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

De Moivre's Theorem

De Moivre's Theorem is a fundamental concept in complex numbers used to solve problems involving powers and roots of complex numbers. It also plays a key role in simplifying trigonometric functions for multiple angles. Also known as De Moivre's Identity or De Moivre's Formula, the theorem is named after the mathematician Abraham De Moivre.

De Moivre’s Formula

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 fundamentally based on the Euler form representation of complex numbers. This theorem establishes a relationship between complex numbers in trigonometric form and their powers.

Statement of the Theorem

For any integer $n \in \mathbb{I}$, real number $\theta \in \mathbb{R}$, and imaginary unit $i = \sqrt{-1}$, the theorem states:

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

Proof Using Euler’s Formula

We start with Euler’s formula:
$e^{i \theta} = \cos \theta + i \sin \theta$

Raising both sides to the integer power $n$, we get:
$(e^{i \theta})^n = (\cos \theta + i \sin \theta)^n$

Therefore,
$e^{i n \theta} = (\cos \theta + i \sin \theta)^n$

Using Euler’s formula again:
$\cos (n \theta) + i \sin (n \theta) = (\cos \theta + i \sin \theta)^n$

This completes the proof for integer powers.

Extension to Multiple Angles

For angles $ \theta_1, \theta_2, \ldots, \theta_n \in \mathbb{R} $, the product of complex numbers in trigonometric form is:
$(\cos \theta_1 + i \sin \theta_1)(\cos \theta_2 + i \sin \theta_2) \cdots (\cos \theta_n + i \sin \theta_n) = \cos (\theta_1 + \theta_2 + \cdots + \theta_n) + i \sin (\theta_1 + \theta_2 + \cdots + \theta_n)$

Proof of the Product Formula

From Euler's formula, for each $\theta_k$, $e^{i \theta_k} = \cos \theta_k + i \sin \theta_k$

Therefore, $ (\cos \theta_1 + i \sin \theta_1)(\cos \theta_2 + i \sin \theta_2) \cdots (\cos \theta_n + i \sin \theta_n) = e^{i \theta_1} \cdot e^{i \theta_2} \cdots e^{i \theta_n} = e^{i (\theta_1 + \theta_2 + \cdots + \theta_n)} = \cos (\theta_1 + \theta_2 + \cdots + \theta_n) + i \sin (\theta_1 + \theta_2 + \cdots + \theta_n)$

Note on Rational Powers: If $n$ is a rational number, $n = \frac{p}{q}$ where $\text{HCF}(p, q) = 1$ and $q > 0$, then the expression $z^n$ (where $z$ is a complex number) will have $q$ distinct values. One of these values is:
$\cos (n \theta) + i \sin (n \theta)$

Additional Variations of the Theorem

$(\cos \theta - i \sin \theta)^n = \cos(n \theta) - i \sin(n \theta)$

$ (\sin \theta + i \cos \theta)^n = i^n \left( \cos(n \theta) - i \sin(n \theta) \right)$ (by factoring out i first)

$(\sin \theta - i \cos \theta)^n = (-i)^n \left( \cos(n \theta) + i \sin(n \theta) \right)$ (by factoring out $i$ first)

Important: $(\cos \theta + i \sin \phi)^n \neq \cos(n \theta) + i \sin(n \phi) \quad \text{unless } \theta = \phi$

Applications of De Moivre’s Theorem

• Finding powers of complex numbers: Simplifies $(\cos \theta + i \sin \theta)^n$ calculations.

• Calculating roots of complex numbers: Essential for extracting nth roots using polar form or Euler form.

