1/z does not have an anti-derivative on $\mathbb C \setminus \{0\}$

Solution 1:

We have the following theorem that may help:

Theorem: Let $f: U \rightarrow \mathbb{C}$ be a complex valued function with $U \subset \mathbb{C}$ an open set. Suppose $\Gamma:[a,b] \rightarrow U$ is continuously differentiable and $F: U \rightarrow \mathbb{C}$ is a holomorphic function where $F'(z)=f(z)$ for all $z \in U$, then $$ \int_\Gamma f(z) dz = F(b) - F(a).$$

Suppose $f(z):= z^{-1}$ does have an anti-derivative $F$ on $\mathbb{C} \setminus \{0\}$, then by the above theorem with $$\Gamma:[0, 2\pi] \rightarrow \mathbb{C} \setminus \{0\} , ~ t \mapsto e^{it}$$ we have that $$\int_\Gamma f(z) dz = F(2\pi) - F(0) = 0.$$ If we calculate this however we note that $$\int_\Gamma f(z) dz = \int_0^{2\pi} e^{-it} . i e^{it} dt = \int_0^{2 \pi} i dt = 2 \pi i \neq 0$$ and we have a contradiction.