Question

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

Answer 1

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

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

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

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

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

$a^2=b$

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

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

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

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

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

$b^2=a$