Abraham De Moivre, the French mathematician came up with the De Moivre’s formula while he was working on normal distribution and probability theory. In this formula complex number and trigonometry is linked together. In this article, we will discuss De Moivre’s theorem in detail.

## Definition

De Moivre’s theorem is used to find out the roots of a complex number raised to power $n$, where n is an integer. De Moivre’s theorem does not stand for non-integer values of $n$. The De Moivre’s formula is as follows:

$Z^{n}$ = $(cos\ x\ +\ i\ sin\ x)^{n}$ = $cos(nx)\ +\ i\ \times\ sin(nx)$

Here, $z$ is a complex number, $x$ is the real number and $i$ stands for the imaginary part. It is also known as $“cis”$. $Cos\ x\ +\ i\ Sin\ x$ is also called $CiS\ (x)$ and $Z^{n}$ = $(cos\ x\ +\ i\ sin\ x)^{n}$ = $cos(nx)\ +\ i\ \times\ sin(nx)$ = $CiS\ (nx)$

## Proof

### Let us prove the De Moivre’s theorem by the following two methods:

1) By Mathematical Induction:

For $n$ = $1$, we get $(Cos\ (\Theta)\ +\ I\ Sin\ (\Theta))^{1}$ = $Cos(1\Theta)\ +\ i\ Sin(1\Theta)$

So, we see that De Moivre’s theorem holds good for $n$ = $1$. Let us now assume that it also holds true for $n$ = $k$, which is

$(Cos\ (\Theta)\ +\ i\ Sin\ (\Theta))^{k}$ = $Cos\ (k\Theta)\ +\ i\ Sin\ (k\Theta)$ ……..  Equation 1

Next, we need to show that it also holds true for $n$ = $k\ +\ 1$

$(Cos\ (\Theta)\ +\ i\ Sin\ (\Theta))^{(k\ +\ 1)}$ = $(Cos\ (\Theta)\ +\ i\ Sin\ (\Theta))\ \times\ (Cos\ (\Theta)\ +\ i\ Sin\ (\Theta))^{k}$

Plugging in equation 1 in the above function, we get

$(Cos\ (\Theta)\ +\ i\ Sin\ (\Theta))^{(k\ +\ 1)}$ = $(Cos\ (\Theta)\ +\ i\ Sin\ (\Theta))\ \times\ (Cos\ (k\ \Theta)\ +\ i\ Sin\ (k\ \Theta))$

$(Cos\ (\Theta)\ +\ i\ Sin\ (\Theta))^{(k\ +\ 1)}$ = $Cos\ (\Theta)\ Cos\ (k\ \Theta)\ +\ i\ Cos\ (\Theta)Sin\ (k\ \Theta)\ +\ i\ Sin\ (\Theta)\ Cos\ (k\ \Theta)\ +\ i^{2}\ Sin\ (\Theta)\ Sin\ (k\ \Theta)$

$(Cos\ (\Theta)\ +\ i\ Sin\ (\Theta))^{(k\ +\ 1)}$ = $Cos\ (\Theta)\ Cos\ (k\ \Theta)\ -\ Sin\ (\Theta)\ Sin\ (k\ \Theta)\ +\ i\ (Cos\ (\Theta)\ Sin\ (k\ \Theta)\ +\ Sin\ (\Theta)\ Cos\ (k\ \Theta))$

Applying trigonometry addition formulas Sin $(A\ +\ B)$ = $SinA\ CosB\ +\ CosA\ SinB$ and $Cos\ (A\ +\ B)$ = $CosA\ CosB\ –\ SinA\ SinB$, we can simplify the above equation as

$(Cos\ (\Theta)\ +\ i\ Sin\ (\Theta))^{(k + 1)}$ = $Cos(\Theta\ +\ k\ \Theta)\ +\ i\ (Sin\ (\Theta\ +\ k\ \Theta)$

Thus, applying induction we have proved that De Moivre’s holds true for all positive integers $n$
2) From Euler’s Formula:

As per Euler’s formula, we know

$Cos\ (\Theta)\ +\ i\ Sin\ (\Theta)$ = $e^{(i\ \Theta)}$

Raising both sides to the power of n, we get

$(Cos\ (\Theta)\ +\ i\ Sin\ (\Theta))^{n}$ = $(e^{(i\ \Theta)})^{n}$

Using De Moivre’s theorem,

$(e^{(i\ \Theta)})^{n}$ = $Cos\ (n\ \Theta)\ +\ i\ Sin\ (n\ \Theta)$

Thus, we deduce the De Moivre’s theorem

$(Cos\ (\Theta)\ +\ i\ Sin\ (\Theta))^{n}$ = $Cos\ (n\ \Theta)\ +\ i\ Sin\ (n\ \Theta)$

Hence, De Moivre is true for any value of n. The values of n which satisfies De Moivre’s theorem could be any number, there is no such restriction that it has to be only positive.

## Theorem

De Moivre’s theorem states that for any real number $\Theta$ and positive integer $n$,

$(cos\ x\ +\ i\ sin\ x)^{n}$ = $cos(nx)\ +\ i\ \times\ sin(nx)$

The usefulness and application of this theorem is that it helps in solving complex numbers expressed in the polar form and raised to large powers. There is no need to carry out multiplication.

