Originally Posted by

**vsywod** Hi,

I'm having trouble with the following exercise:

let $\displaystyle U = - U$ be symmetric and $\displaystyle f: U \to \mathbb{C}$ meromorphic. Let $\displaystyle f $ be even, i.e. $\displaystyle f(z) = f(-z)$. Show that $\displaystyle \text{res}_z(f) = - \text{res}_{-z}f$.

What I did was the following: let $\displaystyle r>0$ such that $\displaystyle z$ is the only singularity in $\displaystyle \overline{D_r(z)}$. Then let $\displaystyle \kappa:[0,2\pi] \to \mathbb{C}, t\mapsto z + {\color{red}r}e^{it}$. Then $\displaystyle \text{res}_z(f) = \frac{1}{2\pi i} \oint_\kappa f(\xi) d\xi$.

Then define $\displaystyle \tilde{\kappa}:= - \kappa(\pi + t)$. Then $\displaystyle \text{res}_{-z}(f) = \frac{1}{2\pi i} \oint_{\tilde{\kappa}} f(\xi) d\xi $$\displaystyle = \frac{1}{2\pi i}\int_0^{2\pi}r\cdot f(-\kappa(\pi+t))dt =

\frac{1}{2\pi i}\int_0^{2\pi}r\cdot f(\kappa(\pi+t))dt = \frac{1}{2\pi i} \oint_{\kappa} f(\xi) d\xi = \text{res}_z f$

But this is not what i was supposed to show. Where is the mistake?