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anonymous
 4 years ago
please help me i have some idea of what to do but i need help !
use de moivre's theorem to solve each of these questions. plot your solutions on the complex plain.
(i) z^3 = 8
anonymous
 4 years ago
please help me i have some idea of what to do but i need help ! use de moivre's theorem to solve each of these questions. plot your solutions on the complex plain. (i) z^3 = 8

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phi
 4 years ago
Best ResponseYou've already chosen the best response.0I think they want you to write 8 as \[ 8=8(\cos(2 \pi) + i \sin(2 \pi))\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well the theorm we have been using is diff cuz u have to add 2n(pi)

phi
 4 years ago
Best ResponseYou've already chosen the best response.0yes, that is even better. 8=8(cos(2nπ)+isin(2nπ))

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0like the way our teacher has been telling us is we have to get this \[\sqrt[3]{8?}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then it is 2 to the power of a third

phi
 4 years ago
Best ResponseYou've already chosen the best response.0See http://en.wikipedia.org/wiki/De_Moivre's_formula#Applications you solve \[ \left( z^3 \right)^{\frac{1}{3}} = 8^{\frac{1}{3}}(\cos(2n\pi)+i \sin(2n\pi)^{\frac{1}{3}}\]

phi
 4 years ago
Best ResponseYou've already chosen the best response.0you get 3 different answers, for n=0,1,2. then they start repeating n=3 gives the same answer as n=0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah i kinda get thanks

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait usually ther is another angle you can't get 0 of 2n pi

phi
 4 years ago
Best ResponseYou've already chosen the best response.0in case it is not clear 8 = 8( cos(0) + i sin(0)) when we add in multiples of 2π we get 8 = 8( cos(2nπ ) + i sin(2nπ))
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