DeMoivre's Theorem - Definition, Proof, Uses and Examples

This theorem was developed by French mathematician Abraham de Moivre. It helps to connect complex numbers with trigonometry. It also helps in finding the powers and the roots of a complex number when they are expressed in Euler form. The main applications of De-moivre's theorem are to find powers of complex numbers, and roots of complex numbers, derive various trigonometric identities, and help in signal processing.

In this article, we will cover the concept of the De-moivre's theorem. This concept falls under the broader category of complex numbers and quadratic equations, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more.

De-moivre's Theorem

One of the fundamental theorems of complex numbers, De Moivre's theorem is utilised to address a variety of complex number problems. The solution of trigonometric functions with numerous angles is another common use of this theorem. "De Moivre's Identity" and "De Moivre's Formula" are other names for DeMoivre's Theorem. The name of this theorem originates from the name of its creator, the renowned mathematician De Moivre.

De Moivre’s Formula for complex numbers is, for any real value of x,

$(\cos x+i \cdot \sin x)^n=\cos (n x)+i \cdot \sin (n x)$

De Moivre’s Theorem Proof

De-moivre’s theorem is based on the Euler form representation of complex numbers.

According to De-moivre’s theorem

1. If $\mathrm{n} \in \mathbb{I}$ (integers), $\theta \in \mathbb{R}$ and $i=\sqrt{-1}$, then
$
(\cos \theta+i \sin \theta)^{\mathbf{n}}=\cos \mathbf{n} \theta+i \sin \mathbf{n} \theta
$

Proof:
We know that, $\mathrm{e}^{i \theta}=\cos \theta+i \sin \theta$
now,
$
\begin{aligned}
& \Rightarrow\left(\mathrm{e}^{i \theta}\right)^{\mathrm{n}}=(\cos \theta+i \sin \theta)^{\mathrm{n}} \\
& \Rightarrow \mathrm{e}^{i(\mathrm{n} \theta)}=(\cos \theta+i \sin \theta)^{\mathrm{n}} \\
& \Rightarrow \cos \mathrm{n} \theta+i \sin \mathrm{n} \theta=(\cos \theta+i \sin \theta)^n
\end{aligned}
$

2. If $\theta_1, \theta_2, \theta_3 \ldots \ldots ., \theta_{\mathrm{n}} \in \mathbb{R}$, then
$
\begin{gathered}
\left(\cos \theta_1+i \sin \theta_1\right)\left(\cos \theta_2+i \sin \theta_2\right)\left(\cos \theta_3+i \sin \theta_3\right) \ldots \ldots \ldots \ldots \ldots\left(\cos \theta_{\mathrm{n}}+i \sin \theta_{\mathrm{n}}\right) \\
=\cos \left(\theta_1+\theta_2+\theta_3+\ldots \ldots .+\theta_{\mathrm{n}}\right)+i \sin \left(\theta_1+\theta_2+\theta_3+\ldots \ldots+\theta_{\mathrm{n}}\right)
\end{gathered}
$

Proof:

$\theta_1, \theta_2, \theta_3 \ldots \ldots \ldots . \theta_{\mathrm{n}} \in \mathbb{R}$ and $\mathrm{i}=\sqrt{-1}$, then
by Euler's formula $\mathrm{e}^{\mathrm{i} \theta}=\cos \theta+\mathrm{i} \sin \theta$

$
\begin{aligned}
& \left(\cos \theta_1+\mathrm{i} \sin \theta_1\right)\left(\cos \theta_2+\mathrm{i} \sin \theta_2\right)\left(\cos \theta_3+\mathrm{i} \sin \theta_3\right) \ldots \ldots \\
& \ldots \ldots \ldots\left(\cos \theta_{\mathrm{n}}+\mathrm{i} \sin \theta_{\mathrm{n}}\right)=\mathrm{e}^{\mathrm{i} \theta_1} \cdot \mathrm{e}^{\mathrm{i} \theta_2} \cdot \mathrm{e}^{\mathrm{i} \theta_3} \cdot \ldots \ldots \ldots . \cdot \mathrm{e}^{\mathrm{i} \theta_{\mathrm{n}}} \\
& \left.=\mathrm{e}^{\mathrm{i}\left(\theta_1+\theta_2+\theta_3+\ldots\right.}+\ldots+\theta_{\mathrm{n}}\right) \\
& =\cos \left(\theta_1+\theta_2+\theta_3+\ldots \ldots+\theta_{\mathrm{n}}\right)+\mathrm{i} \sin \left(\theta_1+\theta_2+\theta_3+\ldots \ldots . .+\theta_{\mathrm{n}}\right)
\end{aligned}
$

Note: If $n$ is a rational number that is not an integer and $n=p / q$, where $\operatorname{HCF}(p, q)=1$ and $q>0$, then $z^n$ will have $q$ values. One of the values will be $\cos n+i \sin n$.

