TITLE: Why doesn't $\frac 1 z$ have an antiderivative in $\mathbb{C}\setminus\{0\}$?
QUESTION [11 upvotes]: Why doesn't $\frac 1 z$ have an antiderivative in $\mathbb{C}\setminus\{0\}$? I understand that the antiderivative could've been $\operatorname{Log}(z)$, but it always has atleast one branch cut. But what if we modify the domain of the $\operatorname{Log}$ function to $-\pi < \theta \leqslant \pi $?
Doesn't this make the anti-derivative of $\frac 1 z$ definable? Although this antiderivative will be discontinuous, but still valid. So, why is this not the case?
Edit: Can we answer this question without invoking the Fundamental Theorem of Contour Integration?

REPLY [6 votes]: Doesn't this make the anti-derivative of $\frac 1 z$ definable? Although this antiderivative will be discontinuous, but still valid. So, why is this not the case?

An antiderivative of a function $f$ on a domain $G$ is a function $F$ on $G$ such that $F'=f$ on $G$, and in particular $F$ must be differentiable at each point in $G$.  If a function is differentiable at a point, it is also continuous at that point.  So your candidate for an antiderivative is not one, being nondifferentiable at each point of discontinuity.  
Other answers have already pointed out a slicker way to see that such an antiderivative is impossible, but one could also use your approach of working with a specific logarithm to see it is impossible to have a global antiderivative.   Suppose there is one, and that $F'(z) = \dfrac1z$ on $\mathbb C\setminus\{0\}$.  Let $\operatorname{Log}$ be defined on $\mathbb C\setminus\{0\}$ by $\operatorname{Log}(z)=\log|z|+i\theta$, with $z=|z|e^{i\theta}$ and $-\pi<\theta\leq \pi$.  Then on $\mathbb C\setminus(-\infty,0]$, $(F-\operatorname{Log})'=0$, from which it follows that $F=\operatorname{Log}+C$ for some constant $C$.  By continuity of $F$ (following from differentiability), $$F(-1)=\lim\limits_{y\searrow 0}F(-1+yi)=\operatorname{Log(-1)}+C = \pi i+C,$$ and $$F(-1)=\lim\limits_{y\searrow 0}F(-1-yi)=\lim\limits_{y\searrow 0}\operatorname{Log}(-1-yi)+C=-\pi i+C.$$  This contradiction shows such an $F$ can't exist.
(I am taking for granted that $\operatorname{Log}'(z)=\frac1z$ on $\mathbb C\setminus(-\infty,0]$, which follows from the appropriate version of the inverse function theorem.  I am also taking for granted that $g'=0$ on a connected open set in $\mathbb C$ implies $g$ is constant.)