I am doing the integral $\oint_C \dfrac{3z^3 + 2}{(z-1)(z^2 + 9)} dz$, and I am trying to find the residue at the pole $3i$;I am unsure how to do this. Could I factor $z^2 + 9$ further?
2026-04-14 07:47:43.1776152863
finding residue with $\oint_C \frac{3z^3 + 2}{(z-1)(z^2 + 9)} dz$
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The pole is simple and so the residue is
$$\begin{align} \lim_{z\rightarrow 3i}(z - 3i) f(z) & = & \lim_{z\rightarrow 3i} (z - 3i) \frac{3z^3 + 2}{(z - 1)(z^3 + 9)} \\& = & \lim_{z\rightarrow 3i}\frac{3z^3 + 2}{(z - 1)(z + 3i)} \\ & = & \frac{-81i + 2}{(3i - 1)(6i)} \\ & = & \frac{2 - 81i}{18 + 6i} \end{align}$$
Warning: I just learnt complex analysis in a few hours the other day so I'm not 100% sure myself, answering for my own education too :)