# De Moivre’s and the nth Root Theorems

## Key Questions

• If the complex number $z$ is

$z = r \left(\cos \theta + i \sin \theta\right)$,

then ${z}^{n}$ can be written as

${z}^{n} = {\left[r \left(\cos \theta + i \sin \theta\right)\right]}^{n} = {r}^{n} {\left[\cos \theta + i \sin \theta\right]}^{n}$

by De Mivre's Theorem,

$= {r}^{n} \left[\cos \left(n \theta\right) + i \sin \left(n \theta\right)\right]$

I hope that this was helpful.

${z}^{\frac{1}{n}} = {r}^{\frac{1}{n}} \left(\cos \left(\frac{\theta}{n}\right) + i \sin \left(\frac{\theta}{n}\right)\right)$

#### Explanation:

Polar form of complex number is $z = r \left(\cos \theta + i \sin \theta\right)$

By De Morvies theorem,

${z}^{\frac{1}{n}} = {r}^{\frac{1}{n}} \left(\cos \left(\frac{\theta}{n}\right) + i \sin \left(\frac{\theta}{n}\right)\right)$

More of the cases, to find expresions for $\sin n x$ or $\cos n x$ as function of $\sin x$ and $\cos x$ and their powers. See below

#### Explanation:

Moivre's theorem says that ${\left(\cos x + i \sin x\right)}^{n} = \cos n x + i \sin n x$

An example ilustrates this. Imagine that we want to find an expresion for ${\cos}^{3} x$. Then

${\left(\cos x + i \sin x\right)}^{3} = \cos 3 x + i \sin 3 x$ by De Moivre's theorem

By other hand applying binomial Newton's theorem, we have

${\left(\cos x + i \sin x\right)}^{3} = {\cos}^{3} x + 3 i {\cos}^{2} x \sin x + 3 {i}^{2} \cos x {\sin}^{2} x + {i}^{3} {\sin}^{3} x = {\cos}^{3} x - 3 \cos x {\sin}^{2} x + \left(3 {\cos}^{2} x \sin x - {\sin}^{3} x\right) i$

Then, equalizing both expresions as conclusion we have

$\cos 3 x = {\cos}^{3} x - 3 \cos x {\sin}^{2} x$
$\sin 3 x = 3 {\cos}^{2} x \sin x - {\sin}^{3} x$