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Let k be the rectangle with corners $-2-2i,2-2i,2+i,-2+i$. Evaluate the integral: $$ \int_k \frac{\cos(z)}{z^4}dz $$ Would the best way to do this problem be to integrate along each contour line using the Cauchy-Goursat theorem: $$ \int_k f(z) \, dz = \int_b^a f(z(t))\frac{dz(t)}{dt} \, dt $$ Is there a way to do one integral and include all of the contour lines on the rectangle. Thanks in advance for the help.

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By Cauchy-Goursat, $$\int_k\frac{f(z)}{(z-z_0)^{n+1}}dz=\frac{2\pi i}{n!}f^{(n)}(z_0)$$ and so $$\int_k \frac{\cos(z)}{z^4}dz=\frac{2\pi i}{3!}\sin(0)=0$$

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  • $\begingroup$ Oh yea thanks, I was over thinking it so much. $\endgroup$ – Sam Houston Aug 25 '14 at 15:48

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