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**manjohn12** Evaluate $\displaystyle \oint_{C} \left(\frac{1}{(z+i)^{3}}- \frac{5}{z+i}+8 \right) \ dz $ where $\displaystyle C $ is the circle $\displaystyle |z+i| = 1 $ with positive orientation.

So we have $\displaystyle \int_{C} \frac{1}{(z+i)^{3}} \ dz + -5 \int_{C} \frac{1}{z+i} \ dz + \int_{C} 8 \ dz $. The last integral is $\displaystyle 0 $ by Cauchy-Goursat. For the two other integrals why can't we say that they are both $\displaystyle 0 $ by Cauchy-Goursat? Because they both have singularities at $\displaystyle z = -i $.

Only the first one is $\displaystyle 0 $ and the second one is $\displaystyle -5(2 \pi i) = -10 \pi i $ by Cauchy Integral Formula. But thet are both analytic within and on $\displaystyle C $. So why are they not both $\displaystyle 0 $?