Question

${\left(\frac{\sqrt{3}+i}{2}\right)}^6+{\left(\frac{i-\sqrt{3}}{2}\right)}^6=$

• A. $-2$
• B. $2$
• C. $-1$
• D. $1$

Answer 1

The correct option is A $-2$

We have $\frac{\sqrt{3}+i}{2}=\frac{i\sqrt{3}+i^2}{2i}$ (Here we multiplied the numerator and denominator by $(-i)$ on the LHS)

We get: $-i\left(\frac{-1+\sqrt{3}i}{2}\right)=i\omega$

Simliarly applying the same idea we get the steps below:

and $\frac{i-\sqrt{3}}{2}=\frac{i^2-i\sqrt{3}}{2i}=-i\left(\frac{-1-\sqrt{3}i}{2}\right)=-i{\omega }^2$

Hence ${\left(\frac{\sqrt{3}+i}{2}\right)}^6+{\left(\frac{i-\sqrt{3}}{2}\right)}^6={(-i\omega )}^6+{(-i{\omega }^2)}^6=i^6({\omega }^6+{\omega }^{12})=-1(1+1)=-2$