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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°).
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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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2you know \(e^{i \theta} = cos \theta + i \ sin \theta\)??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I've seen it before in my lessons but I have never actually applied it to an problems

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2it'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 \(\large \sqrt[3] { 27 \ e^{i \ \frac{11}{6}\pi}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That looks very confusing to solve Im uncertain as to how im supposed to take the cube root of that number

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2there's something related called deMoivre, which might be easier for you right now \((cos \theta + i \ sin \theta)^n = cos \ n \ \theta + i \ sin \ n \ \theta\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, I've heard of DeMoivre's Theorem

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2for you \(n = \frac{1}{3}\) and don't forget the \(\sqrt[3] 27\) !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So \[(\cos \theta+i \sin \theta)^{\frac{ 1 }{ 3 }}=\cos (\frac{ 1 }{ 3 })\theta+i \sin (\frac{ 1 }{ 3 })\theta ?\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2not that simple! you will get roots following this pattern \( 27^{1/3} \ cis(\frac{2π \ m+\theta }{3}) \) for m = 0, 1, 2

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2ie 3 roots for a cubic

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What is \[cis(\frac{ 2pim+\theta }{ 3 }) ???\] Im confused. I know there will be 3 answers to this because its asking for cube roots.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2sorry to confuse you, cis is just a shorthand for \(cos + i sin\) so stuff that argument into both of them

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What does the p stand for in that above formula?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I believe he meant pi times m, it just put pim instead of writing the letter for pi

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2so you will have \(3(cos \frac{\theta}{3} + i sin \frac{\theta }{3})\) \(3(cos \frac{\theta + 2 \pi}{3} + i sin \frac{\theta + 2 \pi}{3})\) \(3(cos \frac{\theta + 4 \pi}{3} + i sin \frac{\theta + 4 \pi}{3})\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2from 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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