Question

Find the value of (${\frac{\sqrt{3}}{2}+\frac{i}{2})}^5$ + (${\frac{\sqrt{3}}{2}-\frac{i}{2})}^5$

• A.
• B. -
• C. i
• D. - i

Answer 1

The correct option is B

-

We know $\omega$ = $\frac{-1+i\sqrt{3}}{2}$ , then ${\omega }^2$ = $\frac{-1-i\sqrt{3}}{2}$
We can see that the terms in equation are similar to $\omega$ and ${\omega }^2$ , but they are not the same . We will try to modify them and express them in terms w and ${\omega }^2$ .
We have $\omega$ = $\frac{-1+i\sqrt{3}}{2}$ ,then ${\omega }^2$ = $\frac{-1-i\sqrt{3}}{2}$
$\frac{\sqrt{3}}{2}$ + $\frac{i}{2}$ = -i ($\frac{-1-i\sqrt{3}}{2}$) = - i$\omega$
and $\frac{\sqrt{3}}{2}$ - $\frac{i}{2}$ = -i$\left(\frac{-1-i\sqrt{3}}{2}\right)$ = - i${\omega }^2$
$\Rightarrow$${(\frac{\sqrt{3}}{2}+\frac{i}{2})}^5+({\frac{\sqrt{3}}{2}-\frac{i}{2})}^5$ = ${(-i\omega )}^5$ +i$({{\omega }^2)}^5$
= ${(-i)}^5$${\omega }^5$ + ${\left(i\right)}^5$ ${\left({\omega }^2\right)}^5$
= -i ${\omega }^2$ + i$\omega$
= -i($\omega$ - ${\omega }^2$)
=i(i$\sqrt{3}$) =-$\sqrt{3}$