If and are the complex cube roots of unity, show that ^2+ ^2+ =0. - Vidyadip

Question

If $\alpha$ and $\beta$ are the complex cube roots of unity, show that $\alpha^2+\beta^2+\alpha \beta=0$.

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Answer

$\alpha$ and $\beta$ are the complex cube roots of unity.
$
\begin{aligned}
& \therefore \quad \alpha=\frac{-i+i \sqrt{3}}{2} \text { and } \beta=\frac{-1-i \sqrt{3}}{2} \\
& \therefore \quad \alpha \beta=\left(\frac{-1+i \sqrt{3}}{2}\right)\left(\frac{-1-i \sqrt{3}}{2}\right) \\
& =\frac{(-1)^2-(\mathrm{i} \sqrt{3})^2}{4}=\frac{1-(-1)(3)}{4} \ldots\left[\because \mathrm{i}^2=-1\right] \\
& =\frac{1+3}{4} \\
& \therefore \quad \alpha \beta=1 \quad \text { } \\
& \text { Also, } \alpha+\beta=\frac{-1+i \sqrt{3}}{2}+\frac{-1-i \sqrt{3}}{2} \\
& =\frac{-1+\mathrm{i} \sqrt{3}-1-\mathrm{i} \sqrt{3}}{2} \\
& =\frac{-2}{2} \\
& \therefore \alpha-\beta=-1 \\
& \text { L.H.S. }=\alpha^2+\beta^2+\alpha \beta \\
& =\alpha^2+2 \alpha \beta+\beta^2+\alpha \beta-2 \alpha \beta \\
& =\left(\alpha^2+2 \alpha \beta+\beta^2\right)-\alpha \beta \\
& =(\alpha+\beta)^2-\alpha \beta \\
& =(-1)^2-1 \\
& =1-1 \\
& =0 \\
& =\text { R.H.S. } \\
&
\end{aligned}
$
[Adding and subtracting $2 \alpha \beta$ ]

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