# Complex analysis Laurent series evaluated on unit circle

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.

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

First of all, you forgot the $n=0$ term, which yields the first summand.
$$\frac{z^n}{e^{in\theta}}=\left(\frac z{e^{i\theta}}\right)^n,$$
and similar for your second one. Collect the two sums, add the forgotten $n=0$ term, and you're done.