Why is there no continuous log function on $\mathbb{C}\setminus\{0\}$?

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.

Solution 1:

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 $\int _\gamma\frac{1}{z} dz=2i\pi$.

Solution 2:

Such a logarithm would be a continuous section of the covering space $exp:\mathbb C\to \mathbb C^*$, hence would be a homeomorphism. But $\mathbb C$ and $\mathbb C^*$ are not homeomorphic .

Solution 3:

Consider the logarithm of $e^{it}$ where $t$ is a real number.

$\ln(e^{it}) = it + 2\pi i k$ where $k$ is some integer. Since for each $t$ this choice of $k$ is a discrete choice, if your logarithm is continuous, $k$ would have to be constant.

So $\ln(e^{it}) = it + 2 \pi i k$ but this is not the same for $t=0$ and $t=2\pi$, so $\ln$ is not well-defined.

So there's one continuity argument I haven't filled in, regarding $k$ having to be constant. But that's not too tricky. I suppose the easiest way to show $k$ is constant is to look at the difference $\ln(e^{it}) - it$, differentiate it and notice that's zero.