First slide

De-moivre's theorem

Question

If _$α$ and _$β$ are the roots of the equation _{$x^{2} - x + 1 = 0,$} then _{$α^{2009} + β^{2009} =$}

Easy

A

2
B

_{$- 2$}
C

_{$- 1$}
D

1

Solution

_{$x^{2} - x + 1 = 0 \Rightarrow x = \frac{1 + \sqrt{1 - 4}}{2}$}
_{$x = \frac{1 + \sqrt{3 i}}{2}$}

_{$α = \frac{1}{2} + i \frac{\sqrt{3}}{2} . β = \frac{1}{2} - \frac{i \sqrt{3}}{2}$}

_{$α = \cos \frac{π}{3} + i \sin \frac{π}{3}, β = \cos \frac{π}{3} - i \sin \frac{π}{3}$}

_{$α^{2009} + β^{2009} = 2 \cos 2009 (\frac{π}{3})$}

_{$= 2 \cos [668 π + π + \frac{2 π}{3}] = 2 \cos (π + \frac{2 π}{3})$}

_{$= - 2 \cos \frac{2 π}{3} = - 2 (- \frac{1}{2}) = 1$}

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App