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Find the roots.

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Find the cube roots of $$4 \sqrt3 + 4i$$

Guest May 7, 2017
#1
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I solved it

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

$$\theta=\frac{\pi}{6}$$

$$r=8$$

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