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Math Help - Complex Analysis: Simplifying Polar Form

  1. #1
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    Complex Analysis: Simplifying Polar Form

    I am having trouble simplifying the following equations which are the solutions to the equation (z+1)^5=z^5.
    \frac{1}{e^{i\frac{2}{5}\pi}-1}

    \frac{1}{e^{i\frac{4}{5}\pi}-1}

    \frac{1}{e^{i\frac{6}{5}\pi}-1}

    \frac{1}{e^{i\frac{8}{5}\pi}-1}

    Any help you can give me will be greatly appreciated!
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  2. #2
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    Re: Complex Analysis: Simplifying Polar Form

    You could expand the fractions with \exp\left[-i \frac{1}{5} \pi\right], \exp\left[-i \frac{2}{5} \pi\right] and so on.
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  3. #3
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    Re: Complex Analysis: Simplifying Polar Form

    I am afraid I really don't understand how this would help.
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  4. #4
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    Re: Complex Analysis: Simplifying Polar Form

    After expanding, the demoninator will be a 2 i Sin[~], which is a multiple of Pi. The numerator is Exp[i~], which could you rewrite in a+ ib if you want to do.

    Just try to expand the first solution with \exp\left[-i \frac{1}{5} \pi\right] and you will see it works.
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  5. #5
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    Re: Complex Analysis: Simplifying Polar Form

    Sorry... I really appreciate your help, but I am just not seeing it. I still have both these items in the denominator and am back to 1 in the numerator.
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    Re: Complex Analysis: Simplifying Polar Form

    I think it might be as simple as it gets for my purpose! Thank you.
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  7. #7
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    Re: Complex Analysis: Simplifying Polar Form

    Quote Originally Posted by tarheelborn View Post
    I am having trouble simplifying the following equations which are the solutions to the equation (z+1)^5=z^5.
    \frac{1}{e^{i\frac{2}{5}\pi}-1}
    What is the original statement of the problem?
    Where do you get those answers?

    It might help to know that
    \frac{1}{\exp\left(i\frac{2}{5}\pi}\right)-1}=\frac{\exp\left(-i\frac{2}{5}\pi\right)-1}{|\exp\left(i\frac{2}{5}\pi}\right)-1|^2}
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  8. #8
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    Re: Complex Analysis: Simplifying Polar Form

    The original problem statement was to find all the solutions of (z+1)^5=z^5. I simplified this equation to equal (1+\frac{1}{z})^5=1 and then wrote \frac{1}{z} in polar form as (\frac{1}{r}e^{-i\theta})^5.

    I then simplified to the equation e^{i(2/5)n\pi} for n=0,\cdots,4.

    Obviously, n=0 isn't a solution, so I came up with the other 4 using that format, but I wasn't able to simplify them. But then I wondered if they need to be simplified. Thanks for your help.
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  9. #9
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    Re: Complex Analysis: Simplifying Polar Form

    Okay, I will explain again

    \frac{1}{e^{i\frac{2}{5}\pi}-1} = \frac{1}{e^{i\frac{2}{5}\pi}-1} \cdot \frac{e^{-i\frac{1}{5}\pi}}{e^{-i\frac{1}{5}\pi}}= \frac{e^{-i\frac{1}{5}\pi}}{e^{i\frac{1}{5}\pi}- e^{-i\frac{1}{5}\pi}} = \frac{e^{-i\frac{1}{5}\pi}}{2 i \sin\left(\frac{1}{5}\pi\right)} = \frac{ e^{-i\frac{1}{10}\pi}}{2 \sin\left(\frac{1}{5}\pi\right)}
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  10. #10
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    Re: Complex Analysis: Simplifying Polar Form

    Oh, sorry, yes I see that. Thank you.
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