1. $(\cos \theta-\mathrm{i} \sin \theta)^{\mathrm{n}}=\cos \mathrm{n} \theta-\mathrm{i} \sin \mathrm{n} \theta$

2. $(\sin \theta+\mathrm{i} \cos \theta)^{\mathrm{n}}=(\mathrm{i})^{\mathrm{n}}(\cos \mathrm{n} \theta-\mathrm{i} \sin \mathrm{n} \theta)$

(by taking i common first)

3. $(\sin \theta-\mathrm{i} \cos \theta)^{\mathrm{n}}=(-\mathrm{i})^{\mathrm{n}}(\cos \mathrm{n} \theta+\mathrm{i} \sin \mathrm{n} \theta)$

(by taking i common first)

4. $(\cos \theta+i \sin \phi)^n \neq \cos \mathrm{n} \theta+\mathrm{i} \sin \mathrm{n} \phi$

Solved Examples Based On the De-Moivre's Theorem

Example 1: Let $a=\cos \alpha+i \sin \alpha, b=\cos \beta+i \sin \beta, c=\cos \gamma+i \sin \gamma$ and $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=1$ then $\cos (2 \alpha-2 \beta)+\cos (23-2 \gamma)+\cos (2 \gamma-2 a)$ equals

1) 0

2) 1

3) 2

4) 3

Solution

As we learned in

De Moivre's Theorem -

$(\cos \theta+i \sin \theta)^n=\cos n \theta+i \sin n \theta \forall n \in I$

Now,

$\begin{aligned} & a^2=\cos 2 \alpha+i \sin 2 \alpha=e^{i 2 a} \\ & b^2=\cos 2 \beta+i \sin 2 \beta=e^{i 2 \beta} \\ & c^2=\cos 2 \gamma+i \sin 2 \gamma=e^{i 2 \gamma}\end{aligned}$

Given
$
\begin{aligned}
& \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=1 \\
& \Rightarrow \frac{e^{12 \gamma}}{e^{128}}+\frac{e^{128}}{e^{12 \gamma}}+\frac{e^{12 \gamma}}{e^{i 2 a}}=1 \\
& \Rightarrow e^{i 2(a-\beta)}+e^{12(3-\gamma)}+e^{12(7-a)}=1
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow \cos (2 \alpha-2 \beta)+i \sin (2 \alpha-2 \beta)+\cos (2 \beta-2 \gamma)+i \sin (2 \beta-2 \gamma)+ \\
& \cos (2 \gamma-2 \alpha)+i \sin (2 \gamma-2 \alpha)=1
\end{aligned}
$

Comparing real and imaginary parts

$\begin{aligned} & \Rightarrow \cos (2 \alpha-2 \beta)+\cos (2 \beta-2 \gamma)+\cos (2 \gamma-2 \alpha)=1 \text { and } \sin (2 \alpha-2 \beta)+ \\ & \sin (2 \beta-2 \gamma)+\sin (2 \gamma-2 \alpha)=0\end{aligned}$

Hence, the answer is the option 2.

Example 2: Let p. $\mathrm{q} \in \mathbb{R}$ and $(1-\sqrt{3} i)^{200}=2^{199}(p+i q), i=\sqrt{-1}$
Then $p+q+q^2$ and $p-q+q^2$ are roots of the equation

1) $x^2-4 x-1=0$

2) $x^2-4 x+1=0$

3) $x^2+4 x-1=0$

4) $x^2+4 x+1=0$

Solution

$\begin{aligned} & (1-\sqrt{3} i)^{200}=2^{199}(p+i q) \\ & \Rightarrow 2^{200} \operatorname{cis}\left(\frac{-\pi}{3}\right)^{200}=2^{199}(p+i q) \\ & \Rightarrow 2^{200}\left(\operatorname{cis}\left(-\frac{200 \pi}{3}\right)\right)=2^{199}(p+i q)\end{aligned}$

