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anonymous

  • one year ago

Someone please help me with this problem, I need to show my work and I have no idea how or where to start : Find the cube roots of 27(cos 330° + i sin 330°).

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  1. IrishBoy123
    • one year ago
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    you know \(e^{i \theta} = cos \theta + i \ sin \theta\)??

  2. anonymous
    • one year ago
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    I've seen it before in my lessons but I have never actually applied it to an problems

  3. IrishBoy123
    • one year ago
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    it's is very useful to know you can write \(27(cos 330° + i sin 330°)\) as \(\large 27 \ e^{i \ \frac{11}{6}\pi}\) so you want \(\large \sqrt[3] { 27 \ e^{i \ \frac{11}{6}\pi}}\)

  4. anonymous
    • one year ago
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    That looks very confusing to solve Im uncertain as to how im supposed to take the cube root of that number

  5. IrishBoy123
    • one year ago
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    there's something related called deMoivre, which might be easier for you right now \((cos \theta + i \ sin \theta)^n = cos \ n \ \theta + i \ sin \ n \ \theta\)

  6. anonymous
    • one year ago
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    Yes, I've heard of DeMoivre's Theorem

  7. IrishBoy123
    • one year ago
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    for you \(n = \frac{1}{3}\) and don't forget the \(\sqrt[3] 27\) !

  8. anonymous
    • one year ago
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    So \[(\cos \theta+i \sin \theta)^{\frac{ 1 }{ 3 }}=\cos (\frac{ 1 }{ 3 })\theta+i \sin (\frac{ 1 }{ 3 })\theta ?\]

  9. IrishBoy123
    • one year ago
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    not that simple! you will get roots following this pattern \( 27^{1/3} \ cis(\frac{2π \ m+\theta }{3}) \) for m = 0, 1, 2

  10. IrishBoy123
    • one year ago
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    ie 3 roots for a cubic

  11. anonymous
    • one year ago
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    What is \[cis(\frac{ 2pim+\theta }{ 3 }) ???\] Im confused. I know there will be 3 answers to this because its asking for cube roots.

  12. IrishBoy123
    • one year ago
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    sorry to confuse you, cis is just a shorthand for \(cos + i sin\) so stuff that argument into both of them

  13. anonymous
    • one year ago
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    What does the p stand for in that above formula?

  14. anonymous
    • one year ago
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    I believe he meant pi times m, it just put pim instead of writing the letter for pi

  15. IrishBoy123
    • one year ago
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    so you will have \(3(cos \frac{\theta}{3} + i sin \frac{\theta }{3})\) \(3(cos \frac{\theta + 2 \pi}{3} + i sin \frac{\theta + 2 \pi}{3})\) \(3(cos \frac{\theta + 4 \pi}{3} + i sin \frac{\theta + 4 \pi}{3})\)

  16. anonymous
    • one year ago
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    \[\pi m\] not pim

  17. anonymous
    • one year ago
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    is what he meant

  18. IrishBoy123
    • one year ago
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    from deMoivre, i think you have already seen how to get the first root ie when you wrote : \((cosθ+isinθ)^{1/3}=cos(1/3)θ+isin(1/3)θ\) the other roots come up only because the sinusoid repeats itself with period \(2 \pi\) and you need to look at all solutions for a cube root, the roots will repeat themselves after you have the first 3. for, say, sixth root, you will get six distinct roots and then they will repeat.... and don;t forget the \(\sqrt[3]{27}\) ....

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