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From what I understand, the main "point" of the Laurent series is that we should be able to derive it easily (e.g. by stitching together known Taylor series), and then exploit its uniqueness to say that these coefficients are the same as $c_n=\frac{1}{2\pi i}\oint_\circ \frac{f(z)}{(z-a)^{n+1}} dz$, from which we can calculate e.g. the residue $2\pi c_{-1}$.

But to actually do this, we need one crucial fact: the series $\sum_{n\in\mathbb{Z}}c_n(z-a)^n$ given by $c_n=\frac{1}{2\pi i}\oint_\circ \frac{f(z)}{(z-a)^{n+1}} dz$ is actually a valid Laurent series for $f(z)$, i.e. where it converges, it converges to $f(z)$.

How does one prove this? It feels like it should follow easily from Cauchy's integral formula, but my brain doesn't seem to be working sensibly at the moment.

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  • $\begingroup$ I believe it has to do with the fact that the function $f(z)$ defined by the convergent series will be holomorphic. And there are ways to check whether this is so...namely whether the Cauchy-Riemann equations are satisfied. $\endgroup$ – JohnColtraneisJC Mar 6 at 15:15
  • $\begingroup$ Compare math.stackexchange.com/a/1200502/42969. $\endgroup$ – Martin R Mar 6 at 15:16
  • $\begingroup$ @JohnColtraneisJC I certainly believe it's holomorphic. It's just "power series is differentiable within its radius of convergence", isn't it? But I don't see how that helps. $\endgroup$ – Abhimanyu Pallavi Sudhir Mar 6 at 15:16
  • $\begingroup$ @MartinR Thanks, that actually answers my question. I guess the best way to think of the proof in the answer you linked, and really even stuff like the uniqueness of Laurent series, is to transform it into a Fourier series. $\endgroup$ – Abhimanyu Pallavi Sudhir Mar 7 at 14:21

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