# Residue calculus for essential singularities

1. Apr 7, 2012

### Drao92

1. The problem statement, all variables and given/known data
Im not sure if i understood correctly how to calculate the residue for functions with essential singularities like:
f(z)=sin(1/z)
h(z)=z*sin(1/z)
j(z)=sin(1/z^2)
k(z)=z*(1/z^2)

2. Relevant equations
So, according to what ive read, when we have a functions with an essential singularity, we expand it in laurent serie and the residue will be the coefficient a-1.
Im a bit confused, if this is for all kind of functions, because in every book ive read, the exemle is always for 1/z(f(z)=e^(1/z) or sin(1/z) etc) and my question is if its 1/z^2 or 1/z^n the residue is also the a-1 coefficient???

3. The attempt at a solution
For all given functions the essential singularity is 0.
f(z)=1/z-1/(z^3*3!)+1/(z^5*5!)+......
the a-1 coefficient is 1 so the residue is 1.
h(z)=z*f(x)=1-1/(z^2*3!)+1/(z^4*5!)+...
In this case the residue is 0.
j(z)=1/(z^2)-1/(z^6*3!)+1/(z^10*5!)+..
For j(z) the residue is 0.
k(z)=z*j(z)=1/(z)-1/(z^5*3!)+1/(z^9*5!)+...
for k(z) the residue is 1.

2. Apr 7, 2012

### Dick

Sure, it's just the coefficient of 1/z. I think you've got it. The answers look right.