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Thread: Complex numbers

  1. #1
    Bar0n janvdl's Avatar
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    Complex numbers

    Explain why z = (1 + i)^n + (1 - i)^n is real for all n \in N
    Thanks in advance for any help
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by janvdl View Post
    Thanks in advance for any help
    Note that (1+i)^2=2i and (1-i)^2=-2i

    Then let n=4k, 4k+1, 4k+2, 4k+3, and use properties of exponents. If you're stuck... well you know what to do
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  3. #3
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    you have a simple case of z + \overline{z}. remember \overline{z}^n = \overline{z^n}

    Bobak
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  4. #4
    Flow Master
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    Quote Originally Posted by bobak View Post
    you have a simple case of z + \overline{z}. remember \overline{z}^n = \overline{z^n}

    Bobak
    And this is especially clear if you consider the polar form of 1 + i and 1 - i and then raise each to the power of n using de Moivre's Theorem.
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  5. #5
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    Finally, you could do this using the binomial theorem, together with the fact that \left(\begin{array}{c}n \\ k\end{array}\right)= \left(\begin{array}{c}n \\ n-k \end{array}\right), and combining all odd powers of i.
    Last edited by HallsofIvy; Jan 11th 2009 at 10:44 AM. Reason: Edited thanks to Opalg
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    Quote Originally Posted by HallsofIvy View Post
    Finally, you could do this using the binomial theorem, together with the fact that \left(\begin{array}{c}n \\ i\end{array}\right)= \left(\begin{array}{c}n \\ n-i \end{array}\right), and combining all odd powers of i.
    ... but preferably don't use i as an index in a question about complex numbers.
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  7. #7
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    Quote Originally Posted by Opalg View Post
    ... but preferably don't use i as an index in a question about complex numbers.
    lol! Or j .....
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