Laurent series for an even function Show that if the Laurent series $\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n$ represents an even function, then $a_{2n+1}=0$ for $n=0,\pm 1,\pm 2,\ldots$, and if it represents an odd function, then $a_{2n}=0$ for $n=0,\pm 1,\pm 2,\ldots$.
where
$a_n=\frac{1}{2\pi i}\int_C \frac{f(z)}{(z-z_0)^{n+1}}dz$
I know the fact that if $f(-z)=f(z)$ then $f$ is even. But I have difficulty applying this to show what i need to have.
 A: This is only true if $z_0 = 0$. Suppose $f(z)$ is even. Then $f(z) - f(-z) = 0$. But 
$$f(z) - f(-z) = \sum_{n = -\infty}^{\infty} a_nz^n - \sum_{n = -\infty}^{\infty} a_n(-z)^n$$
$$= \sum_{n = -\infty}^{\infty} a_nz^n - \sum_{n = -\infty}^{\infty} (-1)^na_nz^n$$
$$= \sum_{n = -\infty}^{\infty} (1 - (-1)^n)a_nz^n$$
The above Laurent series gives the zero function by evenness of $f(z)$. But by uniqueness of Laurent expansions this implies all coefficients are zero. So $(1 - (-1)^n)a_n = 0$ 
for all $n$. When $n$ is odd this gives that $2a_n = 0$ for all odd $n$, or equivalently that $a_n = 0$ for all odd $n$.
A similar argument works when $f(z)$ is odd.
A: You need to have $z_0=0$ here. Then
\begin{align}
  2\pi i a_n &= \int_{|z|=r} \frac{f(z)}{z^{n+1}} \,dz \\
  &= \int_0^{2\pi} \frac{f(re^{i\theta})}{r^{n+1}e^{i(n+1)\theta}} ire^{i\theta} \, d\theta \\
  &= \frac{i}{r^n} \int_0^{2\pi}f(re^{i\theta})e^{-in\theta} \,d\theta
\end{align}
where $r$ lies between the inner and outer radii of the annulus of convergence and we have made the substitution $z=re^{i\theta}$. So
\begin{align}
  2\pi r^na_n &= \int_0^\pi f(re^{i\theta})e^{-in\theta} \,d\theta + \int_\pi^{2\pi} f(re^{i\theta})e^{-in\theta} \,d\theta \\
  &= \int_0^\pi f(re^{i\theta})e^{-in\theta} \,d\theta + \int_0^\pi f(re^{i(\theta+\pi)})e^{-in(\theta+\pi)} \,d\theta \\
  &= \int_0^\pi f(re^{i\theta})e^{-in\theta} \,d\theta + (-1)^n\int_0^\pi f(-re^{i\theta})e^{-in\theta} \,d\theta \\
  &= \int_0^\pi\Big[  f(re^{i\theta}) + (-1)^n f(-re^{i\theta}) \Big]e^{-in\theta} \,d\theta
\end{align}
and now you can see exactly the effect of the symmetry of $f$ and parity of $n$.
