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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.

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  • $\begingroup$ 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
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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...

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  • $\begingroup$ 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

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