# Laurent-series expansion of $1/(e^z-1)$

Find the Laurent series for the given function about the indicated point. Also, give the residue of the function at the point. $$z\mapsto\frac{1}{e^z - 1}$$ The point is $z_0=0$ (four terms of laurent series).

I have wrote $e^z -1$ as $z+z^2/2!+z^3/3!$....

Now i don't know how to proceed with this further.

Thank you.

• For 1/series, you can use long division... Commented Nov 5, 2014 at 1:18
– Pedro
Commented Nov 5, 2014 at 2:57

The other answer has its benefits, but if you want to check your answer or simply bash out the "important" coefficients and ignore all that insignicantly small stuff, then here goes. Just note that we need to be careful about convergence.

We have that $\frac{1}{e^z-1} = \frac{1}{z(1+(z/2!+z^2/3!+...))}$ where we let $(z/2!+z^2/3!+...) = P(z)$. Then:

$$\frac{1}{e^z-1} = 1/z - P(z)/z + P(z)^2/z - P(z)^3/z + ... \\ = 1/z - 1/2 - z/3! + z/(2!)^2 - z^2/4! + 2z^2/(2!\cdot 3!) - z^2/(2!)^3 + O(z^3) \\ = 1/z - 1/2 + z/12 + z^2\cdot (1/6-1/24-1/8) + O(z^3) \\ = 1/z - 1/2 + z/12 + O(z^3).$$

• I think it misses one important point: To use the series for $1 \over (1+z)$, you need to prove that $|z| < 1$, otherwise, the series does not converge. So here, you need to prove that $|P(z)| < 1$ before using the series, which I don't think it's possible in the domain $0 < |z| < 2 \pi$ Commented Apr 23, 2016 at 10:20
• @leducquang late reply, but I think this method is valid: for all small enough $z, |P(z)| < 1$ as $P(z) \rightarrow 0$ as $z \rightarrow 0$. Let's assume we are working in some annulus of sufficiently small radius s.t. this holds. Then we apply Chris Ks argument and generate terms of the Laurent series. But we know the Laurent series is unique on the whole of the annulus $0 < |z| < 2 \pi$ so it must extend. Commented Apr 16, 2018 at 15:40

Let $$f(z) = (e^z - 1)^{-1}$$ which has a simple pole at $$z = 0$$ (easy enough to see). Consider $$h(z) = \frac{e^z - 1}{z} = \sum_{n = 0}^{\infty}\frac{z^n}{(n + 1)!}$$ $$h$$ is an entire function (prove it to yourself). Now let $$g(z) = \frac{1}{h(z)}$$ which is analytic over some area (I leave where as an exercise). Now $$f(z) = \frac{1}{zh(z)} = \frac{g(z)}{z} = \sum_{n = -\infty}^{\infty}a_nz^n$$ Furthermore, $$a_n$$ can be found \begin{align} a_n &= \frac{1}{2\pi i}\int_{|z| = R}\frac{f(z)}{z^{n+1}}dz\\ &= \frac{1}{2\pi i}\int_{|z| = R}\frac{g(z)}{z^{n+2}}dz \end{align} where $$0. $$g$$ is analytic on the inside of $$|z| = R$$. By Cauchy's Theorem, $$a_n = 0$$ for $$n\leq -2$$. $$f(z) = \sum_{k = -1}^{\infty}a_nz^n$$ Now compute the first few $$a_n$$. To find the derivative of $$g$$, we should first find the derivative of $$h$$. $$h^{(k)}(z) = \sum_{n=k}^{\infty}\frac{n!z^{n-k}}{(n-k)!(n+1)!}$$ Therefore, $$h^{(k)}(0) = \frac{1}{k+1}$$ for all $$k\geq 0$$ We can easily see that $$1 = g(z)h(z)$$ so $$0=g'h+gh'$$. In general, $$0 = (gh)^{(k)}(z) = \sum_{i = 0}^k\binom{k}{i}h^{k-i}(z)g^i(z)$$ At $$z = 0$$, $$h^{k-i}(0) = \frac{1}{k - i + 1}$$; therefore, $$0 = (gh)^{(k)}(0) = \sum_{i = 0}^k\binom{k + 1}{i}g^i(0)$$ Going back to the coefficient $$a_n$$, we have $$a_n = \frac{1}{2\pi i}\int_{|z| = R}\frac{g(z)}{z^{n+2}}dz = \frac{g^{(n+1)}(0)}{(n+1)!}$$ for all $$n\geq -1$$. $$0 = \sum_{j = 0}^k\frac{a_{j-1}}{(k-(j-1))!}$$ So $$a_{-1} = 1$$, $$a_0 = -1/2$$, all positive even terms are zero.... Let $$B_k := (-1)^{k-1}(2k)!a_{2k-1}$$ be Bernoulli numbers. Note that $$F(z) = \frac{1}{e^z - 1} -\frac{1}{z} + \frac{1}{2}$$ is an odd function. Therefore, $$f(z) = \frac{1}{z} -\frac{1}{2} +\sum_{k=1}^{\infty}a_{2k-1}z^{2k-1} = \frac{1}{z} -\frac{1}{2} +\sum_{k=1}^{\infty}(-1)^{k-1}\frac{B_k}{(2k)!}z^{2k-1}$$

