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Math Help - Finding and classifying singularities

  1. #1
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    Finding and classifying singularities

    Hi. I have to find and classify the isolated singularities for:

    f(z)=\frac{e^z}{1+e^z}

    So, I've found that e^z=-1\rightarrow{z=i(2k+1)\pi},k\in{Z}

    Now, I think I should make the Laurent series for f(z), but I don't know how to handle this function to get it's Laurent series representation.

    I also know, of course, that: e^z=\sum_0^{\infty}\frac{z^n}{n!}
    Last edited by Ulysses; November 13th 2011 at 12:05 PM.
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  2. #2
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    Re: Finding and classifying singularities

    Quote Originally Posted by Ulysses View Post
    Hi. I have to find and classify the singularities for:

    f(z)=\frac{e^z}{1+e^z}

    So, I've found that e^z=-1\rightarrow{z=i(2k+1)\pi},k\in{Z}

    Now, I think I should make the Laurent series for f(z), but I don't know how to handle this function to get it's Laurent series representation.

    I also know, of course, that: e^z=\sum_0^{\infty}\frac{z^n}{n!}
    Do you know how to prove that a singularity is a pole without finding the Laurent series? (btw the poles are simple).
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    Re: Finding and classifying singularities

    I was looking at my notebook. Should I use a thing called "principle of isolated zeroes" or something like that?

    I was trying to determine the order of the pole

    EDIT: Alright, I did it this way:

    I took \displaystyle\frac{1}{f(z)}=\displaystyle\frac{1+e  ^z}{e^z}=g(z)\Rightarrow{g'(z)=1-\displaystyle\frac{1+e^z}{e^z}}\Rightarrow{g'(z_0)  }=1\neq{0}

    Then it's a pole of order 1.
    Last edited by Ulysses; November 13th 2011 at 01:10 PM.
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    Re: Finding and classifying singularities

    Quote Originally Posted by Ulysses View Post
    I was looking at my notebook. Should I use a thing called "principle of isolated zeroes" or something like that?

    I was trying to determine the order of the pole

    EDIT: Alright, I did it this way:

    I took \displaystyle\frac{1}{f(z)}=\displaystyle\frac{1+e  ^z}{e^z}=g(z)\Rightarrow{g'(z)=1-\displaystyle\frac{1+e^z}{e^z}}\Rightarrow{g'(z_0)  }=1\neq{0}

    Then it's a pole of order 1.
    \lim_{z \to i(2k+1)\pi} \frac{e^z(z - i(2k+1)\pi}{1 + e^z} = ....

    Therefore ....
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