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How are Laurent series and Fourier series related to each other? There is a problem that states that for a periodic function $F(z + 2 \pi ) = F(z)$ that is analytic in finite plane.

$$F(z) = \sum_{n=-\infty}^\infty \alpha_n e^{inz}, \alpha_n=\frac{1}{2\pi}\int_0^{2\pi}F(z)e^{-inz}dz$$

How do I relate $z^n$ to $e^{inz}$ and $\displaystyle \frac{1}{2\pi i}\oint_\gamma\frac{F(z)}{z^{n+1}}dz$ to $\displaystyle \frac{1}{2\pi }\int_0^{2\pi}F(z)e^{-inz}dz$.

Added:: It looks that if $z = e^{i\theta}$ the Laurent series looks like Fourier series. But I not get why $n$ goes from $-\infty$ to $\infty$ since the $F(z)$ is analytic in finite plane shouldn't $n$ start from $0\to \infty$?

Monkey D. Luffy
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    Consider the values of your function $F(z)$ on the unit circle $z=e^{i\phi},\phi\in[0,2\pi]$. Obviously, $F(e^{i\phi})$ is periodic in $\phi$ and therefore can be expanded into Fourier series. The coefficients of this series obviously coincide with the coefficients of Laurent expansion. – Start wearing purple May 09 '13 at 17:46
  • @O.L. I remember your addendum here on this question. Part of it originates from there. What I am not getting is the $n$ starts from $-\infty$ in this question. Since it is analytic, I think it should start from $0$. I, honestly would like to clear my understanding. – Monkey D. Luffy May 09 '13 at 17:50
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    $F(z)$ in your question (not the one in my previous comment) is analytic with respect to $z$, but $\tilde{F}(\xi=e^{2\pi i z})=F(z)$ is not necessarily analytic with respect to $\xi$. – Start wearing purple May 09 '13 at 17:56
  • @O.L. yes that is one of my qualms, do I still get Fourier series if I put $\xi = e^{2 \pi i z}$ since $|\xi|$ is not necessarily unit. – Monkey D. Luffy May 09 '13 at 17:59
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    I don't really understand where do you have problems. Consider an example: take $F(z)=2\cos z$, so that $\tilde{F}(\xi)=\xi+\xi^{-1}$. The first function is analytic the second isn't. Fourier expansion of the first w.r.t. $z$ corresponds to Laurent expansion of the second around $\xi=0$. – Start wearing purple May 09 '13 at 18:31
  • @O.L. could you elaborate it, ... expansion at $\xi = 0 $ => are we expanding series at $\pm \infty$?? – Monkey D. Luffy May 09 '13 at 18:47
  • Yes, $\xi=0$ corresponds to $z=\infty$ if that was the question. – Start wearing purple May 09 '13 at 18:55
  • @O.L. So how do I write answer to this question? Let $F(z) = \sum_{n=0}^\infty a_n z^n = \sum_{n=0}^\infty a_n e^{i \pi nw}$. What do I do to get $\sum_{n=-\infty}^\infty a_n e^{i \pi nw}$ ?? – Monkey D. Luffy May 09 '13 at 19:15

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