• This is many years later but could you possibly expound on some of this? Particularly, I'm not sure what you mean by $h^{(k)} = 1/k+1$. Do you mean $h^{(k)}(0) = 1/k+1$? Also, how are you getting that zero equals the sum of something in terms of derivatives of $h$ and $g$? Commented Feb 16, 2018 at 0:47
• @inkievoyd I don't remember any more and I haven't done any math in awhile. Commented Feb 16, 2018 at 3:59

(This is a rewrite of dustin's answer for my own reference.)

Clearly $$f(z)=\frac{1}{e^z - 1}$$ has a pole at $$z=0$$. Since $$\lim_{z\to0}zf(z)=1$$, the pole is simple. Thus $$f(z)$$ has a Laurent series expansion $$\sum_{n\ge-1}a_nz^n$$ about zero with $$a_{-1}=1$$. Now, as both $$g(z)=\frac{z}{e^z - 1}=\sum_{n\ge0}a_{n-1}z^n$$ and $$\frac{1}{g(z)}=\frac{e^z - 1}{z}=\sum_{n\ge0}\frac{z^n}{(n+1)!}$$ are analytic at zero, we have $$\left(\sum_{n\ge0}a_{n-1}z^n\right)\left(\sum_{n\ge0}\frac{z^n}{(n+1)!}\right)=1.$$ By comparing coefficients on both sides, we see that $$a_0=-\frac12$$ and $$\{a_n\}_{n\ge-1}$$ is given by the recurrence relation $$a_{-1}=1$$ and $$\sum_{k=0}^n\frac{a_{k-1}}{(n-k+1)!}=0\quad(n\ge1).$$ In particular, $$a_0=-\frac12$$. Finally, in a deleted neighbourhood of zero, it is straightforward to verify that $$\sum_{n\ge1}a_nz^n=f(z)-\frac{a_{-1}}{z}-a_0 =\frac{1}{e^z-1}-\frac{1}{z}+\frac{1}{2}$$ is an odd function. Therefore, all positive even terms $$a_2,a_4,\ldots$$ are actually equal to zero and the previous recurrence relation can be rewritten as $$a_{-1}=1,\ a_0=-\frac12$$ and $$\frac{1}{(2n+1)!}-\frac{1}{2(2n)!}+\sum_{k=1}^n\frac{a_{2k-1}}{(2n-2k+1)!}=0\quad(n\ge1).$$

Remark. The coefficients $$B_n$$ of the Taylor expansion $$\frac{z}{e^z-1}=\sum_{n\ge0}\frac{B_nz^n}{n!}$$ are known as Bernoulli numbers. Thus $$a_n=\frac{B_{n+1}}{(n+1)!}$$ for every $$n\ge-1$$.