## De Moivre’s Formula for Multiple Angles

Using De Moivre’s theorem to deduce double angle

De Moivre’s formula $(cos\ x\ +\ i\ sin\ x)^{n}$ = $cos(nx)\ +\ i\ \times\ sin(nx)$

When $n$ = $2$

$(cos\ x\ +\ i\ sin\ x)^{2}$ = $cos(2x)\ +\ i\ \times\ sin(2x)$

$Cos^{2}\ x\ +\ i^{2}\ sin^{2}\ x\ +\ i\ 2sin\ x\ cos\ x$ = $cos(2x)\ +\ i\ \times\ sin(2x)$

Comparing the real part, we get

$Cos^{2}\ x\ +\ i^{2}\ sin^{2}\ x$ = $cos(2x)$

Therefore, $Cos(2x)$ = $cos^{2}\ x\ –\ sin^{2}\ x$ [Double angle formula of cos is derived.]

Comparing the imaginary part, we get

$i\ 2sin\ x\ cos\ x$ = $+\ i\ \times\ sin(2x)$

Therefore, $sin\ (2x)$ = $2sin\ x\ cos\ x$ [Double angle formula of sin is derived]

$Tan\ (2x)$ = $sin$ $\frac{(2x)}{cos}$ $(2x)$ = $2sin\ x\ cos$ $\frac{x}{cos^{2}}$ $x\ –\ sin^{2}\ x$

Dividing both numerator and denominator by $cos^{2}\ x$, we get

$Tan\ (2x)$ = $2tan$ $\frac{x}{1}$ – $tan^{2}\ x$ [Double angle formula of tan is derived]
Using De Moivre’s theorem to deduce triple angle

When $n$ = $3$

$(cos\ x\ +\ i\ sin\ x)^{3}$ = $cos(3x)\ +\ i\ \times\ sin(3x)$

$Cos^{3}\ x\ +\ 3\ cos^{2}\ (x)\ (i\ sin\ x)\ +\ 3\ cos\ x\ (i\ sin\ x)^{2}\ +\ (i\ sin\ x)^{3}$ = $cos(3x)\ +\ i\ sin(3x)$

$Cos^{3}\ x\ +\ i\ 3cos^{2}\ x\ sin\ x\ –\ 3cos\ x\ sin^{2}\ x\ –\ i\ sin^{3}\ x$ = $cos(3x)\ +\ i\ sin(3x)$

Comparing the real part, we get

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

$Cos(3x)$ = $cos^{3}(x)\ –\ 3cos\ x(1\ –\ cos^{2}(x))$

$Cos(3\ x)$ = $cos^{3}(x)\ –\ 3cos\ x\ +\ 3cos^{3}\ x$

$Cos(3x)$ = $4cos^{3}(x)\ -\ 3cos(x)$ [Triple angle formula for cos has been deduced]

Comparing the imaginary part and cancelling the i on both sides, we get

$Sin(3x)$ = $3cos^{2}(x)\ sin\ x\ –\ sin^{3}(x)$

$Sin(3x)$ = $3(1\ -\ sin^{2}(x))\ sin\ x\ –\ sin^{3}(x)$

$Sin(3x)$ = $3sin(x)\ -\ 3sin^{3}(x)\ –\ sin^{3}(x)$

$Sin(3x)$ = $3sin(x)\ –\ 4sin^{3}(x)$ [Triple angle formula for sin has been deduced]

$Tan(3x)$ = $\frac{sin(3x)}{cos(3x)}$

$Tan(3x)$ = $3\ cos^{2}(x)\ sin(x)$ – $\frac{sin^{3}(x)}{cos^{3}(x)}$ – $3cos(x)\ sin^{2}(x)$

Dividing both numerator and denominator by $cos^{3}(x)$, we get

$Tan(3x)$ = $3tan(x)$ – $\frac{tan^{3}(x)}{1}$ - $3tan^{2}(x)$ [Triple angle formula for tan has been deduced]

Similarly, we can formulate other higher multiple angle formulas of trigonometry using De Moivre’s theorem.

## Example

Example:

Given $z$ = $1\ –\ i$, find out $z^{8}$

Solution:

$|z|$ = $root(1^{2}\ +\ (-1)^{2})^{2}$

= $root(2)$

$Arg|z|$ = $\frac{-pi}{4}$

Using De Moivre’s;

$Z$ = $root2\ (cos$ ($\frac{-pi}{4}$) + $i\ sin$($\frac{-pi}{4}$))

$Z^{8}$ = $(root2)^{8}\ (cos$($\frac{-pi}{4}$) $i\ sin$($\frac{-pi}{4}$)$)^{8}$

$Z^{8}$ = $2^{4}\ (cos(8\ \times$ $\frac{-pi}{4}$) + $i\ sin(8\ \times$ $\frac{-pi}{4}$))

$Z^{8}$ = $16\ (cos(-2pi)\ +\ i\ sin(-2pi))$

$Z^{8}$ = $16\ cis(-2pi)$