# Laurent Expansion of sin(1-1/z)

1. Mar 27, 2008

### nicksauce

1. The problem statement, all variables and given/known data
Find the Laurent expansion of $$f(z) = \sin(1-\frac{1}{z})$$ about z = 0, and state the annulus of convergence.

2. Relevant equations

3. The attempt at a solution
I tried doing the regular expansion of sin(z), then applying the binomial expansion on the (1-1/z)^n terms, but I can't help but feel that there's a better way to approach the problem. Any thoughts?

2. Mar 27, 2008

### HallsofIvy

Staff Emeritus
I would suggest that you use the sine sum formula to write it as
$$sin(1-\frac{1}{z})= sin(1)cos(\frac{1}{z})- cos(1)sin(\frac{1}{z})$$
and then use the Taylor's series for sine and cosin.

3. Mar 27, 2008

### nicksauce

Boy don't I feel dumb. Thanks!

4. Nov 18, 2011

### sarazamani

is this sentence correct?
sin(1/z)=1/z-1/z^3*3!+1/z^5*5!+-......
for laurent series?