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

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

[b]2.

lim |f(z)| tends to $$\infty$$ as z tends to singular point

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

[b]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.

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

2. Relevant equations

3. The attempt at a solution

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 Expand out cosh(z) = cos(iz) using its Taylor series.

 Quote by eok20 Expand out cosh(z) = cos(iz) using its Taylor series.
Doing this i get:

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

= - $$\sum^{\infty}_{n = 1}$$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 ?