Question

If $i=\sqrt{-1},$ then $4+5\left[\right(\frac{-1}{2})+i\frac{\sqrt{3}}{2}{]}^{334}+3[\frac{-1}{2}+\left(i\frac{\sqrt{3}}{2}\right){]}^{365}$ is equal to

• A. $1-i\sqrt{3}$
• B. $-1+i\sqrt{3}$
• C. $i\sqrt{3}$
• D. $-i\sqrt{3}$

Answer 1

Answer: (C)

$i\sqrt{3}$

Let $w$ be the cube root of unity.
$\therefore w^3=1\&1+w^2+w=0$
where, $w=\frac{-1+i\sqrt{3}}{2}\&w^2=\frac{1-i\sqrt{3}}{2}$ ...(1)
$z=4+5{\left(\frac{-1+i\sqrt{3}}{2}\right)}^{334}+3{\left(\frac{-1+i\sqrt{3}}{2}\right)}^{365}=4+5w^{334}+3w^{365}$ ...{from (1)}
$⟹z=4+5\left[{\left(w^3\right)}^{111}w\right]+3\left[{\left(w^3\right)}^{121}{w}^2\right]=4+5w+3w^2$ ...{$\because w^3=1$}
$⟹z=w-w^2$ ...{$\because -1=w^2+w$}

$⟹z=\frac{-1+i\sqrt{3}}{2}-\frac{(-1-i\sqrt{3})}{2}=i\sqrt{3}$ ...{from 1}

Ans: C