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Classifying Singularities and the Laurent Series

  1. Jun 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Classify the singularities of ##\frac{1}{z^2sinh(z)}## and describe the behaviour as z goes to infinity

    Find the Laurent series of the above and find the region of convergence

    2. Relevant equations

    N/A
    3. The attempt at a solution
    I thought these two were essentially the same thing, but I'm being given a fair few marks for each question, so I'm convinced I must be over-simplifying it

    For the first part, I used to Laurent expansion to get:

    ##\frac{1}{z^2sinh(z)} = \frac{1}{z^2(z+\frac{z^3}{3!}+\frac{z^5}{5!}+...)}##
    ##\frac{1}{z^2(z+\frac{z^3}{3!}+\frac{z^5}{5!}+...)}= \frac{1}{z^3(1+\frac{z^2}{3!}+\frac{z^4}{5!}+...)}##
    ##\frac{1}{z^3(1+\frac{z^2}{3!}+\frac{z^4}{5!}+...)}=\frac{1}{z^3}(1+\frac{z^2}{3!}+\frac{z^4}{5!}+...)^{-1}##
    ##\frac{1}{z^3}(1+\frac{z^2}{3!}+\frac{z^4}{5!}+...)^{-1}=\frac{1}{z^3}(1-\frac{z^2}{3!}-\frac{z^4}{5!}+(\frac{z^2}{3!}+\frac{z^4}{5!})^2+...)##
    ##= \frac{1}{z^3} -\frac{1}{6z}+\frac{3z}{40}+....##

    Hence, I got there to be an isolated singularity of order 3 at z=0, a simple pole of at z=0 and an essential singularity at ##z=\infinity##

    So the behaviour as z goes to infinity is that the function is undefined? (I showed this by redefining a new variable ##\frac{1}{z}##

    I just can't see what else to do, but for the number of marks it is worth, I don't feel I have done enough work

    Also, Im not sure how to find the region of convergence

    Many thanks

    EDIT: Just had a thought- For the first part I could have written ##\frac{1}{z(i)(iz-\frac{1}{zi})}## ?
     
    Last edited: Jun 10, 2016
  2. jcsd
  3. Jun 11, 2016 #2

    stevendaryl

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    Just a question for you: Is there only one value of [itex]z[/itex] where [itex]sinh(z) = 0[/itex]?
     
  4. Jun 11, 2016 #3

    Oh no of course! There are also poles every ##n \pi i## With order ##\frac{1}{n^2 \pi^2}## ?
     
  5. Jun 11, 2016 #4

    stevendaryl

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    Okay, that needs to be part of the answer to your homework problem. What is the expansion about [itex]z=n\pi\ i[/itex] for [itex]n \neq 0[/itex]?
     
  6. Jun 11, 2016 #5
    Can we not just evaluate it at these points? Since the rest of the function will be analytic?
     
  7. Jun 11, 2016 #6

    stevendaryl

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    You wrote down an expansion in powers of [itex]z[/itex]. You need an expansion in powers of [itex]z-n\pi i[/itex] for the other poles.

    If you have a pole at [itex]z=p[/itex], then the Laurent series is in powers of [itex](z-p)[/itex], not powers of [itex]z[/itex]
     
  8. Jun 11, 2016 #7

    Okay, so we can expand ##\frac {1}{z}= \frac{1}{n \pi i +z- n \pi i} = \frac{1}{n \pi i} (1+\frac{z-n \pi i}{n\pi i})^{-1}##
    ## = \frac{1}{n \pi i}(1-\frac{z-n \pi i}{n\pi i}+\frac{(z-n \pi i)^2}{(n\pi i)^2}+...)##

    How would we expand sinhx such that the brackets of ##z-n \pi i## remain in the denominator

    Many thanks
     
  9. Jun 11, 2016 #8

    stevendaryl

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    Well, since [itex]sinh[/itex] is periodic, the expansion in powers of [itex]z[/itex] is the same as the expansion in terms of [itex]z - n \pi i[/itex].
     
  10. Jun 11, 2016 #9
    So ##sinhz= (z-n\pi i)+ \frac{(z-n \pi i)^3}{3!} +.... ## ?
     
  11. Jun 11, 2016 #10

    stevendaryl

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    Yes. You can see that by letting [itex]q = z - n \pi i[/itex]. Then [itex]sinh(z) = sinh(q + n \pi i) = sinh(q) = q + \frac{q^3}{3!} + ...[/itex]
     
  12. Jun 11, 2016 #11
    Ah okay. Does this mean that strictly speaking this only has a taylor series about ##n \pi i## ? Also when calculating the residue (slightly off topic I'm afraid) about ##n \pi i## do you need to expand ##sinhz## or can we simply 'read off' the residue?
     
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