Existence of complex anti-derivative

So I'm currently reading about complex anti-derivatives. I was given these two questions:

Problem 1:
Does $f(z)= z^n, \quad n\in\mathbb N$ have an anti-derivative on $\mathbb C\setminus \{0\}$?

Solution:

Case 1.1: $n\neq -1$ We find $F(z)=\frac{1}{n+1}z^n$ to be an anti-derivative. (Since $f(z)$ is continuous on $\mathbb C\setminus \{0\}$ we can just integrate it. We also see $F'(z)=f(z)$, so we are good)

Case 1.2: $n=-1$ With $\gamma(t)=e^{2\pi i t}, \quad t\in[0,1]$ we find $\int_\gamma \frac{1}{z} dz = 2\pi i$ so, apparently this does not have an anti-derivative because of Cauchy's integral theorem.

Problem 2:
Does $f(z)= z^n, \quad n\in\mathbb N$ have an anti-derivative on $\mathbb C\setminus (-\infty, 0]$?

Case 2.1: $n\neq-1$ Basically the same argumentation as in Case 1.1.

Case 2.2: $n=-1$ Since $Log(z)$ is continuous on $\mathbb C\setminus (-\infty, 0]$ we find $F(z)=Log(z)$ and we verify with $F'(z)=f(z)$.

Questions:

Question 1: Apparently we use Cauchy's integral theorem to show that there isn't a anti-derivative in case 1.2, but how exactly do we use it?

At the moment I just think of it like that: The anti-derivative of $\frac{1}{z}$ would be $Log(z)$ but $Log(z)$ is not continuous on $\mathbb C \setminus \{0\}$ so it wouldn't fulfill $F'(z)=f(z)$. But this actually just tells me "Log(z)$ isn't the anti-derivative but it doesn't tell me that there isn't any at all.

Question 2: In Case 1.2 we get $\int_\gamma f(z)=2\pi i$ so what exactly does this tell us? Doesn't it mean we actually evaluated the integral? Is it possible to evaluate an integral without it having an anti-derivative?