# Thread: [SOLVED] Calculating residues of an even function, spot the mistake please

1. ## [SOLVED] Calculating residues of an even function, spot the mistake please

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 $\displaystyle \kappa:[0,2\pi] \to \mathbb{C}, t\mapsto z + e^{it}$. Then $\displaystyle \text{res}_z(f) = \frac{1}{2\pi i} \int_\kappa f(\xi) d\xi$.
Then define $\displaystyle \tilde{\kappa}:= - \kappa(\pi + t)$. Then $\displaystyle \text{res}_{-z}(f) = \frac{1}{2\pi i} \int_{\tilde{\kappa}} f(\xi) d\xi $$\displaystyle = \frac{1}{2\pi i}\int_0^{2\pi}r\cdot f(-\kappa(\pi+t))dt = \frac{1}{2\pi i}\int_0^{2\pi}r\cdot f(\kappa(\pi+t))dt = \frac{1}{2\pi i} \int_{\kappa} f(\xi) d\xi = \text{res}_z f But this is not what i was supposed to show. Where is the mistake? Thank you! 2. Originally Posted by vsywod Hi, I'm having trouble with the following exercise: let \displaystyle U = - U be symmetric and \displaystyle f: U \to \mathbb{C} meromorphic. Let \displaystyle f be even, i.e. \displaystyle f(z) = f(-z). Show that \displaystyle \text{res}_z(f) = - \text{res}_{-z}f. What I did was the following: let \displaystyle r>0 such that \displaystyle z is the only singularity in \displaystyle \overline{D_r(z)}. Then let \displaystyle \kappa:[0,2\pi] \to \mathbb{C}, t\mapsto z + {\color{red}r}e^{it}. Then \displaystyle \text{res}_z(f) = \frac{1}{2\pi i} \oint_\kappa f(\xi) d\xi. Then define \displaystyle \tilde{\kappa}:= - \kappa(\pi + t). Then \displaystyle \text{res}_{-z}(f) = \frac{1}{2\pi i} \oint_{\tilde{\kappa}} f(\xi) d\xi$$\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?
When you substitute $\displaystyle \xi = \kappa(t)$ in the integral $\displaystyle \text{res}_{z}(f) = \frac{1}{2\pi i} \oint_{\kappa} f(\xi)\, d\xi$, you have to remember to replace $\displaystyle d\xi$ by $\displaystyle \kappa'(t)dt = ire^{it}dt$. In the integral for the singularity at –z, the corresponding calculation is $\displaystyle d\xi = \tilde{\kappa}'(t)dt = -ire^{it}dt$. That accounts for the change of sign.