Find all complex numbers $z$ such that $z^4 \in \mathbb R $

Question

Find all complex numbers $z$ such that $z^4 \in \mathbb R $

Here is my solution:

We can use the exponential form of a complex number to say that $$z^4 = |z|^4 e^{4\theta i} \quad \mbox{We know that |z| is a real number}$$ $$z^4 \in \mathbb R \iff 4\theta = 2k\pi \quad k \in \mathbb N \\ \theta = \frac {k\pi}{2}$$ And so this means that if the fourth power of a number should be real, then the angle it forms with the horizontal axis must be a multiple of $\pi/2$, and so this will ultimately cover the vertical and the horizontal axis of the complex plane. Therefore, my answer is: $$z^4 \in \mathbb R \iff \Re(z) = 0 \lor \Im(z) = 0$$ But I am not sure if I am not missing something. Besides, I feel this is not the most precise way to solve this problem - I guess that using the standard notation $z = a+bi$ would be more fruitful.
Am I missing something?

Answer 1

You're almost correct. However, your condition $4\theta = 2k\pi$ really should be $4\theta = k\pi$, because we also want to include negative real numbers. This means that the allowed values of $\theta$ are those of the form $k\pi/4$. Therefore, you may write $$ z^4 \in \mathbb R \iff \Re(z) = 0 \lor \Im(z) = 0 \lor \Re(z) = \Im(z) \lor \Re(z) = -\Im(z).$$

Answer 2

I think I would do it the way you did.

Just for fun....

$z = a + bi\\ z^4 = (a^4 - 6a^2b^2 + b^4) + (4a^3b - 4ab^3)i$

$z^4$ is real $\implies 4a^3b - 4ab^3 = 0$

$4ab(a+b)(a-b) = 0$

$a = 0,$ or $b = 0,$ or $a = b,$ or $a = -b.$

Answer 3

There's an error in your initial equivalence: $$r\,\mathrm e^{i\theta}\in\mathbf R\iff 4\theta\equiv 0\mod\color{red}\pi,$$ not $2\pi$, so we have $$\theta\equiv 0\mod\frac\pi4.$$ Also the $k$ ypou mention is in $\mathbf Z$, not $\mathbf N$.

Answer 4

Assume that $z=x+Iy$ then $$ (x+Iy)^4=x^4+y^4-6x^2y^2+4Ixy(x^2-y^2) $$ Now $z^4$ is a real number if and only if $x=y$ or $x=-y$.