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This is part of the question I am currently doing and I am having difficulties calculating the following residues. $$\text{res}\left({\frac{e^{iz}}{1+z^6},e^{\frac{i\pi}{6}}}\right)$$ I have tried using L'hopital to find the limits but couldn't seem to progress anywhere.

J T
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  • Are you still calculating this https://math.stackexchange.com/questions/2594007/complex-integral-with-residuals/ ? – rtybase Jan 10 '18 at 18:07
  • yes, i have tried calculating the residues and got the first one alright but not the second and third one – J T Jan 10 '18 at 18:39

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Since $z=\exp\frac{\pi i}{6}$ is a simple pole for $f(z)=\frac{e^{iz}}{1+z^6}$,

$$\operatorname*{Res}_{z=\exp\frac{\pi i}{6}}f(z) = \lim_{z\to \exp\frac{\pi i}{6}}\frac{e^{iz}(z-\exp\frac{\pi i}{6})}{z^6+1}\stackrel{\text{d.H.}}{=}\lim_{z\to \exp\frac{\pi i}{6}}\frac{e^{iz}}{6z^5}=-\frac{1}{6}\lim_{z\to \exp\frac{\pi i}{6}} ze^{iz}.$$

Jack D'Aurizio
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  • Could you explain how you get from the second to the third equal specifically the one with d.H. – J T Jan 10 '18 at 18:35
  • and is it possible to simplify the final answer further by substituting the limits as z tend to exp $\frac{\pi i}{6}$ – J T Jan 10 '18 at 18:36
  • @JTan: if $\alpha$ is a simple zero of $f(x)$, $$\lim_{x\to \alpha}\frac{g(x)(x-\alpha)}{f(x)}=\lim_{x\to \alpha}\frac{g'(x)(x-\alpha)+g(x)}{f'(x)}=\frac{g(\alpha)}{f'(\alpha)}.$$ – Jack D'Aurizio Jan 10 '18 at 19:15
  • Then I exploited that $z^6+1=0$ implies $$\frac{1}{6z^5}=\frac{z}{6z^6}=-\frac{z}{6}.$$ – Jack D'Aurizio Jan 10 '18 at 19:15