Q. If $i = \sqrt{- 1}$, then $4 + 5\left( - \frac{1}{2} + \frac{i\sqrt{3}}{2} \right)^{334} + 3\left( - \frac{1}{2} + \frac{i\sqrt{3}}{2} \right)^{365}$ is equal to
A) $1 - i\sqrt{3}$
B) $- 1 + i\sqrt{3}$
C) $i\sqrt{3}$
D) $- i\sqrt{3}$
Explanation
(c)
We know, $- \frac{1}{2} + \frac{i\sqrt{3}}{2} = \omega$
$\therefore\ \ \ \ 4 + 5({\omega)}^{334} + 3(\omega)^{365}$
$= 4 + 5\left( \omega^{3} \right)^{111}.\ \omega^{1} + 3\left( \omega^{3} \right)^{121}.\omega^{2}$
$= 4 + 5\omega + 3\omega^{2}$
$= 3\left( 1 + \omega + \omega^{2} \right) + 1 + 2\omega$
$= 1 + ( - 1 + i\sqrt{3})$
$= i\sqrt{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.