Hello I want to solve this $x^5=32$ using De Moivre's theorem

$$z^n=r(\cos(n\theta) + i \sin(n \theta))$$

I want help to find the solution and more specifically to find the missing $\theta$. $\tan\theta = y/x$ couldn't work.

  • $\begingroup$ The $r$ on the right side of the formula should be $r^n$. Then set $n\theta$ to $2\pi$ [using $n=5$] to start. $\endgroup$
    – coffeemath
    Commented Sep 29, 2013 at 1:19
  • $\begingroup$ can you tell us what you have already tried? $\endgroup$
    – user93089
    Commented Sep 29, 2013 at 1:28
  • $\begingroup$ the imaginary part of this is 0 so θ=2π r=1 correct? $\endgroup$
    Commented Sep 29, 2013 at 1:32

1 Answer 1


I suppose that what you are asking is to find the roots of $z^5=32$.

In general, to solve $z$ such that $z^n=z_0$:

Let $z=re^{i\theta}$ and $z_0=r_0e^{i\theta_0}$

We have $({re^{i\theta}})^n=r_0e^{i\theta_0}\implies r^ne^{in\theta}=r_0e^{i\theta_0}$.

So $r=\sqrt[n]{r_0}$ and $\theta=\frac{\theta}{n}+\frac{2k\pi}{n}$ where $k=1,2,3,..,n-1$.

What you have to do is to convert $32$ to its exponential form, find $r_0$ and $\theta_0$ and solve for $r$ and $\theta$ using the equations above. There should be $5$ roots, all with the same $r$ but with different $\theta$.


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