Let $z^{24} = 1$. Is it true that following complex numbers are solving this equation?

  1. $z_1 = \frac{\sqrt{3} - i}{2} $
  2. $z_2 = \frac{1 + i \sqrt{3}}{2} $
  3. $z_3 = \frac{-1 + i}{\sqrt{2}} $

How we can see $|z^{24}| = 1$. But also $|z_1| = |z_2| = |z_3| = 1$. Of course, I can transform these complex numbers to polar form and then use de Moivre's formula to exponent. It isn't difficult but maybe is fast way to solve it problem?

Thanks in advance!

  • 1
    $\begingroup$ $|z^{24}| = 1$ is easy since $x^{24} = 1$. For the others, if you're considering writing them in polar form $e^{a + bi} = r(\cos(b) + i\sin(b))$ remember that to confirm that the absolute value is $1$, you only need to calculate $(e^a)^2 = r^2$ and see if that equals to $1$. $\endgroup$
    – Arthur
    Commented Aug 22, 2013 at 10:46
  • $\begingroup$ I think that de Moivre is probably the fastest way to solve this. At this point you are expected to know $\sin(\pi/4)$, $\sin(\pi/3)$ as well as the cosines, so that shouldn't take long. You could just observe that $z_3^2=-i$, $z_2^3=-1$, $z_1^3=i$ and go from there, but as I did that by mentally placing them on the unit circle myself, I cannot recommend that with clear conscience :-). BUT, there is the trick of calculating a high power using factorization of the exponent. You will be using that elsewhere (congruences and group theory) in math, so it is also "good to know". $\endgroup$ Commented Aug 22, 2013 at 11:10

2 Answers 2


In general, all the solutions to $z^n=1$, where $n$ is a natural number are given by complex numbers of the form $z=e^{2k\pi i/n}$. Also if $w_1$ and $w_2$ are solutions, then $w_1w_2$ is a solution. Now, $z_1=-iz_2, z_2=e^{\pi i/3}$ and $z_3=ie^{\pi i/4}$ $(i=e^{\pi i/2}, -1=e^{\pi i})$.


A pedestrian way to see $|z^{24}|=1$ is to first realize $\left|\frac{a\ +\ i\ b}{c}\right|^2=\frac{a^2+b^2}{c^2}.$

In your case the numbers $a,b$ and $c$ are either $\pm 1,\pm \sqrt{2},\pm 2$ or $\pm \sqrt{3}$ and always in such a way that $\frac{a^2+b^2}{c^2}=1$. For example

$\left|z_2\right|^2 = \left|\frac{1 + i \sqrt{3}}{2}\right|^2 =\left|\frac{1^2+\sqrt{3}^2}{2^2}\right|^2=\left|\frac{1+3}{4}\right|^2=1.$

Now since in this way you see that the absolute value of $z_1,z_2$ and $z_3$ is $1$, and since taking the power of a complex number with absolute value $1$ can't change this, the $24th$ power of such a number also has this absolute value.

To see $z^{24}=1$, which is a stronger statment, you first want to check that $|z|=1$ (we've done that), but you must then also check that the angle of $z$ viewed as a vector in the complex plane is a multiple of $360°$ divided by $24$, i.e. $15°$ or $\tfrac{\pi}{12}$.


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