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Complex Numbers problem

  1. Apr 30, 2014 #1
    If a is a complex number, and a^2-a+1=0, then a^2011=?

    I tried using De Moivre's theorem, Taking a=cosθ+isinθ, but didn't get anywhere, got stuck at
    cos2θ+isin2θ-cosθ-isinθ+1=0. What do I do?
     
  2. jcsd
  3. Apr 30, 2014 #2
    First you should solve the equation for 'a'. After that, you can calculate the argumentum and the abs. value of 'a', resulting in a = r*(cos(theta) + i* sin(theta)). Finally, apply De Moivre for calculating a^2011.
     
  4. Apr 30, 2014 #3

    AlephZero

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    a=cosθ+isinθ assumes that |a| = 1. That is wrong unless you already know that |a| = 1.

    You can just solve the quadratic equation for a (using the standard formula) to get the real and imaginary parts.
     
  5. Apr 30, 2014 #4

    SammyS

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    Do you know the factorization of ##\ a^3+1\ ?##
     
  6. Apr 30, 2014 #5

    Dick

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    It might be difficult to guess that clever trick. Here's another way. You know a^2=a-1. Can you find a simple expression for a^3?
     
  7. May 1, 2014 #6
    Dude, first find the complex number "a" by using quadratic formula. Then you convert it to "cis" notation, and then you finally apply De Moivre's theorem.
     
  8. May 3, 2014 #7

    lurflurf

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    ##a^{2011}=a^{6\cdot 335+1}=(a^6)^{335}a##
    What is ##a^6##?
     
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