Consider the function

$$f(z) = \frac{e^z+1}{e^z-1}$$

This function has a Laurent expansion about $0$ of the form

$$f(z) = \frac{a}{z} + \sum_{n=0}^\infty b_nz^n$$

for constants $a, b_1,\ldots, b_n$. Show that $b_n=0$ for all even $n$.

What is the best way to do this? All I can think to do is find the laurent expansion and then show that a few are zero but obviously this doesn't prove it for all even $n$, just a few.

If we split $f(z)$ into even and odd parts we have

$$f(z) = \frac az +\sum_{k=0}^\infty b_{2k}z^{2k}+\sum_{k=0}^\infty b_{2k+1}z^{2k+1}$$

and how the problem turns into showing the first summation is zero.

  • $\begingroup$ How many coefficients in the power series do you need? it can be done without knowing the whole closed form of the series $\endgroup$
    – DonAntonio
    May 7, 2016 at 21:16
  • $\begingroup$ @Joanpemo whoops I have realised I haven't put the full question down... $\endgroup$ May 7, 2016 at 21:17
  • $\begingroup$ and $\displaystyle\frac{z}{e^z-1} = \sum_{k=0}^\infty \frac{B_k}{k!} z^k$ while $\displaystyle\frac{z e^z}{e^z-1} = \frac{-z}{e^{-z}-1} = \sum_{k=0}^\infty \frac{B_k}{k!} (-z)^k$ hence $\displaystyle\frac{1}{z}\left(\frac{z}{e^z-1} + \frac{z e^z}{e^z-1} \right) = \frac{2}{z} \sum_{k=0}^\infty \frac{B_{2k}}{k!} z^{2k}$ on $0 < |z| < 2 \pi$ en.wikipedia.org/wiki/Bernoulli_number#Generating_function $\endgroup$
    – reuns
    May 7, 2016 at 21:22
  • $\begingroup$ @user1952009 These formulas don't seem correct to me. In your first one you have the LHS which has a Laurent series then on the RHS you have a taylor series with no negative powers. Same for the second. Please check before posting!! $\endgroup$ May 7, 2016 at 21:44
  • $\begingroup$ $\frac{z}{e^z-1}$ is holomorphic (analytic) on $|z| < 2 \pi$ (and at the last one I forgot to write $(2k)!$ instead of $k!$) $\endgroup$
    – reuns
    May 7, 2016 at 21:53

2 Answers 2


Check that $f(-z) = -f(z).$ Thus

$$\frac{a}{-z} + \sum_{n=0}^\infty b_n(-z)^n = -\left (\frac{a}{z} + \sum_{n=0}^\infty b_nz^n\right ) = -\frac{a}{z} + \sum_{n=0}^\infty (-b_n)z^n.$$

Equating coefficients gives the desired result.

  • 1
    $\begingroup$ go through the simple computation -- f is odd $\endgroup$
    – zhw.
    May 7, 2016 at 22:09
  • $\begingroup$ Succinctly written. +1 $\endgroup$
    – Mark Viola
    May 7, 2016 at 22:23

Let $f(z)$ be the function given by


Clearly $f(z)$ is analytic for $0<|z|<2\pi $. Then, note that

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

and $f(z)$ is analytic for all $|z|<2\pi$. Furthermore, we have

$$\begin{align} f(-z)&=\frac{e^{-z}+1}{e^{-z}-1}+\frac{2}{z}\\\\ &=-\frac{e^z+1}{e^z-1}+\frac{2}{z}\\\\ &=-f(z) \end{align}$$

Hence, $f(z)$ is an odd function of $z$ and can be expanded in a Taylor series of odd powers only with $f(z)=\sum_{n=1}^\infty a_nz^{2n-1}$. Finally, we can write

$$\frac{e^{z}+1}{e^{z}-1}=\frac2z+\sum_{n=1}^\infty a_nz^{2n-1}$$

as was to be shown.

  • 1
    $\begingroup$ $f$ is not analytic for $z\ne 0.$ It has poles at $n2\pi i,$ $n$ a non-integer integer. $\endgroup$
    – zhw.
    May 7, 2016 at 22:18
  • $\begingroup$ that should be "nonzero integer" in my first comment $\endgroup$
    – zhw.
    May 7, 2016 at 22:35
  • $\begingroup$ @zhw. No worry. I assumed that you knew (and meant to write) that. $\endgroup$
    – Mark Viola
    May 7, 2016 at 22:40
  • $\begingroup$ @Mark Viola, please provide a formula for $a_n$. $\endgroup$ Feb 20 at 6:33
  • $\begingroup$ @MichaelLevy There is no need to give a "formula" for $a_n$ to show that $f$ has a Laurent series that is expressed with only odd powers of $z$. $\endgroup$
    – Mark Viola
    Feb 20 at 19:28

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