Question

Please solve!!! cube root 64(cos330+isin330) show work please

Answer 1

Hello: 
cube root 64(cos330+isin330) iz the complex number with : 
z^3 = 64(cos330+isin330)   ...(1)
let:    z = r(cos(θ) +isin(θ))
so : z^3 = r^3 (cos(3θ) +isin(3θ))...(2)    ( by moivre)
by(1) and (2) :
 r^3 =64 = 4^3    and    3θ =11π/6 +2kπ.... k=0 , 1 , 2
r =4  and   θ  = 11π/18 +2kπ/3 ..... k=0 , 1 , 2
coclusion : 3 cube root ....calculate if k = 0 , 1 ,  2

Answer 2

$\bf \sqrt[{{ n}}]{z}=\sqrt[{{ n}}]{r}\left[ cos\left( \frac{\theta+2\pi k}{{{ n}}} \right) +i\ sin\left( \frac{\theta+2\pi k}{{{ n}}} \right)\right]\quad k\ roots\\\\ -------------------------------$

$\bf \sqrt[3]{64}[cos(330^o)+i\ sin(330^o)] \\\\\\ \sqrt[3]{64}\left[cos\left( \frac{330+360(0)}{3} \right) + i\ sin\left( \frac{330+360(0)}{3} \right) \right] \\\\\\ 4[cos(110^o)+i\ sin(110^o)]\implies 4(-0.34202014)+4(0.9396926) \\\\\\ \approx -1.37 + 3.76i\impliedby \textit{first root, k = 0}\\\\ -------------------------------$


$\bf \sqrt[3]{64}[cos(330^o)+i\ sin(330^o)] \\\\\\ \sqrt[3]{64}\left[cos\left( \frac{330+360(1)}{3} \right) + i\ sin\left( \frac{330+360(1)}{3} \right) \right] \\\\\\ 4[cos(230^o)+i\ sin(230^o)]\implies 4(-0.6427876)+4(-0.7660444) \\\\\\ \approx -2.57 -3.06i\impliedby \textit{second root, k = 1}\\\\ -------------------------------$


$\bf \sqrt[3]{64}[cos(330^o)+i\ sin(330^o)] \\\\\\ \sqrt[3]{64}\left[cos\left( \frac{330+360(2)}{3} \right) + i\ sin\left( \frac{330+360(2)}{3} \right) \right] \\\\\\ 4[cos(350^o)+i\ sin(350^o)]\implies 4(0.98480775)+4(-0.1736481) \\\\\\ \approx 3.94 -0.695i\impliedby \textit{third root, k = 2}$