Prove that for an odd function, res(f, z0) = res(f, -z0) I am having difficulty proving that, for an odd function, the residue function is symmetric. i.e $res(f, z_0) = res(f, -z_0)$
I am using the Laurent series expansion of a function $$f(z)=\sum_{n=-\infty}^\infty a_n(z-z_0)^n$$
I'm trying to get there by substituting $z$ with $-z$ so that I have the expansion of $-f(x)$ (since $f$ is odd. 
So far I have the residue of $-f$ about $-z_0$ is the $n=-1$ coefficient for  $$f(-z)=\sum_{n=-\infty}^{\infty} a_n(-z+z_0)^n$$ $$=\sum_{n=-\infty}^{\infty} (-1)^na_n(z-z_0)^n$$ so $$res(-f,-z_0)=-a_{-1}$$ Is this correct? If so how does it translate into a proof for $        res(f,z_0)=res(f,-z_0)$. 

The Laurent series about $z=z_0$ of $f$ is, as you point out,
$$f(z) = \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n $$
The Laurent series about $z=-z_0$ of $f$ is, on the other hand,
$$f(z) = \sum_{n=-\infty}^{\infty} b_n (z+z_0)^n $$
Now, $f(z)=-f(-z)$ as $f$ is odd; this implies that $a_n = - (-1)^n b_n$, and specifically that $a_{-1} = b_{-1}$, as was to be shown.
