What is the residue of cot(z) at z=0?

[Dixanadu]

Homework Statement

So guys..the title says it! I need to find the residue of cot(z) at z=0.

Homework Equations

For this situation, since the pole order is 1

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

The Attempt at a Solution

So here's what I am 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 I am doing but what is it? even if i expand sin and cos I still end up with 0...

 

[B-80]

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

0 \times \frac{cos(0)}{sin(0)}=0

 

[Dixanadu]

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

 

[B-80]

textbook didnt really say that

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

 

[Dixanadu]

thanks for the hint tho bro!