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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?

  • 1
    @Jim_CS If you require further feedback it might help to elaborate more specifically on any doubts. If not, please post an answer and accept it so that we know it is resolved.2012-05-17

2 Answers 2

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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?

0

Nothing wrong with your analysis and answer.