Question

${\left[\frac{-1+i\sqrt{3}}{2}\right]}^4+{\left[\frac{-1-i\sqrt{3}}{2}\right]}^4+{\left[\frac{-1+i\sqrt{3}}{2}\right]}^5+{\left[\frac{-1-i\sqrt{3}}{2}\right]}^5$ is equal to

• A. -2
• B. -1
• C. 2
• D. none of these

Answer 1

The correct option is A -2

Simplifying the above expression we get
$w^4+{\overline{w}}^4+w^5+\overline{w^5}$ where $w$ is the cube root of unity.
Now
$\overline{w}=w^2$ and
$1+w+w^2=0$ and $w^3=1$
Therefore
$w^4+w^8+w^5+w^{10}$
$=w^4[1+w^4]+w^5[1+w^5]$
$=w^3.w[1+w^3.w]+w^3.w^2[1+w^3.w^2]$
$=w[1+w]+w^2[1+w^2]$
$=w(-w^2)+w^2(-w)$
$=-2w^3$
$=-2$.