integrating $\oint_C \frac{3z^3 + 2}{(z-1)(z^2 + 9)}dz$ on $|z|=4$ I am doing $\oint_C  \dfrac{3z^3 + 2}{(z-1)(z^2 + 9)}dz$ on $|z|=4$ and I find that there are poles within the contour at $z = 1$ and at $z = 3i$, both simple poles. I find that the integral $I = 2\pi i  \,\text{Res}(1) + 2 \pi i \,\text{Res}(3i)$, or $I = \pi i + 2 \pi i \,\,\,\text{Res}(3i)$, with only that second residue left to find. Is this correct, or should I include  $\text{Res}(-3i)$ as well?

There are 3 poles: $z=1$ and $z=\pm 3 i$.
$$\mathrm{Res}_{z=1} \frac{3 z^3+2}{(z-1)(z^2+9)} = \frac{5}{10} = \frac{1}{2}$$
$$\mathrm{Res}_{z=3 i} \frac{3 z^3+2}{(z-1)(z^2+9)} = \frac{2-i 81}{(-1+3 i)(6 i)}$$
$$\mathrm{Res}_{z=-3 i} \frac{3 z^3+2}{(z-1)(z^2+9)} = \frac{2+i 81}{(-1-3 i)(-6 i)}$$
The integral value is $i 2 \pi$ times the sum of these.
