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Use Cauchy's integral formula to compute the following:

$$\int \limits_{\Gamma} \frac{\cos(z)+i\sin(z)}{(z^2+36)(z+2)}dz$$ where $\Gamma$ is the circle of centre $0$ and radius $3$ traversed in the anticlockwise direction.

I know you can factor the bottom further but I it wont work because when you put it into partial fractions, they don't turn out to be interior points.

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  • $\begingroup$ If they're not interior points, that is a good thing. Note that $\dfrac{a}{bc}=\dfrac{a/b}{c}$. $\endgroup$ – Jonas Meyer Mar 6 '15 at 17:29
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Since the function $f(z) := [\cos(z) + i\sin(z)]/(z^2 + 36)$ is analytic inside and on $\Gamma$, and $-2$ lies inside $\Gamma$, by Cauchy's integral formula, the integral evaluates to

$$2\pi i f(-2) = 2\pi i \frac{\cos(-2) + i\sin(-2)}{40} = \frac{\pi ie^{-2i}}{20}.$$

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The only pole inside the domain defined by $\;\Gamma\;$ is the simple one $\;z=-2\;$ , and

$$\lim_{z\to -2}(z+2)\frac{\cos z+i\sin z}{(z^2+36)(z+2)}=\frac{\cos(-2)+i\sin(-2)}{40}$$

and thus the integral's value is

$$2\pi i\frac{\cos(-2)+i\sin(-2)}{40}=\frac{\sin2+ i\cos2}{20}\pi$$

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