# Questions tagged [laurent-series]

The Laurent series is a generalisation of the power series which allows negative indices and is essential for investigating the behaviour of functions near poles.

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### How to switch to a Laurent series' next convergence ring?

Given the Laurent series $\sum\limits_{k=-\infty}^\infty a_k^{(l)} z^k = f(z)|_{r_l<|z|<R_l}$ of a meromorphic function $f$ on $\mathbb C$ with convergence region $r_l< |z|< R_l$, one can ...
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### An outrageous way to derive a Laurent series: why does this work?

I had to compute a series expansion of $1/(e^{x}-1)$ about $x=0$, and in the course of its derivation, I made a couple of manipulations that are not allowed mathematically. Still, comparing the final ...
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### When proving that f(z) is a polynomial, is it enough to consider just one point instead of keeping z arbitrary?

I think so - but I'd rather ask the MSE community too. Say I am given the bound |f(z)| < $|z|^3$, and that f is entire. Show f must be a polynomial. I used Cauchy's Integral Formula for ...
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### Is there a completion of Laurent series w.r.t. integration?

Given a Laurent series $$f(z) = \sum_{k=-\infty}^\infty a_k(z-z_0)^k,$$ its derivative is obviously also a Laurent series. However, upon integration, the $\frac1{z-z_0}$ term introduces a $\ln(z-z_0)$ ...
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### Calculating Laurent series expansion

I have to calculate the Laurent series expansion of $$f(z) = \frac {2z−2}{(z+1)(z−2)}$$ in $1 < |z| < 2$ and $|z| > 3$. For first annulus, I know I must manipulate the given expression ...
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### Power series expansion of $\frac{z}{(z^3+1)^2}$ around $z=1$

I want to expand $f(z)=\frac{z}{(z^3+1)^2}$ around $z=1$. That is, I want to find the coefficients $c_n$ such that $f(z) = \sum_{n=0}^\infty c_n (z-1)^n$. So far, my first strategy was using long ...