Question

polar form 1 + I into 1 + under root 3 I to the whole square divided by 1 + I​

Answer 1

Answer:

$\frac{(1 + i)(1 + \sqrt{3} \: i) {}^{2} }{1 + i} \\ \\ = {(1 + \sqrt{3}i) }^{2} \\ = 1 - 3 + 2 \sqrt{3} i \\ = - 2 + 2 \sqrt{3} i \\ |z| = \sqrt{ {2}^{2} + {(2 \sqrt{3} )}^{2} } = \sqrt{4 + 12} = 4 \\ \\ \tan( \alpha ) = \frac{2 \sqrt{3} }{ - 2} = - \sqrt{3} \\ \alpha = \frac{2\pi}{3} \\ \\ so \: polar \: \: form \\ \\ z = 4( \cos(\frac{2\pi}{3} + i \sin(\frac{2\pi}{3} ) ) \:$