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

bobmerhebi

[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 [tex]\infty[/tex] as z tends to singular point

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

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



2. Relevant equations



3. The attempt at a solution

eok20

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

bobmerhebi

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 ?