Question

If $\alpha$ and $\beta$ are the complex cube roots of unity, then find ${{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta$
[a] 0
[b] 1
[c] -1
[d] 2

Answer 1

Hint: Use the fact that the complex cube roots of units are the complex roots of the equation ${{x}^{3}}=1$. Subtract 1 on both sides and use the identity ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ to prove that the complex cube roots of units are the roots of the quadratic equation ${{x}^{2}}+x+1=0$. Use the fact that the sum of roots of a quadratic equation $a{{x}^{2}}+bx+c=0$ is given by $\dfrac{-b}{a}$ and the product of the roots is given by $\dfrac{c}{a}$. Use the fact that ${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$ to find the value of ${{\alpha }^{2}}+{{\beta }^{2}}$ and hence find the value of the expression ${{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta$. Alternatively, use the fact that if $\alpha$ is one cube root of unity, then the other root of unity is ${{\alpha }^{2}}$. Hence prove that ${{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta =1+{{\alpha }^{2}}+{{\alpha }^{3}}$. Use the fact that if $\alpha$ is a cube root of unity, then $1+\alpha +{{\alpha }^{2}}=0$. Hence find the value of the given expression.

Complete step-by-step answer:

We know that the complex cube roots of unity are the complex roots of the equation ${{x}^{3}}=1$
Subtracting 1 on both sides, we get
${{x}^{3}}-1=0$
We know that ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$
Hence, we have $\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)$
Since x is complex, we have $x\ne 1$
Hence, we have ${{x}^{2}}+x+1=0$
Hence, we have $\alpha$ and $\beta$ are the roots of the equation ${{x}^{2}}+x+1=0$
We know that the sum of the roots of the equation $a{{x}^{2}}+bx+c=0$ is given by $\dfrac{-b}{a}$ and the product of the roots is given by $\dfrac{c}{a}$.
Hence, we have
$\alpha +\beta =\dfrac{-1}{1}=-1$ and $\alpha \beta =\dfrac{1}{1}=1$
Now, we know that ${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$
Hence, we have ${{\alpha }^{2}}+{{\beta }^{2}}={{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta ={{\left( -1 \right)}^{2}}-2\left( 1 \right)=-1$
Hence, we have ${{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta =-1+1=0$
Hence option [a] is correct.

Note: We know that if $\alpha$ is one cube root of unity, then the other root of unity is ${{\alpha }^{2}}$.
Hence, we have
$\beta ={{\alpha }^{2}}$
Now, we have
${{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta ={{\alpha }^{2}}+{{\alpha }^{4}}+{{\alpha }^{3}}$
Since $\alpha$ is a cube root of unity, we have ${{\alpha }^{3}}=1$
Hence, we have ${{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta ={{\alpha }^{2}}+\alpha +1$
Since $\alpha$ is a cube root of unity, we have $1+\alpha +{{\alpha }^{2}}=0$
Hence, we have ${{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta =0$, which is the same as obtained above.
Hence option [a] is correct.