Question

If $a=\frac{-1+\sqrt{3i}}{2},b=\frac{-1-\sqrt{3i}}{2}$ then show that $a^2=b$ and $b^2=a$.

Answer 1

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

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

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

$=\frac{-2-2\sqrt{3}i}{2^2}$

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

$=b$

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

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

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

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

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

$=a$