So I'm looking at doing a basic contour integral using Cauchy's Residue Theorem. I feel I understand how to do this, and have gone over my work numerous times, yet the webwork system I'm doing this for still says my answer in wrong.
Here it is: let $C$ be the circle $|z| =8$, oriented counter clockwise. Find $\int_C \frac{1}{z^3(z+3i)} dz$.
Right away I see two poles:
- order 1 at $z=-3i$
- order 3 at $z=0$
To find the residue at a pole of order $m$ I use the result: $$\text{Res}(f;z_0) = \lim_{z\to z_0} \frac{1}{(m-1)!} \frac{d^{m-1}}{dz^{m-1}} \left[ (z-z_0)^m f(z) \right] $$
Both poles are inside of $C$, so by Cauchy's Residue Theorem I get: \begin{align} \int_C \frac{1}{z^3(z+3i)} dz &= 2 \pi i \left[ \text{Res}(0) + \text{Res}(-3i) \right]\\ & = 2 \pi i \left[ \lim_{z\to0} \frac{d}{dz} \frac{1}{z+3i} + \lim_{z\to -3i} \frac{1}{z^3} \right] \\ &= 2 \pi i \left[ \frac{i}{27} + \frac{-i}{27} \right] \\ &= 2 \pi i \left( 0 \right) \\ &= 0 \end{align}
Edit: correction: I see I failed to take the double derivative in $\text{Res}(0)$. My current corrected version, which still gives me a wrong answer is: \begin{align} \int_C \frac{1}{z^3(z+3i)} dz &= 2 \pi i \left[ \text{Res}(0) + \text{Res}(-3i) \right]\\ & = 2 \pi i \left[ \lim_{z\to0} \frac{\color{red}1}{\color{red}2}\frac{d^{\color{red}2}}{dz^{\color{red}2}} \frac{1}{z+3i} + \lim_{z\to -3i} \frac{1}{z^3} \right] \\ & = 2 \pi i \left[ \lim_{z\to0} \frac{\color{red}1}{\color{red}2} \frac{2}{(z + 3i)^3} + \lim_{z\to -3i} \frac{1}{z^3} \right] \\ &= 2 \pi i \left[ \frac{i}{27} + \frac{-i}{27} \right] \\ &= 2 \pi i \left( 0 \right) \\ &= 0 \end{align}
Where is my mistake? Thanks.