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Complex Analysis - Removing A Singularity

  1. Apr 1, 2009 #1
    Ok, so I'm suppose to be able to remove the singularity to find the residue of the function

    [tex](z)cos{\frac{1}{z}[/tex]

    I tried to see how "bad" the singularity was by taking the limit, but I can't figure out if

    [tex]
    \lim_{ z \to 0 } (z)cos{\frac{1}{z}
    [/tex]

    goes to 0 or if it is not bounded. If it goes to zero I should be able to remove the singularity. Since I couldn't figure that out I used the Taylor Series of cos(z) to expand the function and I got a Laurent Series.

    [tex](z)cos{\frac{1}{z} = \sum_{n=0}^\infty \frac{(-1)^{(n)}}{(2n)!}z^{(-2n+1)} [/tex]

    This is where I'm stuck. I don't see how I can possibly remove the singularity or if its even removable. Any ideas?
     
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  3. Apr 1, 2009 #2

    Ben Niehoff

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    If you have the Laurent series, then you're done. The residue is the coefficient of the z^-1 term in the series.

    Do you see why?
     
  4. Apr 1, 2009 #3
    Yes, I know that when I compute the residue I'm looking at [tex]a_{-n}[/tex] in the expansion but there are a couple problems. First, I can't figure out the order of the pole. I''m suppose to take the derivative of the function and if i plug in [tex]z_{0}[/tex] and get a singularity then I do it again...until I finally get the order. But because I have [tex]\frac{1}{z}[/tex] inside the cosine won't my order be infinite?

    According to my professor there's no way to remove the singularity but I've seen similar versions of this problem and it always says to compute the integral. If I'm computing the integral then I'll have

    [tex]
    \int_{C} (z)cos{\frac{1}{z} = 2\pi iRes(f(z), 0)
    [/tex]

    So, to compute the Residue I need to figure out what coefficient I'm looking for. But the coefficient depends on the order.
     
  5. Apr 1, 2009 #4

    Ben Niehoff

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    No, the residue is always the [itex]a_{-1}[/itex] term, even if the order of the pole is infinite.

    The whole thing about finding the order of the pole is just a shortcut for when you don't want to compute the Laurent series. If the integral is nonzero, then the Laurent series must have a non-zero [itex]a_{-1}[/itex] term. The residue formula is just a way of obtaining it if you can't get it any other way.

    Write out an arbitrary Laurent series and try integrating it around a closed contour - you'll see why only the [itex]a_{-1}[/itex] term is important.
     
  6. Apr 2, 2009 #5
    I see how the [itex]a_{-1}[/itex] term is the only important term. Like, in this case how would I compute the [itex]a_{-1}[/itex]? I had the impression that if the order of a pole is infinite then there is no way to remove the singularity unless the function has a limit that can be defined. I'm not sure if this is correct. And do you have any examples of functions where we can't remove the singularity and why it can't be removed?
     
  7. Apr 2, 2009 #6

    Ben Niehoff

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    The [itex]a_{-1}[/itex] term is already contained in the series you wrote down:

    [tex](z)cos{\frac{1}{z} = \sum_{n=0}^\infty \frac{(-1)^{(n)}}{(2n)!}z^{(-2n+1)} [/tex]

    Look carefully.
     
  8. Apr 2, 2009 #7
    I know the [itex]a_{-1}[/itex] term is inside the Laurent Series. So your saying because plugging in n=1 gives me the
    [itex]
    a_{-1}
    [/itex] term I just use that to compute the residue? So, I'd have [itex]a_{-1} = -1/2[/itex] and [tex]\int_{C} (z)cos{\frac{1}{z} = 2\pi iRes(f(z), 0)=-\pi i[/tex]?
     
  9. Apr 3, 2009 #8
    This is just to help clarify some things. You're given the function [itex]f(z) = z\cos \frac{1}{z}[/itex]. You say you are suppose to be able to remove the singularity at [itex]z=0[/itex] to determine the residue. This is incorrect language. We know [itex]z=0[/itex] is an isolated singularity. There are three types of singularities it can be: (1) removable singularity, (2) pole, or (3) essential singularity. You can only "remove" a removable singularity.

    Theorem: If [itex]f(z)[/itex] has an isolated singularity at [itex]z=0[/itex], then given [itex]f(z)[/itex]'s Laurent expansion in an annulus centered at [itex]z=0[/itex], [tex]f(z) = \sum_{n=-\infty}^\infty a_n z^n[/tex] the function has:

    (1) a removable singularity iff there are no terms with negative powers of [itex]z[/itex] in the Laurent expansion,
    (2) a pole iff there are a finite amount of terms with negative powers of [itex]z[/itex], and
    (3) an essential singularity iff there is an infinite amount of terms with negative powers of [itex]z[/itex].

    You have the Laurent expansion. So, just expand it out, writing out at least the first 4 terms and then you will be able to determine what type of singularity [itex]z=0[/itex] is by just looking at the singular part of the Laurent expansion, i.e. the part with negative powers of [itex]z[/itex].

    Also, as mentioned, the residue of [itex]f(z)[/itex] at [itex]z=0[/itex] is defined as the coefficient [itex]a_{-1}[/itex] in the Laurent expansion in an annulus centered at [itex]0[/itex], i.e. the coefficient of the [itex]\frac{1}{z}[/itex] term. This is true for any type of isolated singularity. You found the coefficent correctly, [itex]a_{-1}=-1/2[/itex], and yes, this means [itex]\text{Res}(f;0) = -1/2[/itex]. You computed the integral correctly, as long as [itex]C[/itex] is a closed, simple curve (i.e. doesn't self-intersect). If it isn't simple, then you have to multiply by the index of [itex]C[/itex] with respect to [itex]z=0[/itex].
     
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