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F(z) = [1-cosh(z)] / z^3. its pole, order & residue

  1. May 31, 2009 #1
    1. Show that f(z) = [1 - cosh(z)] / z3 has a pole as its singular point. Determine its order m & find the residue B.


    2.

    lim |f(z)| tends to [tex]\infty[/tex] as z tends to singular point

    bm + bm+1(z-z0)+...+ b1(z-z0)m-1 + [tex]\sum[/tex][tex]^{\infty}_{n = 0}[/tex]an(z-z0)n= (z-z0) m f(z)


    3. f(z) = [1 - cosh(z)] / z3 = 1/z3 - cos(iz)/z3

    that's all i could do. I don't know how to do the limit of the cos part. does it approach infinity ? I think NOt right ?

    plz help asap.

    thx in advnace
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 31, 2009 #2
    Expand out cosh(z) = cos(iz) using its Taylor series.
     
  4. May 31, 2009 #3
    Doing this i get:

    f(z) = 1 - (1 + z2/2 + z4 / 4! + z6 / 6! + ....)

    = - [tex]\sum^{\infty}_{n = 1}[/tex]z2n/(2n)! tends to 0 as z tends to 0

    but this is not the limit we have for a pole. but its for a removable singular point. isn't it ?
     
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