# Complex integration with Cauchy's Integral Formula

Calculate$$\int_\gamma \frac{(z+27i)(z+16)}{z(z+81)^2}dz$$ where $\gamma$ is the triangle whose vertices are the third roots of $z = -8i$, oriented counterclockwise.

I calculated the third roots of $-8i$ and they all have modulus $2$. This tells me that the maximum distance of $\gamma$ from the origin will be $2$.

There are singularities at $z=0, z=-81$. As $81 > 2$, this singularity falls outside $\gamma$ so the only one that matters is $z = 0.$

I then applied Cauchy's Integral Formula $$\int_\gamma \frac{(z+27i)(z+16)}{z(z+81)^2}dz = 2\pi i [\frac{(z+27i)(z+16)}{(z+81)^2}] |_{z=0}$$

And I got a final result of $\displaystyle\frac{-32\pi}{243}$.

Is my analysis and final result correct?

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Yep, completely spot on! Only thing I can see you could have improved slightly was to notice the cube roots of $-8i$ should have modulus 2 without calculating them, saves you some work. –  Ragib Zaman Apr 18 '12 at 14:00
(+1) for showing work. –  The Chaz 2.0 Apr 18 '12 at 14:01
You should copy this to an answer and accept it :) –  Neal Apr 18 '12 at 15:34
@Jim_CS If you require further feedback it might help to elaborate more specifically on any doubts. If not, please post an answer and accept it so that we know it is resolved. –  Bill Dubuque May 17 '12 at 15:54