# Using De Moivre's theorem

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.

• 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. Commented Sep 29, 2013 at 1:19
• can you tell us what you have already tried? Commented Sep 29, 2013 at 1:28
• the imaginary part of this is 0 so θ=2π r=1 correct? Commented Sep 29, 2013 at 1:32

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 $\theta$.