Question

Find all the complex roots. Write the answer in the indicated form. The complex cube roots of 27(cos 234° + i sin 234°) (polar form)
a. 3(cos 78° + i sin 78°), 3(cos198° + i sin 198°), 3(cos 318° + i sin 318°)

b. 3(cos 78° + i sin 78°), 3(cos 118° + i sin 118°), 3(cos 158° + i sin 158°)

c. -3(cos 78° + i sin 78°), 3(cos 118° + i sin 118°), -3(cos 158° + i sin 158°)

d. -3(cos 78° + i sin 78°), 3(cos 198° + i sin 198°), -3(cos 318° + i sin 318°)

Answer 1

$\bf \textit{ roots of complex numbers, DeMoivre's theorem} \\\\ \sqrt[n]{z}=\sqrt[n]{r}\left[ cos\left( \cfrac{\theta+2\pi k}{n} \right) +i\ sin\left( \cfrac{\theta+2\pi k}{n} \right)\right]\quad \begin{array}{llll} k\ roots\\ 0,1,2,3,... \end{array}\\\\ -------------------------------$

$\bf z=27[cos(234^o)+i~sin(234^o)] \\\\\\ \sqrt[3]{z}\implies \sqrt[3]{27}\left[ cos\left( \cfrac{234^o+360^o(0)}{3} \right) +i\ sin\left( \cfrac{234^o+360^o(0)}{3} \right)\right] \\\\\\ 3\left[ cos\left( \cfrac{234^o}{3} \right) +i\ sin\left( \cfrac{234^o}{3} \right)\right]\implies \stackrel{k=0~~1st~root}{3[cos(78^o)+i~sin(78^o)]}\\\\ -------------------------------$

$\bf \sqrt[3]{z}\implies \sqrt[3]{27}\left[ cos\left( \cfrac{234^o+360^o(1)}{3} \right) +i\ sin\left( \cfrac{234^o+360^o(1)}{3} \right)\right] \\\\\\ 3\left[ cos\left( \cfrac{594^o}{3} \right) +i\ sin\left( \cfrac{594^o}{3} \right)\right]\implies \stackrel{k=1~~2nd~root}{3[cos(198^o)+i~sin(198^o)]}\\\\ -------------------------------$

$\bf \sqrt[3]{z}\implies \sqrt[3]{27}\left[ cos\left( \cfrac{234^o+360^o(2)}{3} \right) +i\ sin\left( \cfrac{234^o+360^o(2)}{3} \right)\right] \\\\\\ 3\left[ cos\left( \cfrac{954^o}{3} \right) +i\ sin\left( \cfrac{954^o}{3} \right)\right]\implies \stackrel{k=2~~3rd~root}{3[cos(318^o)+i~sin(318^o)]}$