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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How can we tell unless you include your attempt? The actual Laurent series involves Bernoulli numbers. – Angina Seng Apr 20 '19 at 11:32
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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. – Kavi Rama Murthy Apr 20 '19 at 11:53
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@Kavi Rama Murthy.. thanks sir, got my mistake. – ogirkar Apr 20 '19 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...
DonAntonio
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@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. – DonAntonio Apr 20 '19 at 12:06
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