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Simple Complex Number Review Question

  1. Sep 2, 2013 #1
    1. The problem statement, all variables and given/known data
    z=1 + e^(iθ) calculate z^2 and lzl^2



    2. Relevant equations



    3. The attempt at a solution

    for z^2

    (1+e^(iθ))(1+e^(iθ)) = 1 + 2e^(iθ) + e^(i2θ).. is that the final answer? i expanded it into cosines and sines as well but that doesn't simplify anymore i don't believe.

    for lzl^2 = (1+e^(iθ))*(1+e^(-iθ))= 1 + e^(iθ) + e^(iθ) + e^0 = 2 + 2e^(iθ) = 2(1+e^(iθ)) = 2z

    but did I do something wrong because the book says the final answer should not have an imaginary number in it? but z has e^(iθ) in it? thanks for the help
     
  2. jcsd
  3. Sep 2, 2013 #2

    Dick

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    It doesn't have just an e^(iθ) in it. It should also have an e^(-iθ). Check your expansion. Put them together and the result is not imaginary.
     
  4. Sep 2, 2013 #3
    Do you mind showing me the e^(-i@) + e^(-i@) cancellation?
     
  5. Sep 2, 2013 #4
    Do u mind showing me the cancellation written on?
     
  6. Sep 2, 2013 #5

    Dick

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    Express the sum e^(iθ)+e^(-iθ) (which is what you should have had in your expansion) in terms of sines and cosines.
     
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