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Math Help - Complex Analysis singularities question

  1. #1
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    Complex Analysis singularities question

    question:
    ------------------------------------------

    locate each of the isolated singularities of the given function and tell whether it is removable singularity, a pole, or an essential singularity. If the singularity is removable, give the value of the function at the point; if the singularity is a pole, give the order of the pole.

    (e^z-1)/(e^2z-1)

    ------------------------------------------

    What I first tried to do was factoring the bottom.

    (e^z-1)/((e^z+1)(e^z-1))

    but realized

    e^z-1 =0 for z = i(2(pi)n) n = 0, 1 , 2 ...
    e^z+1=0 for z= i((pi)+2(pi)n) n = 0, 1, 2...

    so basically, denominator = 0 whenever z = i(pi)n , n = 0, 1, 2....

    I tried to make this into An(z-z0)^n , but was unsuccessful.

    But can't really tell if this really is a removable singularity at z = i(pi)n, n= 0, 1, 2.... (I think |f(z)| remains bounded as z --> any of those points, so it is removable?)

    I thought about essential singularities, but I don't think that's the case...

    If anyone can solve this problem/ explain, that would be of great help. Thanks.
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  2. #2
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    Quote Originally Posted by hohoho00 View Post
    question:
    ------------------------------------------

    locate each of the isolated singularities of the given function and tell whether it is removable singularity, a pole, or an essential singularity. If the singularity is removable, give the value of the function at the point; if the singularity is a pole, give the order of the pole.

    (e^z-1)/(e^2z-1)

    ------------------------------------------

    What I first tried to do was factoring the bottom.

    (e^z-1)/((e^z+1)(e^z-1))

    but realized

    e^z-1 =0 for z = i(2(pi)n) n = 0, 1 , 2 ...
    e^z+1=0 for z= i((pi)+2(pi)n) n = 0, 1, 2...

    so basically, denominator = 0 whenever z = i(pi)n , n = 0, 1, 2....

    I tried to make this into An(z-z0)^n , but was unsuccessful.

    But can't really tell if this really is a removable singularity at z = i(pi)n, n= 0, 1, 2.... (I think |f(z)| remains bounded as z --> any of those points, so it is removable?)

    I thought about essential singularities, but I don't think that's the case...

    If anyone can solve this problem/ explain, that would be of great help. Thanks.
    Removable singularities at z such that e^z - 1 = 0 (since the limit exists and is finite).

    Essential singularities at z such that e^z + 1 = 0 (since the Laurent series doesn't terminate).
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  3. #3
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    Thank you ^^

    I was wondering though, after removing (e^z -1), we get the equation:

    1/ (e^z+1)

    but is it possible to express this in laurent series? Or can we still say it is a laurent series despite all the series is at the denominator?
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  4. #4
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    Quote Originally Posted by hohoho00 View Post
    Thank you ^^

    I was wondering though, after removing (e^z -1), we get the equation:

    1/ (e^z+1)

    but is it possible to express this in laurent series? Or can we still say it is a laurent series despite all the series is at the denominator?
    Yes, this function does have a Laurent series in the neighbourhood of any singularity. In fact, the singularities are poles of order 1, because the denominator has zeros of order 1.

    To see that there is a simple pole at z-i\pi, for example, you can use l'H˘pital's rule to check that \lim_{z\to i\pi}\frac{z-i\pi}{e^z+1} = \frac1{e^{i\pi}}=-1.
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  5. #5
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    Quote Originally Posted by Opalg View Post
    Yes, this function does have a Laurent series in the neighbourhood of any singularity. In fact, the singularities are poles of order 1, because the denominator has zeros of order 1.

    To see that there is a simple pole at z-i\pi, for example, you can use l'H˘pital's rule to check that \lim_{z\to i\pi}\frac{z-i\pi}{e^z+1} = \frac1{e^{i\pi}}=-1.
    My mistake. Thanks for the catch.
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  6. #6
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    Hmm.... does that mean it just has a simple pole at all i(pi)n? and there's no essential singularities?
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  7. #7
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    Quote Originally Posted by hohoho00 View Post
    Hmm.... does that mean it just has a simple pole at all i(pi)n? and there's no essential singularities?
    My chance at redemption.

    e^z + 1 = 0 \Rightarrow z = (\pi + 2n\pi)i = \pi (2n+1) i where n is an integer.

    Now apply the same argument that Opalg has already given to show that these are simple poles.
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