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

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

• Does the expression in the title have anything to do with the question at all? – MPW Apr 13 '15 at 19:06
• No :-), thank you for warning. Now is it corrected. – user227317 Apr 13 '15 at 19:09
• 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. – MPW Apr 13 '15 at 20:07
• Yes, sorry. I copy it from $z_2$ and not change result. $z_3$ is in circle. [fast and furious] – user227317 Apr 13 '15 at 20:11