1. ## Integration along curve

This kind of problem is new to me, and I am having a hard time to approach it it. Can someone help please?
1) Evaluate the integrals $\int_\gamma z^n dz$ for all integers n, where $\gamma$ is any circle centered at the origin with counterclockwise orientation.
2) Same question but with $\gamma$ being any circle not containing the origin.

2. Originally Posted by namelessguy
This kind of problem is new to me, and I am having a hard time to approach it it. Can someone help please?
1) Evaluate the integrals $\int_\gamma z^n dz$ for all integers n, where $\gamma$ is any circle centered at the origin with counterclockwise orientation.
2) Same question but with $\gamma$ being any circle not containing the origin.
Case 1: n > 0.

You can appeal to the Cauchy-Goursat Theorem which says that if f(z) is analytic inside and on a closed contour C then $\oint_C f(z) \, dz = 0$. Alternatively you can easily modify the argument given in case 2: n < 0.

Case 2: n < 0.

Let $C_{\rho}: \, |z| = \rho$, that is, the circle with radius $\rho$ and centre at $z = 0$. On $C_{\rho}$:

$z = \rho e^{i \theta}, ~ 0 \leq \theta < 2 \pi$.

$dz = i \, \rho e^{i \theta} \, d \theta$.

Then $\oint_{C_{\rho}} \frac{1}{z^m} \, dz = \int_{0}^{2 \pi} \frac{i \, \rho e^{i \theta}}{\rho^m e^{i m \theta}} \, d \theta = \frac{i}{\rho^{m-1}} \int_{0}^{2 \pi} e^{i (1 - m) \theta} \, d \theta$.

$m = 1$ (that is, $n = -1$): $\oint_{C_{\rho}} \frac{1}{z} \, dz = 2 \pi i$.

$m \neq 1$: $\oint_{C_{\rho}} \frac{1}{z^m} \, dz = \frac{1}{\rho^{m-1}(1 - m)} \left[ e^{i(1 - m) \theta} \right]_{0}^{2 \pi} = \frac{1}{\rho^{m-1}(1 - m)} \left( e^{2 \pi i (1 - m)} - 1 \right)$ $= \frac{1}{\rho^{m-1}(1 - m)} (1 - 1) = 0$.

Case 3: n = 0.

Left for you to do.

These results readily follow from Cauchy's Integral Theorem (which you might not have met yet).

It follows from the Cauchy-Goursat Theorem that when the circle does not contain the origin, the integral will equal zero. Alternatively, the case 2 argument is readily modified.

3. You can also do this without Cauchy's theorem. You just need to find an anti-derivative which exists throught the entire disk. That can be done since we stay away from the origin, thus, $\tfrac{z^{n+1}}{n+1}$ is the anti-derivative for $n\not = -1$. If $n=-1$ then it is possible to define $\log z$ in an analytic way so that it does hit the origin and is therefore anaytic on the disk.

4. We're not supposed to use the Cauchy-Goursat's Theorem on this problem because we haven't learned it yet. So when the circle doesn't contain the origin, we still have to do some work to show that the integral equals 0.

5. Originally Posted by namelessguy
We're not supposed to use the Cauchy-Goursat's Theorem on this problem because we haven't learned it yet. So when the circle doesn't contain the origin, we still have to do some work to show that the integral equals 0.
You can do it by finding an anti-derivative.
That is my above post.