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

  1. #1
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    [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!
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  2. #2
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    Quote Originally Posted by vsywod View Post
    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.
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