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Math Help - solutions to relative simple expression

  1. #1
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    solutions to relative simple expression

    I could need a hint on how to find all solutions to this equation

    \left( \frac{1+z^{-1}}{1+z} \right)^N = C

    where C is a real constant, N is a positive integer, and and z is the complex variable. I'm not sure I can just write

    \frac{1+z^{-1}}{1+z} \right = C^{1/N}. It should probably look more like \frac{1+z^{-1}}{1+z} \right = C^{1/N}e^{2\pi i k/N}, for k=0,1,...N-1.

    But then I'm unsure as how to proceed. I have tried writing z in polar form but it didn't give me a break through.
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    Re: solutions to relative simple expression

    Quote Originally Posted by niaren View Post
    I could need a hint on how to find all solutions to this equation

    \left( \frac{1+z^{-1}}{1+z} \right)^N = C

    where C is a real constant, N is a positive integer, and and z is the complex variable. I'm not sure I can just write

    \frac{1+z^{-1}}{1+z} \right = C^{1/N}. It should probably look more like \frac{1+z^{-1}}{1+z} \right = C^{1/N}e^{2\pi i k/N}, for k=0,1,...N-1.

    But then I'm unsure as how to proceed. I have tried writing z in polar form but it didn't give me a break through.
    \displaystyle \begin{align*} \left(\frac{1 + z^{-1}}{1 + z}\right)^N &= C \\ \left(\frac{1 + \frac{1}{z}}{1 + z}\right)^N &=C \\ \left(\frac{\frac{z + 1}{z}}{1 + z}\right)^N &= C \\ \left(\frac{1 + z}{z(1 + z)}\right)^N &= C \\ \left(\frac{1}{z}\right)^N &= C \\ z^{-N} &= C \\ z &= C^{-\frac{1}{N}} \end{align*}
    Thanks from niaren
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    Re: solutions to relative simple expression

    Dammit Don't know how I could miss that. Thanks.
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    Re: solutions to relative simple expression

    Quote Originally Posted by niaren View Post
    Dammit Don't know how I could miss that. Thanks.
    I expect there are other solutions too though...
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    Re: solutions to relative simple expression

    Do you mean that there is another way to arrive at the N solutions or do you mean that there are more than N solutions?
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    Re: solutions to relative simple expression

    Quote Originally Posted by niaren View Post
    \frac{1+z^{-1}}{1+z} \right = C^{1/N}.
    Let p = 1/N
    [1 + z^(-1)] / (1 + z) = C^p

    (z + 1) / z = C^p + zC^p

    z^2 C^p + z(C^p - 1) - 1 = 0

    z = {1 + C^p +- SQRT[C^(2p) + 2C^p + 1]} / (2C^p)

    I'm sure you can see your way to wrap that up, getting same as Prove It
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    Re: solutions to relative simple expression

    Quote Originally Posted by Wilmer View Post
    Let p = 1/N

    z = {1 + C^p +- SQRT[C^(2p) + 2C^p + 1]} / (2C^p)
    I actually had that expression written down but gave up because it looked too complicated. I will look at it tomorrow.
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    Re: solutions to relative simple expression

    Quote Originally Posted by niaren View Post
    I actually had that expression written down but gave up because it looked too complicated. I will look at it tomorrow.
    Why?
    C^(2p) + 2C^p + 1 = (C^p + 1)^2
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    Re: solutions to relative simple expression

    \displaystyle{ \left(\frac{1}{z}\right)^{N}=C\angle 2k\pi}

    \frac{1}{Z}=C^{1/N}\angle (2k\pi/N),\qquad k=0,1,2,...(N-1).
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    Re: solutions to relative simple expression

    Do something similar with this one, Bob? :

    [(1 + 1/a) / (1 + a)]^(1/m) + [(1 + b) / (1 + 1/b)]^n = k

    Solve for b.

    b = [(ka^(1/m) - 1) / a^(1/m)]^(1/n) ... unless I goofed!
    Last edited by Wilmer; June 28th 2012 at 07:34 AM.
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