = 0, a = \pm 1$

That is $z= \pm 1$ or $z=\pm i$ or $z=0$

Answer 3

One way of doing this is to use the polar form of $z$, and say $z = r e^{i \theta}$. Take the modulus of the original equation to get

$$|z|^3 = |z|$$ so $|z|=r$ must be either $0$ or $1$. In the $r=1$ case, substitute in again $$e^{3i\theta}=e^{-\theta}$$ so $e^{4i\theta}=1$, and $\theta = 0, \frac{\pi}2, \pi, -\frac{\pi}2$ so $z = 0, \pm 1, \pm i$.