Double Math

De Moivre’s Theorem

In the field of complex numbers, De Moivre’s Theorem is one of the most important and useful theorems which connects complex numbers and trigonometry. Also helpful for obtaining relationships between trigonometric functions of multiple angles. DeMoivre’s Theorem is also known as “De Moivre’s Identity” and “De Moivre’s Formula”. The name of the theorem is after the name of great Mathematician Abraham De Moivre, who made many contributions to the field of mathematics, mainly in the areas of theory of probability and algebra.

Unlike real numbers, complex numbers have both imaginary and real components. You can represent it by the expression (a+bi), where a and b are real numbers, and i is imaginary. i in the definition of the function is defined as √(-1). Complex numbers, z = a+bi, can be represented in polar form along with being able to be represented as points (a,b) on a graph, as illustrated below:

Contents

• Formula
• Proof
• Uses
• Problems

De Moivre’s Formula

For any real number x and integer n.

({\cos\theta\;+\;i\;\sin\theta)}^n\;=\;\cos\;n\theta\;+i\;\sin\;n\theta

Proof:

we prove it by using the mathod of mathematical induction

Step. 1

For n=1,

({\cos\;\theta\;+\;i\;\sin\theta)}^1\;=\;\cos\;1\theta\;+i\;\sin\;1\theta.

({\cos\;\theta\;+\;i\;\sin\theta)}^{}\;=\;\cos\;\theta\;+i\;\sin\;\theta

so, Statement is true for n=1

Step.2

Let us suppose that statement is true for n=k\; i.e,

({\cos\theta\;+\;i\;\sin\theta)}^k\;=\;\cos\;k\theta\;+i\;\sin\;k\theta\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(ii)

Step.3

Now we are to prove statement is true for n=k+1.

for this put n=k+1 on left hand side of the above equation

L.H.S=({\cos\theta\;+\;i\;\sin\theta)}^{k+1}

({\cos\theta\;+\;i\;\sin\theta)}^{k+1}=({\cos\theta\;+\;i\;\sin\theta)}^k({\cos\theta\;+\;i\;\sin\theta)}

({\cos\theta\;+\;i\;\sin\theta)}^{k+1}=({\cos k\theta\;+\;i\;\sin k\theta)}({\cos\theta\;+\;i\;\sin\theta)}.

by using (ii)

({\cos\theta\;+\;i\;\sin\theta)}^{k+1}=\cos k\theta\;\cos\theta\;+i\;\cos k\theta\;\sin\theta\;+i\;\sin k\theta\;\cos\theta+i^2\sin k\theta\sin\theta.

({\cos\theta\;+\;i\;\sin\theta)}^{k+1}=\cos k\theta\;\cos\theta\;+i\;\cos k\theta\;\sin\theta\;+i\;\sin k\theta\;\cos\theta+(-1)\sin k\theta\sin\theta.

({\cos\theta\;+\;i\;\sin\theta)}^{k+1}=\cos k\theta\;\cos\theta\;+i\;\cos k\theta\;\sin\theta\;+i\;\sin k\theta\;\cos\theta-\sin k\theta\sin\theta.

({\cos\theta\;+\;i\;\sin\theta)}^{k+1}=\cos k\theta\;\cos\theta\;-\sin k\theta\sin\theta+i\;\cos k\theta\;\sin\theta\;+i\;\sin k\theta\;\cos\theta.

({\cos\theta\;+\;i\;\sin\theta)}^{k+1}=\cos k\theta\;\cos\theta\;-\sin k\theta\sin\theta+i\;(\cos k\theta\;\sin\theta\;+\;\sin k\theta\;\cos\theta).

({\cos\theta\;+\;i\;\sin\theta)}^{k+1}=\cos(k\theta+\theta)+i\;\sin(k\theta+\theta).

({\cos\theta\;+\;i\;\sin\theta)}^{k+1}=\cos(k+1)\theta+i\;\sin(k+1)\theta.

Hence Statement is true for n=k+1.