Question

How do you use de Moivre's theorem to find roots$?$

Answer 1

Find the root by De Moivre’s theorem:

De Moivre’s theorem states that:

Suppose $z=r\cos \theta +i(r\sin \theta )$ is a complex number and $n$ be a positive integer such that

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

$\begin{array}{rcl}{\mathit{z}}^{\mathbf{n}}\mathbf{}& \mathbf{=}& \mathbf{}{\mathbf{r}}^{\mathbf{n}}\mathbf{}\mathbf{(}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{n}\mathbf{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathbf{i}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{n}\mathbf{\theta }\mathbf{)}\end{array}$

For example;

Let a complex number be ${\left(\cos x+i\sin x\right)}^3$

Here, $n=3$ and $r=1$

Now, Apply De Moivre’s theorem to find the roots

$\begin{array}{rcl}{\left(r\cos x+ir\sin x\right)}^n& =& r^n\left(\cos nx+i\sin nx\right)\end{array}$

$\Rightarrow$$\begin{array}{rcl}{\left(\cos x+i\sin x\right)}^3& =& 1\left(\cos 3x+i\sin 3x\right)\\ & =& \left(\cos 3x+i\sin 3x\right)\end{array}$

Hence, De Moivre’s theorem is used to find the roots of complex numbers.