Question

If z = 2 ( cos 2¥/3 + isin 2¥/3 ), then z^3=

i. 2 - i8√3

ii. -4 + i4√3

iii.-8-i√2

iv. -8 - i4

Answer 1

If $Z=2(cos(\frac{2¥}{3})+isin(\frac{2¥}{3}))$

we know de moivre's formula

:-

$Z^n=r(re^{r\theta})^n=r^n(cosn\theta+isinn\theta)\\orZ^n=r^n[cos(n¥)+isin(n¥)]\\so,\space Z^3=2^3(cos(\frac{2¥}{3})3+isin(\frac{2¥}{3}.3))\\=8(cos(2¥)+isin(2¥))\\=8(\frac{-1}{2}+i.\frac{\sqrt3}{2})$