TITLE: Why is there no continuous log function on $\mathbb{C}\setminus\{0\}$?
QUESTION [12 upvotes]: Over the years, I've often heard that there is no logarithm function which is continuous on $\mathbb{C}\setminus\{0\}$.
The usual explanation is usually some handwavey argument about following such a function around the unit circle, and getting a contradiction at $e^{2\pi i}$ or something.
I've been a little unsatisfied with these. What is a more formal, rigorous proof that there is no continuous log function on $\mathbb{C}\setminus\{0\}$ that is understandable to a nonexpert? Thanks.

REPLY [19 votes]: If a holomorphic function $f(z)$ on a domain $D\subset \mathbb C$ has a primitive $F(z)$ on $D$ , that is $F'(z)=f(z) $, then its integral over any piecewise differentiable  loop $\gamma$ in $D$ is zero :  $\int _\gamma f(z)dz=0$.  
If $\log (z)$ existed on $\mathbb C^*$, it would be a primitive of  $1/z$  [ to  get $\log'(z)=1/z$ just differentiate $\exp(\log(z))=z$ ]. Hence for the circle $\gamma (t)= e^{2i\pi t} \; (0\leq t \leq 1)$ we would have $\int _\gamma\frac{1}{z} dz=0$.    
However a trivial calculation shows that $\int _\gamma\frac{1}{z} dz=2i\pi$.