I am reviewing for a complex analysis final and this was a question on the review sheet. No answers were provided so I attempted it on my own. Using Cauchy's Integral Formula, I have that $$2\pi i \cdot f(z) = \int_C \dfrac{1}{z^3+4z^2+3z}dz = -\dfrac{\pi i}{3}.$$ If anyone wants to do this problem, the denominator factors into $z(z+1)(z+3).$
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1What is your question? – Berci Dec 13 '13 at 00:31
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As long as $\frac13 - \frac12 = - \frac16$, I agree with your result. – Daniel Fischer Dec 13 '13 at 00:32
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@Berci Sorry I wasn't entirely clear in asking my question. I basically just wanted someone else to double-check my work. I got the answer that I wrote up there. I was just hoping that someone else could verify that I was correct (which apparently I am). – Bark Jr. Dec 13 '13 at 00:39
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@Daniel Fischer - And thank you. – Bark Jr. Dec 13 '13 at 00:39
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What is the $\large C$ definition ?. – Felix Marin Dec 13 '13 at 00:41
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@FelixMarin $C$ denotes a circle that is positively oriented (counter-clockwise) with its center at 0 and it has a radius of 2. That was just the notation given for this particular problem. I've seen several other notations. – Bark Jr. Dec 13 '13 at 00:46
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@BarkJr. 0 k. Thanks. – Felix Marin Dec 13 '13 at 00:46
2 Answers
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Hints:
i) locate and classify the poles.
ii) use partial fraction.
iii) use Cauchy's formula.
See a related problem.
Mhenni Benghorbal
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$\int_C \frac{1}{z(z+1)(z+3)} \, \mathrm{d}z = 2 \pi i \times Res\left(\frac{1}{z(z+1)(z+3)}; z=0, z=-1\right) = 2 \pi i \times (1/3 - 1/2) = - \frac{\pi i}{3}$. I agree with your answer.
Sourav D
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