Residue calculus for essential singularities

[Drao92]

Homework Statement

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)

Homework Equations

So, according to what I've 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 I've 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?

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.

 

[Dick]

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