Determining order of the poles in z/(e^z-1)

[Ratpigeon]

Homework Statement

Find the order of the poles of

f(z)=z/(e^z-1)

Homework Equations

The poles are at z=2 π i k k$\in$Z\0
(Because at z=0 f(z) has a removable singularity -set f(0)=1)

The Attempt at a Solution

I tried using the Taylor series of e^z - $\sum$z^n/n!
But I just got
f(z)=1/$\sum$z^n/(n+1)!
and I somehow need to take out a factor of (z-2 π i k)^j for some j>0 out of that...

 

[jackmell]

Homework Statement

Find the order of the poles of

f(z)=z/(e^z-1)

Homework Equations

The poles are at z=2 π i k k$\in$Z\0
(Because at z=0 f(z) has a removable singularity -set f(0)=1)

The Attempt at a Solution

I tried using the Taylor series of e^z - $\sum$z^n/n!
But I just got
f(z)=1/$\sum$z^n/(n+1)!
and I somehow need to take out a factor of (z-2 π i k)^j for some j>0 out of that...

Given f(z), what is the minumum power of (z-z_0) would I have to multiply the function by so that:

$$\lim_{z\to z_0} (z-z_0)^n f(z) \neq \infty$$

Take for example at zero:

$$\lim_{z\to 0} z^0 \left(\frac{z}{e^z-1}\right)\neq \infty$$

thus the order of the pole at zero is zero, i.e., it's removable. Ok, now you try the pole at $2n\pi i$.

 

[Ratpigeon]

Right - got them. I got my brain stuck on taylor series, when I needed l'hospital's rule. Thanks.
They're simple poles with limits z->z_0=z_0, right? :)