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; $$\underset{z=0}{res}\frac{2z+i}{z^2(z^2+4)}=\frac{1}{2}$$ $$\underset{z=2i}{res}\frac{2z+i}{z^2(z^2+4)}=-\frac{5}{16}$$ $$\int_\Gamma\frac{2z+i}{z^2(z^2+4)}dz=2\pi i(\underset{z=0}{res}f(z)+\underset{z=2i}{res}f(z))=2\pi i(\frac{1}{2}-\frac{5}{16})=\frac{3}{8}\pi i$$

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.