1. 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 $\displaystyle 0$?

yeah you had to use Cauchy Integral formula for derivatives because the function was "too singular."

2. Originally Posted by 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 $\displaystyle 0$?
The integrands of the other two integrals are not analytic inside and on $\displaystyle C$ (there is a singularity at $\displaystyle z = -i$) so the Cauchy-Goursat Theorem cannot be applied.

You know that $\displaystyle f^{(n)}(\alpha) = \frac{n!}{2 \pi i} \oint \frac{f(z)}{(z - \alpha)^{n+1}} \, dz$. Therefore:

1. $\displaystyle \oint_{C} \frac{1}{(z+i)^3} \ dz = 0$ since $\displaystyle n = 2$ and $\displaystyle f(z) = 1$.

2. $\displaystyle \oint_{C} \frac{1}{(z+i)} \ dz = 2 \pi i$ since $\displaystyle n = 0$ and $\displaystyle f(z) = 1$.