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samnatha

  • 2 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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  1. phi
    • 2 years ago
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    I think they want you to write 8 as \[ 8=8(\cos(2 \pi) + i \sin(2 \pi))\]

  2. samnatha
    • 2 years ago
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    well the theorm we have been using is diff cuz u have to add 2n(pi)

  3. phi
    • 2 years ago
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    yes, that is even better. 8=8(cos(2nπ)+isin(2nπ))

  4. samnatha
    • 2 years ago
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    like the way our teacher has been telling us is we have to get this \[\sqrt[3]{8?}\]

  5. samnatha
    • 2 years ago
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    then it is 2 to the power of a third

  6. phi
    • 2 years ago
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    See http://en.wikipedia.org/wiki/De_Moivre's_formula#Applications you solve \[ \left( z^3 \right)^{\frac{1}{3}} = 8^{\frac{1}{3}}(\cos(2n\pi)+i \sin(2n\pi)^{\frac{1}{3}}\]

  7. phi
    • 2 years ago
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    you get 3 different answers, for n=0,1,2. then they start repeating n=3 gives the same answer as n=0

  8. samnatha
    • 2 years ago
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    yeah i kinda get thanks

  9. samnatha
    • 2 years ago
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    wait usually ther is another angle you can't get 0 of 2n pi

  10. phi
    • 2 years ago
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    in 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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