Complex Analysis: Proving a function is equivalent to its series representation

[Grothard]

Homework Statement

Compare the function f(z) = (pi/sin(pi*z))^2 to the summation of g(z) = 1/(z-n)^2 for n ranging from negative infinity to infinity. Show that their difference is
1) pole-free, i.e. analytic
2) of period 1
3) bounded in the strip 0 < x < 1

Conclude that they are equivalent

The Attempt at a Solution

Part 2 is easy to show; I don't need any help with that one.

I'm working on part 1 right now. I noticed that in both equations there exist poles whenever z is an integer, and those are the only poles. This means that the two functions have exactly the same poles. I'm not sure how to use that to prove that their difference has no poles, though.

 

[Dick]

The poles have order two at integer n. You want to show the Laurent series expansion of f(x) around x=n takes the form 1/(z-n)^2+analytic stuff. So the poles cancel.

 

[Grothard]

Alright, that makes sense. How would I go about expanding 1/sin(z)^2 with the Laurent series? Do I make it 1/(1-cos(z)^2)? I'm having a hard time putting it into the form 1/(z-n)^2

 

[Dick]

Alright, that makes sense. How would I go about expanding 1/sin(z)^2 with the Laurent series? Do I make it 1/(1-cos(z)^2)? I'm having a hard time putting it into the form 1/(z-n)^2

Just worry about the pole at z=0. You can show the others are the same. Use the taylor series for sin(z).