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Math Help - [SOLVED] Calculating residues of an even function, spot the mistake please

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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 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 = <br />
\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 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 = <br />
\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.
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