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Let $$ f(z) = \frac{e^{az}}{1 + e^z} \qquad (0 < a < 1) $$ I know this function has poles at $z = (2k + 1) i \pi$ with $k \in \mathbb{Z}$. Let's say I need to find the residue at the pole $z = \pi i$. How would I compute this? I tried using the formula $$ Res(f(z), z= \pi i) = \lim_{z \to \pi i} (z - z_0) f(z)$$ but that doesn't seem to work.

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  • $\begingroup$ You can use the series expansion of $f(z)$ about $z=i\pi$ and take the coefficient of the $(z-i\pi)^{-1}$ term. $\endgroup$ – user170231 Jan 12 '16 at 20:45
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Let $\zeta=z-i \pi$. Then the function of $\zeta$ is

$$\frac{e^{i a (\zeta+i \pi)}}{1+e^{(\zeta+ i \pi)}} = e^{i \pi a} \frac{e^{i a \zeta}}{1-e^{\zeta}}$$

Now take the Laurent expansion of this about $\zeta=0$ and see that the coefficient of $\zeta^{-1}$ is clearly $-e^{i \pi a}$.

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  • $\begingroup$ I tried taking the Laurent expansion, but still I don't see that the coefficient is $-e^{i \pi a}$. I expand the denominator as a geometric series right? And the numerator is just complex exponential expansion? $\endgroup$ – Kamil Jan 12 '16 at 21:04
  • $\begingroup$ @Kamil: That's right. The $1$'s cancel in the denominator, leaving a $\zeta$ as the leading term. The rest should be simple. $\endgroup$ – Ron Gordon Jan 12 '16 at 21:05
  • $\begingroup$ Ah, I see it now! Thank you $\endgroup$ – Kamil Jan 12 '16 at 21:07

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