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Is the following an explicit expression for the Laurent series:

$$f(z)=\sum_{n=0}^{\infty}(z-z_0)^n\frac{f^{(n)}(z_0)}{n!}?$$

The reason I ask is because this is what I have seen being derived when studying Laurent series, but then I struggle to find this as the definitive Laurent series when searching the internet and books.

This would imply that the Laurent series is directly analogous to the Taylor series except being applied to complex numbers.

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  • $\begingroup$ This is a power series. A Laurent series extends this to negative $n$. The Laurent series discussion in Wikipedia is typical of most books, I am not sure what you mean by definitive. $\endgroup$ – copper.hat Apr 30 '16 at 21:15
  • $\begingroup$ in general, if $f(z)(z-z_0)^K$ is holomorphic on $|z-z_0| < C$ then $$ f(z) = \sum_{n=0}^\infty \frac{(z-z_0)^{n-K}}{n!} \frac{d[f(z) (z-z_0)^K]}{dz}(z_0)$$ for every $|z-z_0| < C$. but when $z_0$ is an essential singularity, there is no such $K$, and we can't write any such generalized Taylor series, and we need more work for defining the Laurent series. $\endgroup$ – reuns Apr 30 '16 at 21:46
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    $\begingroup$ You have $f^{(0)}$, but it should be $f^{(n)}$. $\endgroup$ – DisintegratingByParts May 1 '16 at 22:31

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