+0  
 
0
277
1
avatar

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

Guest May 7, 2017
 #1
avatar
+1

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

18 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.