Prove that $ \int_{\gamma} \frac{1}{z} dz = 0$

Question

Let $\displaystyle\gamma$ be a closed curve exactly located in $A =\mathbb C \setminus\{z\in\mathbb C: Re(z)\leq 0\}$.

I found a similar problem here : Find $\int_{\gamma}\frac{dz}z$, but they concluded that the value of the contour integral is $\displaystyle i\pi$.

How does the result changes for this problem?

Answer 1

This answer is quite topological. If you are familiar with differential forms, you’d notice $\frac {1}{z}$ is holomorphic in $A$. Hence $\frac {\mathrm{d}z}{z}$ is a closed form. Further notice, $A$ is simply connected, hence the the closed form $\frac {\mathrm{d}z}{z}$ is actually an exact form. So it must have $0$ integral on any closed path.

Answer 2

If you know the Cauchy's integral theorem you can use that $\gamma$ is closed and $\frac{1}{z}$ is holomorphic in $A$, so the theorem say that the integral is zero.

However I think the easy way is notice that in $A$ the principal branch of the complex logarithm is well defined, so you can use that $\frac{1}{z} = \frac{\mathrm{d}}{\mathrm{d}z} \log(z)$, so the integral is

$$ \int_{\gamma} \frac{1}{z} \,\mathrm{d}z = \log(\gamma(0)) - \log (\gamma(1)) = 0$$ because the curve is closed, so $\gamma(0)=\gamma(1)$.

Edit: Here I'm thinking in $\gamma$ parameterized in the unit interval, i.e. $\gamma: [0,1] \rightarrow A$.