$\begin{aligned} & \Rightarrow 2\left(\operatorname{cis}\left(-66 \pi-\frac{2 \pi}{3}\right)\right)=(p+i q) \\ & \Rightarrow 2\left[\operatorname{cis}\left(\frac{-2 \pi}{3}\right)\right]=(p+i q) \\ & \Rightarrow 2\left[\frac{-1}{2}-\frac{\sqrt{3} i}{2}\right]=(p+i q) \\ & \Rightarrow p=-1, q=-\sqrt{3}\end{aligned}$
$
\begin{aligned}
& \text { Now } \\
& \alpha=p+q+q^2=2-\sqrt{3} \\
& \beta=p-q+q^2=2+\sqrt{3}
\end{aligned}
$

req. quad is $x^2-4 x+1=0$

Hence, the answer is the option 2.

Example 3: The value of $\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9} i \cos \frac{2 \pi}{9}}\right)^3$ is

1)$-\frac{1}{2}(\sqrt{3}-i)$

2) $-\frac{1}{2}(1-i \sqrt{3})$

3) $\frac{1}{2}(1-i \sqrt{3})$

4) $\frac{1}{2}(\sqrt{3}+i)$

Solution

$\begin{aligned} & \frac{\pi}{2}-\frac{2 \pi}{9} \\ & =\frac{95-4 \pi}{18}=\frac{5 \pi}{18} \\ & \Rightarrow \frac{1+\cos \frac{5 \pi}{18}+i \sin \frac{5 \pi}{18}}{1+\cos \frac{5 \pi}{18}-i \sin \frac{5 \pi}{18}}\end{aligned}$

$\begin{aligned} & =\frac{2 \cos ^2 \frac{5 \pi}{36}+2 i \sin \frac{5 \pi}{36} \cdot \cos \frac{5 \pi}{36}}{2 \cos ^2 \frac{5 \pi}{36}-2 i \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36}} \Rightarrow\left(\frac{\mathrm{e}^{i \frac{5 \pi}{36}}}{\mathrm{e}^{-i \frac{5 \pi}{36}}}\right)^3 \\ & =e^{\left(\frac{5 \pi}{18}\right)^3}=e^{i\left(\frac{5 \pi}{6}\right)} \\ & \left(\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}\right)=-\frac{\sqrt{3}}{2}+\frac{i}{2}\end{aligned}$

Hence, the answer is the option 1.

Example 4: The complex number $z=\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}$ is equal to :

1) $\sqrt{2} i\left(\cos \frac{5 \pi}{12}-i \sin \frac{5 \pi}{12}\right)$
2) $\sqrt{2}\left(\cos \frac{\pi}{12}+\sin \frac{\pi}{12}\right)$
3) $\sqrt{2}\left(\cos \frac{5 \pi}{12}+i \sin \frac{5 \pi}{12}\right)$
4) $\cos \frac{\pi}{12}-1 \sin \frac{\pi}{12}$

Solution

$Z=\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}$
$\mathrm{i}-1=\sqrt{2}\left(\frac{-1}{\sqrt{2}}+\frac{\mathrm{i}}{\sqrt{2}}\right)=\sqrt{2} \cdot \mathrm{e}^{\mathrm{i} \frac{3 \pi}{4}}$
$z=\frac{\sqrt{2} \cdot e^{i \frac{3 \pi}{4}}}{e^{i \cdot \frac{3 \pi}{4}}}$
$=\sqrt{2} \cdot \mathrm{e}^{\mathrm{i}\left(\frac{3 \pi}{4}-\frac{\pi}{3}\right)}$
$=\sqrt{2} \mathrm{e}^{\frac{5 \pi}{12} \mathrm{i}}$
$=\sqrt{2}\left(\cos \left(\frac{5 \pi}{12}\right)+\mathrm{i} \sin \left(\frac{5 \pi}{12}\right)\right)$
Hence, the answer is the option (3).

Summary

One of the fundamental theorems of complex numbers, De Moivre's theorem is utilized to address a variety of complex number problems. The solution of trigonometric functions with numerous angles is another common use of this theorem. "De Moivre's Identity" and "De Moivre's Formula" are other names for DeMoivre's Theorem.