# 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 $res(f,z_0)=res(f,-z_0)$.

• – Git Gud May 19 '14 at 17:37
• The "so $\text{res}(-f,-z_0)=-a_{-1}$" part is wrong. The residue is the $-1^\text{th}$ coefficient of the power series around the residue. What you got is a power series around $z_0$, not $-z_0$. – Git Gud May 19 '14 at 18:36

## 1 Answer

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.