Question

If z = 1 - square root 3i, what is z^3 Answers -8 8 -8i 8i

Answer 1

Answer:

$z = 1 - \sqrt{3} i \\ {z}^{3} = (1 - \sqrt{3} i) ^{3} \\ =(1 - \sqrt{3} i)(1 - \sqrt{3} i)(1 - \sqrt{3} i) \\ open \: brackets \\ = (1 - 2 \sqrt{3} i + ( { \sqrt{3} i})^{2} )(1 - \sqrt{3} i) \\ but \: {( \sqrt{3}i) }^{2} = - 3 \: since \: {i}^{2} \: is \: - 1 \: in \: complex \: numbers \\ = (1 - 2 \sqrt{3} i - 3)(1 - \sqrt{3} i) \\ = ( - 2 - 2 \sqrt{3} i)(1 - \sqrt{3} i) \\ = ( - 2 + 2 \sqrt{3} i - 2 \sqrt{3} i + 2( { \sqrt{3} i)}^{2} ) \\ = - 2 - 6 \\ = - 8$