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

• May 14th 2009, 05:12 AM
vsywod
[SOLVED] Calculating residues of an even function, spot the mistake please
Hi,

I'm having trouble with the following exercise:
let $U = - U$ be symmetric and $f: U \to \mathbb{C}$ meromorphic. Let $f$ be even, i.e. $f(z) = f(-z)$. Show that $\text{res}_z(f) = - \text{res}_{-z}f$.

What I did was the following: let $r>0$ such that $z$ is the only singularity in $\overline{D_r(z)}$. Then let $\kappa:[0,2\pi] \to \mathbb{C}, t\mapsto z + e^{it}$. Then $\text{res}_z(f) = \frac{1}{2\pi i} \int_\kappa f(\xi) d\xi$.
Then define $\tilde{\kappa}:= - \kappa(\pi + t)$. Then $\text{res}_{-z}(f) = \frac{1}{2\pi i} \int_{\tilde{\kappa}} f(\xi) d\xi$ $= \frac{1}{2\pi i}\int_0^{2\pi}r\cdot f(-\kappa(\pi+t))dt =
\frac{1}{2\pi i}\int_0^{2\pi}r\cdot f(\kappa(\pi+t))dt = \frac{1}{2\pi i} \int_{\kappa} f(\xi) d\xi = \text{res}_z f$

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

Thank you!
• May 14th 2009, 10:49 AM
Opalg
Quote:

Originally Posted by vsywod
Hi,

I'm having trouble with the following exercise:
let $U = - U$ be symmetric and $f: U \to \mathbb{C}$ meromorphic. Let $f$ be even, i.e. $f(z) = f(-z)$. Show that $\text{res}_z(f) = - \text{res}_{-z}f$.

What I did was the following: let $r>0$ such that $z$ is the only singularity in $\overline{D_r(z)}$. Then let $\kappa:[0,2\pi] \to \mathbb{C}, t\mapsto z + {\color{red}r}e^{it}$. Then $\text{res}_z(f) = \frac{1}{2\pi i} \oint_\kappa f(\xi) d\xi$.
Then define $\tilde{\kappa}:= - \kappa(\pi + t)$. Then $\text{res}_{-z}(f) = \frac{1}{2\pi i} \oint_{\tilde{\kappa}} f(\xi) d\xi$ $= \frac{1}{2\pi i}\int_0^{2\pi}r\cdot f(-\kappa(\pi+t))dt =
\frac{1}{2\pi i}\int_0^{2\pi}r\cdot f(\kappa(\pi+t))dt = \frac{1}{2\pi i} \oint_{\kappa} f(\xi) d\xi = \text{res}_z f$

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

When you substitute $\xi = \kappa(t)$ in the integral $\text{res}_{z}(f) = \frac{1}{2\pi i} \oint_{\kappa} f(\xi)\, d\xi$, you have to remember to replace $d\xi$ by $\kappa'(t)dt = ire^{it}dt$. In the integral for the singularity at –z, the corresponding calculation is $d\xi = \tilde{\kappa}'(t)dt = -ire^{it}dt$. That accounts for the change of sign.