I try to translate a statement that I've read in a translated „Introduction to Complex Analysis” book...
Let me define $f$ as a continuous complex valued function on a set $D\subset\mathbb{C}$, where $g'=f$ (so the complex derivative of $g$ is $f$). If $\Gamma\subset D$, then
$$\int_{\Gamma}f\left(z\right)dz=g\left(\gamma\left(b\right)\right)-g\left(\gamma\left(a\right)\right),$$
where $\gamma$ is a representation of $\Gamma$. Specially, if $\Gamma$ is a simple closed curve, then
$$\int_{\Gamma}f\left(z\right)dz=0.$$
The proof was the following:
$$\int_{\Gamma}f\left(z\right)dz\dot{=}\int_{a}^{b}f\left(\gamma\left(t\right)\right)\gamma'\left(t\right)dt=\int_{a}^{b}g'\left(\gamma\left(t\right)\right)\gamma'\left(t\right)dt=\int_{a}^{b}\left(g\left(\gamma\left(t\right)\right)\right)'dt=g\left(\gamma\left(b\right)\right)-g\left(\gamma\left(a\right)\right),$$
where (I guess) we used the definition of complex integral, the chain rule and as a last step the Newton-Leibniz formula. In this case if $\Gamma$ is a closed curve, then $\gamma\left(a\right)=\gamma\left(b\right)$, so
$$\int_{\Gamma}f\left(z\right)dz=g\left(\gamma\left(b\right)\right)-g\left(\gamma\left(a\right)\right)=g\left(\gamma\left(b\right)\right)-g\left(\gamma\left(b\right)\right)=0.$$
However, the „specially” part feels strange to me. I would say the following counter example, which is definitely not a counter example, but I dont know what I miss from the statement...
If $\Gamma\dot{=}C\left(0,1\right)$ is the complex unit circle, then $$\int_{\Gamma}\frac{1}{z}dz=\int_{0}^{1}\frac{1}{e^{2\pi it}}2\pi ie^{2\pi it}dt=2\pi i\neq0.$$
So why isn't it a counter example? Is the proof not correct at some point? $C\left(0,1\right)$ is a simple closed curve, $f$ is continuous on e.g. $D\dot{=}\left\{ z:z\in\mathbb{C},\frac{1}{2}<\left|z\right|<\frac{3}{2}\right\}$, and $\Gamma\subset D$.
$f:z\mapsto\frac{1}{z}$ does not have a primitive $g$ on any circular neighbourhood. You might be thinking that $g=\log $ works, but it doesn't, because you will always run into the following problem: let's say, we choose the argument of a complex number to be between $-\pi$ and $\pi$, choosing $\log(-1)=\pi i$. As $z\to-1$ from above the real axis, $\log z\to\pi i$, but as $z\to-1$ from below we have $\log z\to-\pi i$. So $g$ isn't even continuous, never mind differentiable.