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I need to compute the residue of $$f(z)=\frac{\sinh(z)\sin(\omega z)}{\cosh^2(z)},$$ where $\omega\in\mathbb{R}$ is a parameter, at $z=\frac{i\pi}{2}$, but can I do it without computing the integral given in the residue's definition? Is there any other way?

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  • $\begingroup$ You can try finding the Laurent series expansion of $f(z)$. The residue will be the $a_{-1}$ term in the series. See page 8 here. $\endgroup$
    – user170231
    Jan 22, 2015 at 17:57

2 Answers 2

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Since $\cosh(iz)=\cos(z)$, $z_0=\frac{i\pi}{2}$ is a double pole for the given function. This gives: $$R=\operatorname{Res}\left(f(z),z=z_0\right)=\left.\frac{d}{dz}(z-z_0)^2 f(z)\right|_{z=z_0} $$ and by playing a bit with Laurent expansions, or just carrying out the full calculation, we have: $$ R = -i\omega \cosh\frac{\pi\omega}{2}.$$

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  • $\begingroup$ @ Jack D'Aurizio: It would be interesting to see the full calculation done by hand. $\endgroup$
    – user64494
    Jan 22, 2015 at 18:49
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To find the residue of a function $f$ at $z=z_0$, we find the coefficient of $1/z$ in the expansion of $f(z+z_0)$ about $z=0$.

$$\cosh{\left ( z+i \frac{\pi}{2}\right)} = \cosh{z} \cosh{\left ( i \frac{\pi}{2} \right )} + \sinh{z} \sinh{\left ( i \frac{\pi}{2} \right )} = i \sinh{z}$$

$$\implies \cosh^2{\left ( z+i \frac{\pi}{2}\right)} =-\sinh^2{z}$$

$$\sinh{\left ( z+i \frac{\pi}{2}\right)} = \sinh{z} \cosh{\left ( i \frac{\pi}{2} \right )} + \cosh{z} \sinh{\left ( i \frac{\pi}{2} \right )} = i \cosh{z}$$

$$\sin{\left [ \omega \left ( z+i \frac{\pi}{2}\right)\right ]} = \sin{\omega z} \cos{\left ( i \frac{\pi}{2} \omega \right )} + \cos{\omega z} \sin{\left ( i \frac{\pi}{2} \omega \right )}$$

We now want to find the coefficient of $1/z$ in the Laurent expansion about $z=0$ of

$$\frac{\sinh{\left ( z+i \frac{\pi}{2}\right)} \sin{\left [ \omega \left ( z+i \frac{\pi}{2}\right)\right ]}}{\cosh^2{\left ( z+i \frac{\pi}{2}\right)}} = \frac{i \cosh{z} \left [\sin{\omega z} \cos{\left ( i \frac{\pi}{2} \omega \right )} + \cos{\omega z} \sin{\left ( i \frac{\pi}{2} \omega \right )} \right ]}{-\sinh^2{z}}$$

which is

$$-i \omega \cos{\left ( i \frac{\pi}{2} \omega \right )} = -i \omega \cosh{\left ( \frac{\pi}{2} \omega \right )} $$

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