I wanted to find Laurent series representation of function $1/(e^{z} -1)$. So I took minus common and apply the series formula of $1/(1-z)$ and then I use series formula for each $e^{zn}$. But I am getting very different answer what book have provided. What's wrong in my attempt.

  • $\begingroup$ How can we tell unless you include your attempt? The actual Laurent series involves Bernoulli numbers. $\endgroup$ – Lord Shark the Unknown Apr 20 at 11:32
  • $\begingroup$ You cannot use the series for $\frac 1 {1-z}$ and then change $z$ to $e^{z}$. This is because you would require $|e^{z}| <1$ or $\Re z <0$ for this to be valid. $\endgroup$ – Kavi Rama Murthy Apr 20 at 11:53
  • $\begingroup$ @Kavi Rama Murthy.. thanks sir, got my mistake. $\endgroup$ – Believer Apr 20 at 12:11

Assuming you mean around $\;z=0\;$ , observe we need to worry only for small values of $\;|z|\;$ , so we can assume $\;|z|<1\;$ and thus:

$$e^z-1=z+\frac{z^2}2+\mathcal O(z^3)\implies$$

$$\frac1{e^z-1}=\frac1{z\left(1+\frac z2+\mathcal O(z^2)\right)}=\frac1z\cdot\left(1-\frac z2+\frac{z^2}4-\ldots\right)=\frac1z-\frac12+\frac z4-\ldots$$

If you need more addends just add them...

  • $\begingroup$ I don't get $+\frac{z}4$. $\endgroup$ – Lord Shark the Unknown Apr 20 at 11:58
  • $\begingroup$ @LordSharktheUnknown I think that what you don't get is $\;+\frac12\;$ but $\;-\frac12\;$, and then $\;+\frac z4\;$ . Anyway, your comment made me see the mistake. Thanks. $\endgroup$ – DonAntonio Apr 20 at 12:06
  • $\begingroup$ I still don't get $+\frac z4$. I get $+\frac z{12}$. $\endgroup$ – Lord Shark the Unknown Apr 20 at 12:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.