Residue of cot(z)

1. Nov 28, 2013

1. The problem statement, all variables and given/known data
So guys..the title says it! I need to find the residue of cot(z) at z=0.

2. Relevant equations
For this situation, since the pole order is 1

$Residue=\lim_{z \to z_{0}}(z-z_{0})f(z)$

3. The attempt at a solution
So here's what Im doing in steps:

First, the singularity is at z=0. So $z_{0}=0$.

Then I multiply both sides by $(z-z_{0})=z$...to get $(z-z_{0})f(z)=zcot(z)$

Now taking the limit of this is as z = 0 is $0 \times \frac{cos(0)}{sin(0)}=0$...but this is wrong, the residue is 1...

I know its something stupid that im doing but what is it? even if i expand sin and cos I still end up with 0...

2. Nov 29, 2013

B-80

Study this arguement carefully and see if that doesn't remind you of some elementary calculus:

3. Nov 29, 2013

Yea I guess you're supposed to use L'Hopital's rule to find the behaviour of the function towards a limit....textbook didnt really say that

4. Nov 29, 2013

B-80

For most problems you encounter, there won't be a textbook to tell you anything at all. Take a minute to understand the trick.

5. Nov 29, 2013