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.

  • 1
    $\begingroup$ The pole at $z=0$ has order $m=3$ (as you already noticed), but your computation of the residue at $z=0$ is done for a simple pole. $\endgroup$ – Martin R May 13 '16 at 2:18
  • 1
    $\begingroup$ forget about $\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] $ and remember that $\text{Res}(f;z_0)$ is the coefficient $c_{-1}$ in the Laurent series of $f(z)$ around $z_0$ : $f(z) = \sum_{k=-m}^\infty c_k (z-z_0)^k$. why do we care only of $c_{-1}$ ? because $\int_{|z-z_0| = r} c_k (z-z_0)^k dz = 0$ whenever $k \ne -1$. $\endgroup$ – reuns May 13 '16 at 2:28
  • $\begingroup$ You have to differentiate twice. $\endgroup$ – kmitov May 13 '16 at 2:33
  • $\begingroup$ For the pole of order $m=3$, one has that $m-1=2$. You used $1$. $\endgroup$ – MPW May 13 '16 at 2:34

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