Question

If $a=\frac{-1+\sqrt{3}i}{2},\ \ b=\frac{-1-\sqrt{3}i}{2}$ then show that a² = b and b² = a.

Answer 1

To prove
${a}^{2} = b \: \: \: \: \: and \: \: \: {b}^{2} = a$
let square a

${a}^{2} =\bigg(\frac{-1+\sqrt{3}i}{2}\bigg) ^{2} \\ \\ = \frac{1 + 3 {i}^{2} - 2 \sqrt{3} i }{4} \\ \\ as \: {i}^{2} = - 1 \\ \\ so \\ \\ = \frac{1 - 3 - 2 \sqrt{3}i }{4} \\ \\ = \frac{2( - 1 - \sqrt{3}i) }{4} \\ \\ = \frac{( - 1 - \sqrt{3}i) }{2} = b \\ \\ hence \: prove$
Now for

${b}^{2} =\bigg(\frac{-1 - \sqrt{3}i}{2}\bigg) ^{2} \\ \\ = \frac{1 + 3 {i}^{2} + 2 \sqrt{3} i }{4} \\ \\ as \: {i}^{2} = - 1 \\ \\ so \\ \\ = \frac{1 - 3 + 2 \sqrt{3}i }{4} \\ \\ = \frac{2( - 1 + \sqrt{3}i) }{4} \\ \\ = \frac{( - 1 + \sqrt{3}i) }{2} = a \\ \\ hence \: prove$
Hope it helps you.