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.

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?

Answering my own question as Neal advised me to.

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

There are singularities at z=0,z=−81. As 81>2, this singularity falls outside γ so the only one that matters is z=0. I then applied Cauchy's Integral Formula ∫γ(z+27i)(z+16)z(z+81)2dz=2πi[(z+27i)(z+16)(z+81)2]|z=0 And I got a final result of −32π243.

Is my analysis and final result correct?