Complex Analysis Question

[PowerClean]

This one is pretty involved so mad props to whoever can help me figure it out. I've been thinking about this for more than an hour and it's bugging me.

Consider the function f(z) = 1/(sin(pi/z)).

It has singular points at z=0 and z=1/n (where n is an integer). However, my book says each singular point except z=0 is isolated. (An isolated singular point is one where every epsilon neighborhood around that point is analytic.) Anyways, the argument is that if we find any epsilon neighborhood around z=0, then we can always find a 1/n (where n is an integer) inside this neighborhood. As we know, 1/n is a singular point hence there doesn't exist a neighborhood around z=0 that is analytic. This I understand.

However, I don't get why this same argument couldn't be applied to the other singular points z=1/n. Namely, for any epsilon neighborhood around z=1/n, we can find some other z=1/m inside this neighborhood... can't we?

My OCD prevents me from moving on forward in the chapter until I've figured this out exactly. It's driving me nuts. My prof doesn't have office hours until Tuesday of next week and he's notorious for not answering emails. If any of yall can help me figure this out before then, it'd be great.

 

[AKG]

Uh, consider 1/3. Then the closest 1/n's around it are 1/4 and 1/2. Choose the neighbourhood about 1/3 with radius less than min(|1/4 - 1/3|, |1/2 - 1/3|), so we could pick 1/30, for example. We're dealing with (9/30, 11/30). Find a 1/n in there other than 1/3. You really really shouldn't have spent an hour on this...

Are you sure you understand why 0 is not isolated?

 

[matt grime]

Power clean, are you sure you've got the correct definition? Isolated means there is *an* open neighbourhood of the point on which it is analytic (except for the singularity), not for every neighbourhood. (That would mean that every analytic function with tow singularties would never have isolated singularities.)

It is not an isolated singulairty if *for all* nbds there is another singularity in that nbd. The negation of "for all" is "there exists"