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

Question

Does $z^n, n\in\mathbb Z$ on $\mathbb C \{0\}$ have an anti derivative?

For $n\neq 1$ it's $\frac{1}{n+1}z^{n+1}$ but if $n=-1$ we get $1/z$, so:

$\int_\gamma 1/z dz = 2\pi i \int_0^1 dt = 2\pi i$

Since the integral isn't zero, we can say, that there is no anti-derivative.

Question: I don't get it. I know that there is Cauchys Theorem which states that the integral should be zero if there exists an anti-derivative (why?), but that doesn't help me. My argumentation would be, that $\mathbb C \{0\}$ isn't null-homotop so by cauchy the integral can't be zero and thus there isn't an anti derivative. But that's again, just technical "talking" and no real knowledge.

So, what's the problem here and why isn't there a anti-derivative of $1/z$ on $\mathbb C\setminus \{0\}$? On what set would there be an anti-derivative of $1/z$?

Answer 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 $$ \int_\Gamma f(z) dz = F(\Gamma(b)) - F(\Gamma(a)).$$

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 $$\int_\Gamma f(z) dz = F(\Gamma(2\pi)) - F(\Gamma(0)) = 0.$$ 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.

Answer 2

@S. Deward answered incredibly to your first question, so I'll just answer your second question: "On what set would there be an anti-derivative of $\frac{1}{z}$?".

If you look at the answer, you will realize how the problem arises from being able to draw a circle around 0 (the answer uses a circle of radius 1, but you can see that it's the same if we consider radius $r>0$ with the parametrization $re^{it}$). So basically the (maximal) domain we are looking is the whole complex plane minus a ray.

To see that is indeed correct, notice how in that domain, which under rotation we may assume WLOG $\mathbb{C}\setminus(-\infty,0]$ we can define an holomorphic logarithm (for example the principàl branch $Log$). Noticing that function satifies $z=e^{Log(z)}$ and deriving on both sides, we see that $$1=Log(z)'e^{Log(z)}=Log(z)'z\implies Log(z)'=\frac{1}{z}$$ as in the real case, so $Log(z)$ is an antiderivative for $\frac{1}{z}$ on $\mathbb{C}\setminus(-\infty,0]$, and any other larger domain that this one which is not equivalent (i.e., a domain which has no path connecting $0$ and $\infty$) to it won't have a antiderivative of $\frac{1}{z}$.

Hope this helped clarifying your mind about it.