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Let $f(z)$ be a function analytic on an annulus that includes the unit circle $z=e^{i\theta}$. By taking that circle as the path of integration for the coefficients in the Laurent series, show that $$ f(z)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(e^{i\theta})d\theta+\frac{1}{2\pi}\sum_{n=1}^{\infty}\int_{-\pi}^{\pi}f(e^{i\theta})\left[\left(\frac{z}{e^{i\theta}}\right)^{n}+\left(\frac{e^{i\theta}}{z}\right)^{n}\right] d\theta $$

Attempt: By using the expressions for the coefficeints of the Laurent series, I have made it to $$ f(z)=\sum_{n=1}^{\infty}\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{f(e^{i\theta})}{e^{in\theta}}d\theta \cdot z^{n}+\sum_{n=1}^{\infty}\int_{-\pi}^{\pi}f(e^{i\theta})e^{in\theta}d\theta\cdot z^{-n} $$

Any thoughts on how to continue would be very much appreciated.

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  • $\begingroup$ For explicit multiplication, use $\cdot$ (\cdot) instead of *, which denotes a convolution, or omit the $\cdot$ if not really necessary $\endgroup$ Jul 16, 2013 at 8:47

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First of all, you forgot the $n=0$ term, which yields the first summand.

Second, note that in your first term you have

$$\frac{z^n}{e^{in\theta}}=\left(\frac z{e^{i\theta}}\right)^n,$$

and similar for your second one. Collect the two sums, add the forgotten $n=0$ term, and you're done.

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