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.
Answer:
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?