The Residues of an even function or an odd function on $U$ subset open symmetric I have to proof that for $f$ even function holomorphic with singularities isolated then $$res_{z}f=-res_{-z}f$$ an simmetric for $f$ odd, $i.e.$ $$res_{z}f=res_{-z}f$$ My hint is proof that laurent series of $f$ have coefficient even o odd severally but I can't see how help. since $$res_{z_0}f=\int_{\vert \zeta-z_0\vert=r} f(\zeta)d\zeta$$ Because for $-z_0$ I have another integration domain, then How do I compare the expression integral for the residues in $z_0$ y $-z_0$?

Let $f$ be even and $\gamma(t)=z+re^{2i\pi t}$ so

$$res_{z}f = \frac{1}{2i\pi}\int_{0}^{1}f(\gamma(t))\gamma'(t)dt = -\int_{0}^{1}f(-\gamma(t))(-\gamma'(t))dt$$

but since $\beta(t)=-\gamma(t)=-z-re^{2i\pi t}$ is a parametrization of $S_r(-z)$, so $$res_{z}f = -\int_{0}^{1}f(\beta(t))(\beta'(t))=-res_{-z}f$$

the odd case is the same.