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

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. $\gamma: [0,1] \rightarrow A$.
