De-Moivre’s Theorem

Mathematics Class XI Chapter 2: Complex Numbers and Quadratic Equations De-Moivre’s Theorem

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Statement:

$(\cos \theta+i \sin \theta)^n=\cos n~ \theta+i \sin~ n ~\theta$ for any integer $n$ .

De-Moivre’s Theorem (for Rational Index)

Let $P \text { and } q$ be two integers with $q > 0$ , $\left( \cos \dfrac{P}{q}\theta+i \sin \dfrac{P}{q}\theta \right)^q=(\cos P ~\theta+i \sin~ P~ \theta)=(\cos \theta+i \sin \theta)^P$

$\begin{aligned}\to \left[ \cos \left( \dfrac{P ~\theta}{q}+\dfrac{2 \pi}{q} \right)+ i \sin\left( \dfrac{P~ \theta}{q}+\dfrac{2 \pi}{q} \right) \right]^q&=\cos (P ~\theta+2 \pi) +i \sin (P~ \theta+2\pi) \\&=\cos (P ~\theta)+ i \sin (P ~\theta)\\&=(\cos \theta +i \sin \theta)^P \end{aligned}$

Example 1:

$\text { Value of }\begin{aligned}\left( \cos \dfrac{\pi}{5}+i \sin \dfrac{\pi}{5} \right)^5&=\cos \left( 5\times \dfrac{\pi}{5} \right)+i \sin \left( 5 \times \dfrac{\pi}{5} \right) \\&=\cos \pi+i \sin \pi=-1+i \times 0=-1 \end{aligned}$

Example 2:

Solve: $\left( \cos \dfrac{3 \pi}{7}+i \sin \dfrac{3 \pi}{7} \right)^7$

$=(\cos 3 \pi+i \sin 3 \pi)^{\frac{1}{7}\times7}$

$=(\cos 3 \pi+i \sin 3 \pi)$

$=(\cos \pi+i \sin \pi)^3$

$=(-1+i \times 0)^3=(-1)^3=-1$

Principal Argument of $z$ in the interval $0 \leq \theta < 2 \pi$

(i) In the interval $0 \leq \theta < \dfrac{\pi}{2}$

(ii) In the interval $\dfrac{\pi}{2} < \theta \leq \pi$

(iii) In the interval $\pi \leq \theta < \dfrac{3\pi}{2}$

(iv) In the interval $\dfrac{3 \pi}{2} < \theta \leq 2\pi$

All of the angles ( $\theta$ ) in the interval $0\leq \theta < 2 \pi$ are in anti-clockwise direction.

Principal Argument of $z$ in the interval (-\pi < \theta \leq \pi)

(i) In the interval $-\pi < \theta < -\dfrac{\pi}{2}$

(ii) In the interval $-\dfrac{\pi}{2}< \theta \leq \pi$

All of the angles in the interval $-\pi < \theta \leq \pi$ are in clockwise direction.