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Find the cube roots of \(4 \sqrt3 + 4i\)

Guest May 7, 2017
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I solved it smiley

 

\(tan(\theta)=\frac{y}{x}=\frac{b}{a}=\frac{4}{4\sqrt{3}}\)

\(\theta=\frac{\pi}{6}\)

\(r=8\)

 

In polar form

\(8[cos(\frac{\pi}{6})+isin(\frac{\pi}{6})] \)

 

Now to find the roots.

\(\sqrt[n]{8}[cos(\frac{\frac{\pi}{6}+2\pi k}{n})+isin(\frac{\frac{\pi}{6}+2\pi k}{n})]\)

This question asks for cube roots so n=3

\(\sqrt[3]{8}[cos(\frac{\frac{\pi}{6}+2\pi k}{3})+isin(\frac{\frac{\pi}{6}+2\pi k}{3})]\)

 

Now that this is all filled in, lets change the k to each root we need.

\(k=0: 2[cos(\frac{\frac{\pi}{6}}{3})+isin(\frac{\frac{\pi}{6}}{3})]\)

\(k=1: 2[cos(\frac{\frac{\pi}{6}+2\pi}{3})+isin(\frac{\frac{\pi}{6}+2\pi}{3})]\)

\(k=2: 2[cos(\frac{\frac{\pi}{6}+4\pi }{3})+isin(\frac{\frac{\pi}{6}+4\pi }{3})]\)

Guest May 8, 2017

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