Question

If $\alpha$ and $\beta$ are the complex cube roots of unity, then find ${\alpha }^2+{\beta }^2+\alpha \beta =0$.

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is A $0$

If $\alpha ,\beta$ are the complex root of unity than,

$\alpha =\frac{-1+\sqrt{3}i}{2}$

$\beta =\frac{-1-\sqrt{3}i}{2}$

so,

${\alpha }^2+{\beta }^2+\alpha \beta$

$\to (\frac{-1+\sqrt{3}i}{2}{)}^2+(\frac{-1-\sqrt{3}i}{2}{)}^2+\left(\frac{-1+\sqrt{3}i}{2}\right)\left(\frac{-1-\sqrt{3}i}{2}\right)$

$\to \frac{1}{4}(1+3i^2-2\sqrt{3}i+1+3i^2+2\sqrt{3}i+1-3i^2)$. ($i^2=-1$)

$\to \frac{1}{4}(3+3i^2)$. put the value of $i^2$

$\to \frac{3-3}{4}$

$\to 0$

so option $A$ will be the right answer