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

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

Komal MiglaniUpdated on 02 Jul 2025, 08:06 PM IST

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.

This Story also Contains

  1. De-moivre's Theorem
  2. De Moivre’s Theorem Proof
  3. Solved Examples Based on De-Moivre's Theorem
DeMoivre's Theorem - Definition, Proof, Uses and Examples
DeMoivre's Theorem - Definition, Proof, Uses and Examples

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 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.

Frequently Asked Questions (FAQs)

Q: How does De Moivre's Theorem relate to the concept of phase in signal processing?
A:
In signal processing, De Moivre's Theorem is crucial for understanding phase relationships. It shows how the phase of a complex signal changes when the signal is raised to a power, which is important in many signal processing applications, including modulation and demodulation techniques.
Q: What role does De Moivre's Theorem play in the study of fractals?
A:
In the study of fractals, particularly those generated by complex functions, De Moivre's Theorem is useful in understanding how certain transformations affect the fractal shape. It helps explain the self-similarity and rotational symmetry observed in many complex-plane fractals, such as the Mandelbrot set.
Q: What's the significance of De Moivre's Theorem in the theory of Riemann surfaces?
A:
In the theory of Riemann surfaces, De Moivre's Theorem helps in understanding the behavior of multi-valued complex functions. It's particularly useful in visualizing how these functions behave under analytic continuation around branch points, which is crucial in defining and working with Riemann surfaces.
Q: How can De Moivre's Theorem be used to separate real and imaginary parts of complex powers?
A:
De Moivre's Theorem provides a straightforward way to separate the real and imaginary parts of complex powers. For (cos θ + i sin θ)^n, the real part is cos(nθ) and the imaginary part is sin(nθ). This separation is often crucial in solving problems involving complex numbers.
Q: How does De Moivre's Theorem help in understanding the periodicity of complex exponentials?
A:
De Moivre's Theorem helps explain why complex exponentials are periodic. Since cos θ + i sin θ = cos(θ + 2π) + i sin(θ + 2π), raising both sides to the nth power shows that the complex exponential function has a period of 2π/n. This insight is fundamental in many areas of mathematics and physics.
Q: What role does De Moivre's Theorem play in Fourier analysis?
A:
De Moivre's Theorem is fundamental to Fourier analysis. It allows us to express periodic functions as sums of sines and cosines (or complex exponentials), which is the basis of Fourier series. The theorem helps in understanding how these components combine and how the coefficients relate to the original function.
Q: What's the significance of De Moivre's Theorem in the theory of angular momentum in quantum mechanics?
A:
In quantum mechanics, De Moivre's Theorem is crucial for understanding angular momentum. It's used in the mathematical description of spin states and in calculating how these states change under rotations. The theorem helps explain the quantization of angular momentum in quantum systems.
Q: How does De Moivre's Theorem relate to the concept of winding numbers in complex analysis?
A:
De Moivre's Theorem is closely related to the concept of winding numbers in complex analysis. The theorem essentially describes how many times a complex number winds around the origin when raised to a power. This connection is important in understanding the behavior of complex functions and in the theory of contour integrals.
Q: What's the connection between De Moivre's Theorem and the roots of unity filter in signal processing?
A:
De Moivre's Theorem is fundamental to the roots of unity filter in signal processing. This filter uses the complex roots of unity (given by De Moivre's Theorem) to separate a signal into its frequency components. Understanding the theorem helps in grasping how this filter operates and why it's effective.
Q: How can De Moivre's Theorem be used to derive formulas for trigonometric functions of multiple angles?
A:
De Moivre's Theorem provides an elegant way to derive formulas for trigonometric functions of multiple angles. By expanding (cos θ + i sin θ)^n using the binomial theorem and equating real and imaginary parts, you can obtain expressions for cos(nθ) and sin(nθ) in terms of powers of cos θ and sin θ.