Laurent Series Expansion of \frac{1}{e^z - 1}

[FeDeX_LaTeX]

Homework Statement
Determine the Laurent series expansion of

$$\frac{1}{e^z - 1}$$

The attempt at a solution

I've spotted that

$$\frac{1}{e^z - 1} = \frac{1}{2}\left( \coth{\frac{z}{2}} - 1\right)$$

but I don't know what to do next. WolframAlpha gives the series centred at 0 as:

$$\frac{1}{z} -\frac{1}{2} + \frac{z}{12} - \frac{z^3}{720} + \frac{z^5}{30240} + ...$$

but I don't know how they arrived at this. How are they evaluating f(0), f'(0), etc.? I'm getting an undefined answer for f(0) and f'(0) too.

I'm defining f(z) as

$$f(z) = \frac{1}{2}\left( \coth\frac{z}{2} - 1 \right)$$

Any help?

 

[Mandelbroth]

Did the question specify where to center the Laurent series?

At a point a, the Laurent series around a for your function is given by $\displaystyle \sum_{n=-\infty}^{\infty}\left[\oint_{A}\frac{dz}{2\pi i(e^z-1)(z-a)^{n+1}}(z-a)^n\right]$, where A is a contour centered at a.

 

[FeDeX_LaTeX]

Did the question specify where to center the Laurent series?

At a point a, the Laurent series around a for your function is given by $\displaystyle \sum_{n=-\infty}^{\infty}\left[\oint_{A}\frac{dz}{2\pi i(e^z-1)(z-a)^{n+1}}(z-a)^n\right]$, where A is a contour centered at a.

No, it didn't specify -- although the answer given gives the Laurent series centred at z = 0.

Where did you get that from?

I simplified that down to

$\displaystyle \frac{1}{2 \pi i}\sum_{n=-\infty}^{\infty}\left[\oint_{A}\frac{dz}{z(e^z-1)}\right]$

when a = 0. What do I do from here?

 

[jackmell]

I think you need to do long division and equate coefficients. We know it has a simple pole at the origin so we can write:

$$\frac{1}{z+z^2/2+z^3/3!+\cdots}=\frac{d_0}{z}+d_1+d_2z+\cdots$$

or:

$$-1+d_0+d_1z+d_2 z^2+\cdots+z d_0/2+\cdots+d_0/3! z^2+\cdots=0$$

and so forth. Now equate coefficients.

 

[Dick]

I think you need to do long division and equate coefficients. We know it has a simple pole at the origin so we can write:

$$\frac{1}{z+z^2/2+z^3/3!+\cdots}=\frac{d_0}{z}+d_1+d_2z+\cdots$$

or:

$$-1+d_0+d_1z+d_2 z^2+\cdots+z d_0/2+\cdots+d_0/3! z^2+\cdots=0$$

and so forth. Now equate coefficients.

Or factor the z out the denominator of your first expression getting it in the form $\frac{1}{z(1+f(z))}$ and use the taylor series expansion $\frac{1}{1+f(z)}=1-f(z)+f(z)^2-f(z)^3...$. Only keep the powers of z up to the highest power you need.

 

[vela]

How are they evaluating f(0), f'(0), etc.? I'm getting an undefined answer for f(0) and f'(0) too.

They're not evaluating f(0), f'(0), … because f(z) is singular at z=0. If the function didn't blow up at the origin, then the series for f(z) is just its Taylor series about $z=0$, and you find it the usual way. But as you saw, you're running into problems because f(z) blows up at z=0. In cases like these, you need to find the Laurent series.

So the fact that this problem asked you to find the Laurent series suggests you're looking for where the function blows up, which in this case is z=0, so you want to expand the function about z=0.

Personally, I think Dick's suggestion is the most straightforward way to find the series. I prefer to avoid long division when possible.

 

[vela]

I simplified that down to

$\displaystyle \frac{1}{2 \pi i}\sum_{n=-\infty}^{\infty}\left[\oint_{A}\frac{dz}{z(e^z-1)}\right]$

when a = 0. What do I do from here?

Mandelbroth meant
$$\sum_{n=-\infty}^{\infty}\left[\left(\oint_{A}\frac{dz}{2\pi i(e^z-1)(z-a)^{n+1}}\right)(z-a)^n\right].$$ You can't cancel the factors of (z-a).