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Singularity points + Laurent series

  1. Mar 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Find and determine the type of singularity points for ##f(z)=\frac{\sin(3z)-3z}{z^5}##. Also calculate the regular and main part of Laurent series around those points.


    2. Relevant equations



    3. The attempt at a solution

    I am already having troubles with the first part.

    Singularity point is of course ##z=0##. For ##z=0## I can use Taylor expansion for ##\sin## which than gives me

    ##f(z)=\frac{\sin(3z)-3z}{z^5}\doteq \frac{(3z-\frac{(3z)^3}{3!}+\frac{(3z)^5}{5!}-\cdots )-3z}{z^5}=\frac{z^3(-\frac{3^3}{3!}+\frac{(3^5z^2}{5!}-\cdots )}{z^5}=\frac{-\frac{3^3}{3!}+\frac{3^5z^2}{5!}-\cdots }{z^2}##

    Now I have no idea how to determine the type of singularity point. However, if I would have to guess I would say that in ##z=0## is pole of second order.

    If that is ok, I would be happy to get a hint on the second part of this problem - Laurent series.
     
  2. jcsd
  3. Mar 27, 2014 #2

    vela

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    That's right. It's a pole of order 2, and you've essentially derived the Laurent series. Just simplify what you have to
    $$f(z) = -\frac{3^3}{3!}\frac{1}{z^2} + \frac{3^5}{5!} - \frac{3^7}{7!} z^2 + \cdots.$$ Now you just have to identify the regular and principal parts of the series.
     
  4. Mar 27, 2014 #3
    ##f(z)=-\frac{3^3}{3!}\frac{1}{z^2}+\frac{3^5}{5!}-\frac{3^7}{7!}z^2+\frac{3^9}{9!}z^4-\cdots ##

    Where principal part is ##-\frac{3^3}{3!}\frac{1}{z^2}## and regular part is ##\frac{3^5}{5!}-\frac{3^7}{7!}z^2+\frac{3^9}{9!}z^4- \cdots ##

    And this should be it.
     
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