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Math Help - complex equation with exponential

  1. #1
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    complex equation with exponential

    How can i solve exp(\frac{1}{z})=z_0, where z_0 is any complex number?
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  2. #2
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    Re: complex equation with exponential

    z=\frac{1}{\log{z_0}}.

    The log is the complex log, which is multiple-valued: \log(re^{i\theta})=\log{r}+i(\theta+2n\pi), where n is any integer.

    And \frac{1}{z}=\frac{1}{x+iy}=\frac{x-iy}{x^2+y^2}.

    Is that what you were asking?

    - Hollywood
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  3. #3
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    Re: complex equation with exponential

    Quote Originally Posted by hollywood View Post
    z=\frac{1}{\log{z_0}}.

    The log is the complex log, which is multiple-valued: \log(re^{i\theta})=\log{r}+i(\theta+2n\pi), where n is any integer.

    And \frac{1}{z}=\frac{1}{x+iy}=\frac{x-iy}{x^2+y^2}.

    Is that what you were asking?

    - Hollywood
    my problem, in its completeness, is the following: i'm looking for singularities of a function whose denominator is e^\frac{1}{z}-3. So i look for solutions of e^\frac{1}{z}=3. I put e^\frac{1}{z}=e^{Log(3)} where Log is the principal branch of complex logarithm and by periodicity of complex exp i get \frac{1}{z}=log(3)+2n\pi i, where log in RHS is the real log. So z=(log(3)+2n\pi i)^{-1} and i was looking for an easy way to write the RHS in order to compute residues at those points
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  4. #4
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    Re: complex equation with exponential

    Since \log{3} and 2n\pi are both real, you can do what I suggested in my previous post:

    \frac{1}{\log{3}+2n\pi{i}}=\frac{\log{3}-2n\pi{i}}{(\log{3})^2+4n^2\pi^2}

    Other than that, I don't really see how to simplify it.

    - Hollywood
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