• Solving trigonometric equations: Provides a method for multiple-angle identities

Solved Examples Based on 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 have learned in De Moivre's Theorem:
$(\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n \theta \quad \forall n \in \mathbb{I}$

Now, $a^2 = \cos 2\alpha + i \sin 2\alpha = e^{i 2\alpha}$
$b^2 = \cos 2\beta + i \sin 2\beta = e^{i 2\beta}$
$c^2 = \cos 2\gamma + i \sin 2\gamma = e^{i 2\gamma}$

Given, $\frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} = 1$

$\Rightarrow \frac{e^{i 2\alpha}}{e^{i 2\beta}} + \frac{e^{i 2\beta}}{e^{i 2\gamma}} + \frac{e^{i 2\gamma}}{e^{i 2\alpha}} = 1$

$\Rightarrow e^{i 2(\alpha - \beta)} + e^{i 2(\beta - \gamma)} + e^{i 2(\gamma - \alpha)} = 1$

$\Rightarrow \cos (2\alpha - 2\beta) + i \sin (2\alpha - 2\beta) + \cos (2\beta - 2\gamma) + i \sin (2\beta - 2\gamma) + \cos (2\gamma - 2\alpha) + i \sin (2\gamma - 2\alpha) = 1$

Comparing real and imaginary parts,

$\cos (2\alpha - 2\beta) + \cos (2\beta - 2\gamma) + \cos (2\gamma - 2\alpha) = 1$ and $\sin (2\alpha - 2\beta) + \sin (2\beta - 2\gamma) + \sin (2\gamma - 2\alpha) = 0$

Hence, the answer is option 2.

Example 2: Let $p,q \in \mathbb{R}$ 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:

$(1 - \sqrt{3} i)^{200} = 2^{199} (p + i q)$

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

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

$\Rightarrow 2 \operatorname{cis}\left(-66 \pi - \frac{2 \pi}{3}\right) = p + i q$

$\Rightarrow 2 \operatorname{cis}\left(\frac{-2 \pi}{3}\right) = p + i q$

$\Rightarrow 2 \left(-\frac{1}{2} - \frac{\sqrt{3} i}{2}\right) = p + i q$

$\Rightarrow p = -1, \quad q = -\sqrt{3}$

Now, $\alpha = p + q + q^2 = 2 - \sqrt{3}$

$\beta = p - q + q^2 = 2 + \sqrt{3}$

Required quadratic is $x^2 - 4x + 1 = 0$

Hence, the answer is 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:

$\frac{\pi}{2} - \frac{2 \pi}{9} = \frac{9 \pi - 4 \pi}{18} = \frac{5 \pi}{18}$

$\Rightarrow \frac{1 + \cos \frac{5 \pi}{18} + i \sin \frac{5 \pi}{18}}{1 + \cos \frac{5 \pi}{18} - i \sin \frac{5 \pi}{18}}$

$= \frac{2 \cos^2 \frac{5 \pi}{36} + 2 i \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36}}{2 \cos^2 \frac{5 \pi}{36} - 2 i \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36}} \Rightarrow \left(\frac{e^{i \frac{5 \pi}{36}}}{e^{-i \frac{5 \pi}{36}}}\right)^3$

$= e^{i \left(\frac{5 \pi}{18}\right) 3} = e^{i \frac{5 \pi}{6}}$

$= \cos \frac{5 \pi}{6} + i \sin \frac{5 \pi}{6} = -\frac{\sqrt{3}}{2} + \frac{i}{2}$

Hence, the answer is 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).

List of topics related to the De-Moivre's Theorem

Below is the list of topics related to De Moivre’s Theorem that are essential for understanding its applications and concepts.

NCERT Resources

This section offers essential solutions, exemplar problems, and notes for Class 11 Maths Chapter 5 on Complex Numbers and Quadratic Equations. These materials support clear understanding and effective exam preparation.

NCERT Solutions for Class 11 Maths Chapter 5 -Complex Numbers and Quadratic Equations

NCERT Exemplar Class 11 Maths Solutions for Chapter 5 - Complex Numbers and Quadratic Equations

NCERT Class 11 Maths Notes for Chapter 5 - Complex Numbers and Quadratic Equations

Practice Questions based on the De-Moivre's Theorem

Below are practice questions based on De Moivre's Theorem to strengthen understanding and application skills. These help in finding powers and roots of complex numbers effectively.

De Moivre's Theorem - Practice Question MCQ

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