Question

if x and y are complex cube root of unity then show that x^2+y^2+xy=0​

Answer 1

Given data:

$x$ and $y$ are complex cube roots of unity

To show: $x^{2}+y^{2}+xy=0$

Step-by-step solution:

We must remember that if $x$ and $y$ are the two complex roots of unity, then we can write

• $x=\frac{-1+\sqrt{3}i}{2}$ and

• $y=\frac{-1-\sqrt{3}i}{2}$.

Now, left hand side

$=x^{2}+xy+y^{2}$

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

$=\frac{1-2\sqrt{3}i-3}{4}+\frac{1+3}{4}+\frac{1+2\sqrt{3}i-3}{4}$ since $i^{2}=-1$

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

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

$=\frac{0}{2}$

$=0$ = right hand side

Conclusion:

Thus, $x^{2}+xy+y^{2}=0$ ( proved )