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Calculate $\int_\Gamma \frac{2z+i}{z^2(z^2+4)}$ with residue theory. Where $\Gamma:|z-3i|=4$ is positively oriented circle.

Pls, for check my solution.

poles: $z_1=0$ (order 2 pole) $z_2=-2i$ (simple) $z_3=2i$ (simple)

$z_1:|0-3i|=3<4$ => in circle; $z_2:|-2i-3i|=5>4$ => out of circle; $z_3:|2i-3i|=1<4$ => in circle; $$\underset{z=0}{res}\frac{2z+i}{z^2(z^2+4)}=\frac{1}{2}$$ $$\underset{z=2i}{res}\frac{2z+i}{z^2(z^2+4)}=-\frac{5}{16}$$ $$\int_\Gamma\frac{2z+i}{z^2(z^2+4)}dz=2\pi i(\underset{z=0}{res}f(z)+\underset{z=2i}{res}f(z))=2\pi i(\frac{1}{2}-\frac{5}{16})=\frac{3}{8}\pi i$$

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  • $\begingroup$ Does the expression in the title have anything to do with the question at all? $\endgroup$ – MPW Apr 13 '15 at 19:06
  • $\begingroup$ No :-), thank you for warning. Now is it corrected. $\endgroup$ – user227317 Apr 13 '15 at 19:09
  • $\begingroup$ I also point out that $z_3$ is in the circle, not outside of it. This is surely just a typo since you are properly including it in the list of points for calculating residue. $\endgroup$ – MPW Apr 13 '15 at 20:07
  • $\begingroup$ Yes, sorry. I copy it from $z_2$ and not change result. $z_3$ is in circle. [fast and furious] $\endgroup$ – user227317 Apr 13 '15 at 20:11
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The computation of the residues looks correct to me, working it out quickly on the back of a napkin. So I would say your answer is correct.

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