Determine the singularity type of the given function (Theo. Phys)

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SUMMARY

The discussion centers on determining the singularity type of the function f(z) = cos(z + 1/z). The user identifies a singular point at z = 0, noting that as z approaches 0, the limit of cos(z + 1/z) diverges due to the oscillatory nature of the cosine function between -1 and 1. The analysis reveals that the term 1/z introduces a pole of order 1 at z = 0, leading to the conclusion that cos(z + 1/z) also possesses a pole of order 1 at this point.

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Homework Statement
Given the function
f(z) = cos(z+1/z) determine the function's singular point(s), and whether it is a pole, essential singularity, or removable singularity. Determine the residue if it is an essential singularity or a pole.
Relevant Equations
A pole if:
lim f(z) = infinity
z->a
Order of a pole:
Order P that satisfies
lim (z-a)^P * f(z) = finite non-zero number
z->a

Possibly use Taylor expansion or L'Hopital's rule depending on the type of limits obtained
NOTE: Was not sure where to post this as it is a math question, but a part of my "Theoretical Physics" course.

I have no idea where to start this and am probably doing this mathematically incorrect.

given the function f(z) = cos(z+1/z) there should exist a singular point at z=0 as at z = 0; lim as z-> 0 cos(z+1/z) diverges since

lim as z->0 (z+1/z) -> infinity and so cos (z+1/z) oscillates between -1 and 1 (?).

I'm not sure how to do this and I think what I am doing here is definitely wrong, so any help/explanation would be appreciated.
Since the function f(z) depends mainly on z+1/z, that is the part I will deal with:

lim z-> 0 (z + 1/z) = lim z->0 (z) + lim z -> 0 (1/z)

= 0 + lim z-> 0 (1/z)
Where 1/z has a pole at z= 0 of order p =1
so cos(z+1/z) should also have a pole at z-0 of order P =1 ??

I'm really confused, and completely lost on how to approach this/why.
 
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Hi.
It is enough to invetigate
z\ cos(1/z)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!z^{2n-1}}
 

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