Question

Find all cube roots of the complex number 64(cos (219°) + i sin (219°)). Leave answers in polar form

and show all work

Answer 1

`\sqrt[n]{z}=\sqrt[n]{r}\left[ \cos\left( \cfrac{\theta+2\pi k}{n} \right) +i\sin\left( \cfrac{\theta+2\pi k}{n} \right)\right]\quad \begin{array}{llll} k\ roots\\ 0,1,2,3,... \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \boxed{k=0}\hspace{5em} \sqrt[ 3 ]{64} \left[ \cos\left( \cfrac{ 219^o + 360^o( 0 )}{3} \right) +i \sin\left( \cfrac{ 219^o + 360^o( 0 )}{3} \right)\right]`

`\sqrt[ 3 ]{64} \left[ \cos\left( \cfrac{ 219^o }{3} \right) +i \sin\left( \cfrac{ 219^o }{3} \right)\right]\implies \boxed{4[\cos(73^o)+i\sin(73^o)]} \\\\[-0.35em] ~\dotfill\\\\ \boxed{k=1}\hspace{5em} \sqrt[ 3 ]{64} \left[ \cos\left( \cfrac{ 219^o + 360^o( 1 )}{3} \right) +i \sin\left( \cfrac{ 219^o + 360^o( 1 )}{3} \right)\right]`

`\sqrt[ 3 ]{64} \left[ \cos\left( \cfrac{ 579^o }{3} \right) +i \sin\left( \cfrac{ 579^o }{3} \right)\right]\implies \boxed{4[\cos(193^o)+i\sin(193^o)]} \\\\[-0.35em] ~\dotfill\\\\ \boxed{k=2}\hspace{5em} \sqrt[ 3 ]{64} \left[ \cos\left( \cfrac{ 219^o + 360^o( 2 )}{3} \right) +i \sin\left( \cfrac{ 219^o + 360^o( 2 )}{3} \right)\right] \\\\\\ \sqrt[ 3 ]{64} \left[ \cos\left( \cfrac{ 939^o }{3} \right) +i \sin\left( \cfrac{ 939^o }{3} \right)\right]\implies \boxed{4[\cos(313^o)+i\sin(313^o)]}`