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lora16 [44]
3 years ago
13

Find all the complex roots. Write the answer in the indicated form. The complex cube roots of 27(cos 234° + i sin 234°) (polar f

orm)
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°)
Mathematics
1 answer:
Black_prince [1.1K]3 years ago
3 0
\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)]}
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