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How do I use Residue Theorem to evaluate $ \ \oint_{C_3 (0)} \frac{z+7}{z^4 + z^3 - 2 z^2}\,dz \ $ where $C_3(0)$ is the circle of radius 3 centered at the origin, oriented in the counter- clockwise direction.

This seems to be a very complicated case, any ideas?

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up vote 1 down vote accepted

Well, let's begin by identifying the poles of the function. First the denominator factors as $$z^2(z^2 + z - 2) = z^2(z+2)(z-1)$$ We can partial fraction to decompose the integral into $$\oint_{C_3(0)}-\frac{7}{2z^2} -\frac{9}{4z} - \frac{5}{12(z+2)} + \frac{8}{3(z-1)}\ \rm dz$$ We can easily read off the residuals here $$\sum\operatorname{Res}(f,\ \alpha)=-\frac{9}{4}-\frac{5}{12} + \frac{8}{3} = 0$$

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There are simple poles at $z=-2$ and $z=1$. The residues at these points can be found by evaluating $\frac{f(z_{0})}{g'(z_{0})}$ where $f,g$ are the numerator and the denominator of the integrand respectively.

There is a double pole at 0 and one way to find the residue there is to evaluate the derivative of $\frac{z+7}{z^{2}+z-2}$ at 0.

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