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I am doing $\oint_C \dfrac{3z^3 + 2}{(z-1)(z^2 + 9)}dz$ on $|z|=4$ and I find that there are poles within the contour at $z = 1$ and at $z = 3i$, both simple poles. I find that the integral $I = 2\pi i \,\text{Res}(1) + 2 \pi i \,\text{Res}(3i)$, or $I = \pi i + 2 \pi i \,\,\,\text{Res}(3i)$, with only that second residue left to find. Is this correct, or should I include $\text{Res}(-3i)$ as well?

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2 Answers 2

up vote 2 down vote accepted

Let $\,C_1,\,C_2,\,C_3\,$ be little circles (say, of radius $\,0.1\,$) around each of the poles $\,1,\,-3i,\,3i\,$ resp., of the function, so by Cauchy's Integral Theorem:

$$\int\limits_{|z|=4}\frac{3z^3+2}{(z-1)(z^2+9)}dz=\int\limits_{C_1}\frac{\frac{3z^3+2}{(z^2+9)}}{z-1}dz+\int\limits_{C_2}\frac{\frac{3z^3+2}{(z-1)(z-3i)}}{z+3i}dz+\int\limits_{C_3}\frac{\frac{3z^3+2}{(z-1)(z+3i)}}{z-3i}dz=$$

$$=2\pi i\left(\left.\frac{3z^3+2}{(z^2+9)}\right|_{z=1}+\left.\frac{3z^3+2}{(z-1)(z-3i)}\right|_{z=-3i}+\left.\frac{3z^3+2}{(z-1)(z+3i)}\right|_{z=3i}\right)=$$

$$=2\pi i\left(\frac{1}{2}+\frac{2+81i}{-18+6i}+\frac{2-81i}{-18-6i}\right)=$$

$$=2\pi i\left(\frac{1}{2}-\frac{1}{6}\left(-5\right)\right)=\frac{8\pi i}{3}$$

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Okay thanks, I was coming up short whether to take both 3i and -3i as residues –  user1535776 Feb 20 '13 at 3:26
    
Any time, and yes: we must take both since they both are contained within the domain enclosed by $\,\{|z|=4\}\,$ –  DonAntonio Feb 20 '13 at 3:27

There are 3 poles: $z=1$ and $z=\pm 3 i$.

$$\mathrm{Res}_{z=1} \frac{3 z^3+2}{(z-1)(z^2+9)} = \frac{5}{10} = \frac{1}{2}$$

$$\mathrm{Res}_{z=3 i} \frac{3 z^3+2}{(z-1)(z^2+9)} = \frac{2-i 81}{(-1+3 i)(6 i)}$$

$$\mathrm{Res}_{z=-3 i} \frac{3 z^3+2}{(z-1)(z^2+9)} = \frac{2+i 81}{(-1-3 i)(-6 i)}$$

The integral value is $i 2 \pi$ times the sum of these.

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Thanks, not sure how I missed that...old age I guess. –  Ron Gordon Feb 20 '13 at 3:27
    
Yeah...we all are deep into that...*sigh*...yet the alternative isn't that appealing, either, so... –  DonAntonio Feb 20 '13 at 3:29

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