$\int_{\gamma}\frac{1}{z^2-2z}dz$

Question

Let $\gamma(t)=2+e^{it}, t \in[0,2\pi]$ find $\int_{\gamma}\frac{1}{z^2-2z}dz$

I did the following

$\int_{\gamma}\frac{1}{z^2-2z}dz=-\frac{1}{2}\int_{\gamma}\frac{1}{z}dz+\frac{1}{2}\int_{\gamma}\frac{1}{z-2}dz$

The integral $\int_{\gamma}\frac{1}{z-2}dz$ is equal to $2\pi i$ because $D(2,1)\setminus 2$ is not simple connected I don't know how to justify properly the fact that $\int_{\gamma}\frac{1}{z}dz$ is equal to $0$ I can't say the image of the curve is simple connected because we don't include $2$ and if I consider another curve- disc it has to have 0 as center which we can't include, and again it won't be a simple connected, so we can't use Cauchy theorem.

How can I justify properly that $\int_{\gamma}\frac{1}{z}dz=0$ without calculating the integral?

Answer 1

$z=0$ is outside of the contour, so $\frac{1}{z}$ is holomorphic throughout. What does that tell you?

Answer 2

Rewrite $\int_{\gamma} \frac{1}{z^2 - 2z} dz$ as $\int_{\gamma} \frac{1/z}{z -2} = \int_{\gamma} \frac{f(z)}{z - 2}.$

$f(z) = \frac{1}{z}$ is holomorphic everywhere on the complex plane except $z = 0,$ which is exterior to $\gamma,$ so $f(z)$ is holomorphic everywhere interior to $\gamma$ and we can directly apply the Cauchy Integration Formula: $\int_{\gamma} \frac{f(z)}{z - 2} = 2\pi i \cdot \frac12 = \pi i.$