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Math Help - Complex numbers question (Addition)

  1. #1
    Senior Member I-Think's Avatar
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    Complex numbers question (Addition)

    (\frac{1+i}{1-i})^{16}+(\frac{1-i}{1+i})^8

    Progress
    (\frac{1+i}{1-i})^{16}+(\frac{1-i}{1+i})^8=\frac{(1+i)^{16}(1+i)^8+(1-i)^{16}(1-i)^8}{(1-i)^{16}(1+i)^8}

    \frac{(1+i)^{24}+(1-i)^{24}}{(1-i)^{16}(1+i)^8}

    \frac{(1+i)^{24}+(1-i)^{24}}{2^8(1-i)^{8}}

    Any hints on continuing, other than ugly expanding.
    Thanks in advance
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by I-Think View Post
    (\frac{1+i}{1-i})^{16}+(\frac{1-i}{1+i})^8

    Progress
    (\frac{1+i}{1-i})^{16}+(\frac{1-i}{1+i})^8=\frac{(1+i)^{16}(1+i)^8+(1-i)^{16}(1-i)^8}{(1-i)^{16}(1+i)^8}

    \frac{(1+i)^{24}+(1-i)^{24}}{(1-i)^{16}(1+i)^8}

    \frac{(1+i)^{24}+(1-i)^{24}}{2^8(1-i)^{8}}

    Any hints on continuing, other than ugly expanding.
    Thanks in advance


    Hints:

    (1+i)^2=?
    (1-i)^2=?
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  3. #3
    MHF Contributor
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    Exponentiation of Complex Numbers is always easiest in polar form...

    Note that 1 + i = e^{\frac{\pi i}{4}} and 1 - i = e^{-\frac{\pi i}{4}}.


    So \left(\frac{1 + i}{1 - i}\right)^{16} + \left(\frac{1 - i}{1 + i}\right)^8 = \frac{(1 + i)^{16}}{(1 - i)^{16}} + \frac{(1 - i)^8}{(1 + i)^8}

     = \frac{\left(e^{\frac{\pi i}{4}}\right)^{16}}{\left(e^{-\frac{\pi i}{4}}\right)^{16}} + \frac{\left(e^{-\frac{\pi i}{4}}\right)^{8}}{\left(e^{\frac{\pi i}{4}}\right)^{8}}

     = \frac{e^{4\pi i}}{e^{-4\pi i}} + \frac{e^{-2\pi i}}{e^{2\pi i}}

     = e^{8\pi i} + e^{-4\pi i}

     = \left(e^{\pi i}\right)^8 + \left(e^{\pi i}\right)^{-4}

     = (-1)^8 + (-1)^{-4}

     = 1 + 1

     = 2.
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