Complex root

  • Thread starter Valhalla
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  • #1
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I just bombed a quiz because it was 2 questions and this was one of them:

Find all three complex roots of the following equation (give answers in polar and rectangular form)

[tex]z^3+8=0[/tex]

Looks easy enough,

[tex] z=2e^{-i\frac{\theta}{3}} [/tex]

This is where I think I completely realized I wasn't sure what I was doing. My roommate suggested I look for the roots of unity which I know that:

[tex]r^n(cos(n\theta)+isin(n\theta))=1+i*0[/tex]

so if I want to consider mine it should be:

[tex]8^{1/3}(cos(\frac{\theta}{3})+isin(\frac{\theta}{3})=-8 [/tex]

so then

[tex]\theta=\frac{k2\pi}{3}[/tex]

is this the right track?
 
Last edited:

Answers and Replies

  • #2
StatusX
Homework Helper
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I'm not sure what you're trying to do (eg, your second to last latex line doesn't look right). Just keep in mind that there will be a unique solution for r, and then you're left with finding the cube roots of -1 (not 1). If it helps, these are the sixth roots of 1 that aren't also cube roots of 1.
 

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