# Thread: Cauchy Integral

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

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

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

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

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

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

Only the first one is $0$ and the second one is $-5(2 \pi i) = -10 \pi i$ by Cauchy Integral Formula. But thet are both analytic within and on $C$. So why are they not both $0$?
The integrands of the other two integrals are not analytic inside and on $C$ (there is a singularity at $z = -i$) so the Cauchy-Goursat Theorem cannot be applied.

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

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

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