Why doesn't $\frac 1 z$ have an antiderivative in $\mathbb{C}\setminus\{0\}$? 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?
 A: 
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.)
A: There cannot be an anti-derivative defined everywhere (except at $0$). 
Suppose it  would exist. The integral of $1/z$ over any closed curved would be $0$. This is however not the case, for example, for circles around the origin. Contradiction. 
A: You can show that $1/z$ does not have an antiderivative on $\mathbb{C}-0$ by integrating around the unit circle.  If it did have an antiderivative by the fundamental theorem of calclus the integral would be zero, but the Cauchy integral formula (or an explicit calculation) tells you it is $2\pi i$.  
As others have mentioned attempting to explicitly define an antiderivative as $\log z$ fails because you need to take a branch cut.  
