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Math Help - Complex roots - Equation

  1. #1
    Junior Member utopiaNow's Avatar
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    Complex roots - Equation

    Show that the real part of any solution of (z + 100)^{100} = (z - 1)^{100} must be 0.

    Now in the book they solve it by starting: (z + 100)^{100} = (z - 1)^{100} \Rightarrow (z +1) = (z - 1)e^{\frac{2\pi ki}{100}}.

    I understand the steps that follow it, however I don't understand that step. I referred to the definition of m-th roots of complex numbers, which says: z^{1/m} = |z|^{1/m}e^{i(\theta + 2k\pi)/m}

    What am I missing? Thanks in advance.
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  2. #2
    Moo
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    Hello,

    Recall that e^{2ik\pi}=1

    We have :
    (z + 100)^{100} = (z - 1)^{100} e^{2ik\pi}

    Take the 100-th root on both side to get :
    (z+100)=(z-1) e^{\frac{2ik\pi}{100}}

    What I don't understand is why they wrote (z+{\color{red}1})=\dots
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  3. #3
    Junior Member utopiaNow's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    Recall that e^{2ik\pi}=1

    We have :
    (z + 100)^{100} = (z - 1)^{100} e^{2ik\pi}

    Take the 100-th root on both side to get :
    (z+100)=(z-1) e^{\frac{2ik\pi}{100}}

    What I don't understand is why they wrote (z+{\color{red}1})=\dots
    Hi,
    Yes, thanks for the tip, I see it now. That was a typo on my part it should say (z + 1)^{100}, where I accidentally typed (z + 100)^{100}, then I copy-pasted the typo.

    Thanks again.

    Now I'm having trouble understand a subsequent step, which I thought I understood.
    They get all the way to z = \frac{e^{2\pi k i/100} + 1}{e^{2\pi k i/100} - 1} = \frac{e^{\pi k i/100} + e^{-\pi k i/100}}{e^{\pi k i/100} - e^{\pi k i/100}} = -i\frac{\cos{(\pi k/100)}}{\sin{(\pi k/100)}}. To me the it seems the denominator is 0 in the middle step, that's where my problem is, as I don't see how they made that substitution either.
    Last edited by utopiaNow; September 24th 2009 at 03:04 PM.
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