Logarithm over complex numbers The logarithm function is not certainly defined for every $\text{Re}(z)\leq0$, but the question is where is it defined?
I also know $\displaystyle \int_{C} \frac{1}{z} dz\neq0$ where $C$ is the unit circle defined by $ \gamma(t)=e^{it} $ for $0\leq t\leq 2\pi$, which implies that $\frac{1}{z}$ has no antiderivative.
Is this true because any set $U\supset C$ contains $z\in \mathbb{C}:Re(z)\leq0$ ?
If $U$ does not contain $z\in \mathbb{C}:Re(z)\leq0$ , is $\log(z)$ "well behaved" in $U$?
 A: One thing you should do is to remove from your mind the concept of THE logarithm function, and replace it with A logarithm function. Once you've done that, then you can formulate a sensible question:

On what (open) subsets $U \subset \mathbb C - \{0\}$ does there exist a logarithm function $l : U \to \mathbb C$?

And you can even say precisely what you mean by this, namely that $l$ is holomorphic on $U$ and the composition $z \mapsto \exp(l(z))$, when restricted to $U$, is the identify function on $U$.
To start, here's some examples: for each angle $\theta \in \mathbb{R}$ a logarithm is defined on the domain $U_\theta$ obtained by removing the ray of angle $\theta$ from $\mathbb C$, which is given by the formula
$$U_\theta = \{r e^{2 \pi i \phi} \mid r > 0, \,\, \theta - 2\pi < \phi < \theta\}
$$
For more examples, every simply connected open subset $U \subset \mathbb C - \{0\}$ has a logarithm.
And here is a reasonably simple general answer to this question: a logarithm is defined on $U$ if and only if for every smooth Jordan curve $C$ in $U$, the curve $C$ does not wind around the origin, equivalently $\int_C \frac{dz}{z}=0$.
