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Homework Help: Isolated singularity (complex analysis)

  1. Oct 10, 2007 #1
    1. The problem statement, all variables and given/known data

    1) [tex]\frac{e^{z}-1}{z}[/tex]

    Locate the isolated singularity of the function and tell what kind of singularity it is.

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

    z_0 = 0

    find the laurant series for the given function about the indicated point. Also, give the residue of the function at the point.

    2. Relevant equations

    3. The attempt at a solution

    1) I thought that it first was a removable singularity. Because of only z in the denominator. But I don't get why the value at 0 is 1 ? I attempted to use g(z) = z, calculate with H(z)/(z - z_0) But that didn't help me in the understanding of the question. So my question really is how do you know what the value is at the 0. And how do you calculate the singularity with more rigour!

    2) First thing I saw was [tex]\frac{1}{1 - z}[/tex] and that is [tex]e^{z}[/tex]
    But then I thought that maybe you take the taylor expansion of cos z and [tex]e^{z}[/tex] and multiply them. But from there it went downhill :confused:
    Last edited: Oct 10, 2007
  2. jcsd
  3. Oct 11, 2007 #2


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    Science Advisor

    What is the Taylor's series, about 0, for ez? Subtract 1 from the Taylor's series and then divide by z. What is the value of that when z= 0?

    I don't understand this at all! In what sense is [tex]\frac{1}{1 - z}[/tex] the same as ez?

    You're confusing me! There is no ez in problem 2!

    If you multiply both numerator and denominator by [itex]1+ cos(z)[/itex] you get
    [tex]\frac{1+ cos(z)}{sin^2(z)}[/itex]
    You might find that simpler to work with.
  4. Oct 11, 2007 #3
    Hallsofivy: first of all, thanks for the quick response! I am very grateful for this :smile:

    on problem #1; isn't the taylor series [tex]e^{z}[/tex] for [tex]\frac{z^{n}}{n!}[/tex]?

    then you get z + [tex]\frac{z^{2}}{2!}[/tex] + [tex]\frac{z^{3}}{3!}[/tex] + ... + [tex]\frac{z^{n}}{n!}[/tex] so if you divide by z you get 1 + z + [tex]\frac{z^{2}}{2!}[/tex] with all of the z:s being zero and a 1 left! thanks for that one, when you explained, it became rather straightforward.

    on problem #2: I thought that [tex]e^{z}[/tex] equals [tex]\frac{1}{1 - z}[/tex] because if you take the taylor series to the limit it becomes that. But it was apparantly wrong. And getting to know that is rather nice. And maybe I should just get the [tex]e^{z}[/tex] thinking out of this problem. I will come back to you with my solution on this problem tomorrow.
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