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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.

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    Very closely related: http://math.stackexchange.com/questions/91131/why-there-is-no-continuous-argument-function-on-mathbbc-setminus-02012-02-01

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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.

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    Thank you Professor. Plugging in, $\ln(e^{i0})=0$ and $\ln(e^{i2\pi})=2\pi i+2\pi i k$. This implies $(2\pi i)(k+1)=0$, but what if $k=-1$? Does a further case of $t=4\pi$ need to be added to see that $4\pi i+2\pi i k=(2\pi i)(2+k)=0$ to see that $\ln$ is not well defined, or am I missing something?2012-02-01
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    $\ln(e^{i0}) = 2\pi i k$, not zero, well, it depends on your choice of $k$.2012-02-01
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    Here is a simpler way to look at this argument: Suppose that $\ln$ is defined on $C \0$. Then $\ln$ is continuous there, but so is $it$. It follows from here that the function $\ln*z-it$ is continuous and takes only values of the type $\{2 \pi ki\}$. But any such function must be constant...2012-02-01
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    Oh I see. Also, in differentiating $\ln(e^{it})-it$, does that assume $\ln$ is already differentiable? Is that is a problem if one is trying to show no such continuous function exists?2012-02-01
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    Well there is no need to differentiate, is you are familiar with the notion of connected sets. Or if you know which are the only open and closed subsets of $C - 0$... The point is that the point $2 \pi ki$ is closed, but you can also put a small ball around it so it meets no other point of this type.... And then the preimage has to be both open and closed...2012-02-01
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    @N.S. Thanks, I understand then why $\ln*z-it$ is continuous, but why would it only take values of type $\{2\pi ki\}$? I know it has that form for inputs of from $e^{it}$, but why should the outputs be of that form for all nonzero complex numbers?2012-02-01
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    That any continuous logarithm function has to be differentiable, this comes from the inverse function theorem (multi-variable calculus).2012-02-01
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    That any two values of log differ by an integer multiple of $2\pi i$, this comes from thinking about what the exponential map does -- write $e^{x+iy}$ in polar coordinates, for example.2012-02-01
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    They shouldn't be. $\:$ $\operatorname{ln}(\operatorname{exp}(i\cdot (-\pi)))$ and $\operatorname{ln}(\operatorname{exp}(i\cdot \pi))$ will be of that form with different values of $k$. $\:\:$ By the contrapositive of the intermediate value theorem, $\: t\mapsto (\operatorname{ln}(\operatorname{exp}(i\cdot t))+(-(i\cdot t))) \:$ is not continuous. $\:\:$ Therefore $\operatorname{ln}$ is not continuous. $\:\:\:$2012-02-01
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    Ok, lets make it more formal. Suppose by contradiction ln is continuous. Define a new function $g: R \rightarrow C$ by $g(t)= \ln(e^{it})-it$. This function is continuous on reals. Also, if $\ln(e^{it})=z$ it follows that $e^{z}=e^{it}$ and thus $e^{z-it}= 1$. This shows that $z-it$ is of the form $2k \pi i$. But since $z=\ln(e^{it})$, we just proved that this function only takes values of that type. Now, using the continuity of $g$, it follows that $g$ is constant... Thus, we proved that $\ln(e^{it})-it$ is constant which implies that $\ln(e^{i0})=\ln(e^{2 \pi i})-2 \pi i$. COntradiction2012-02-01
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    The contradiction being the fact that $e^{i0}=e^{2 \pi i}$.2012-02-01
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    Dear @N.S., I very much like your formal comment, thank you. Do you mind giving a quick explanation of why continuity of $g$ implies $g$ is constant here?2012-02-01
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    If $g$ takes two different values, when you multiply it by $i$, you get a continuous function from $R$ to $R$ whose image is only points of the type $2k \pi$. Now, if this function takes two different values, by the IVT it takes all the values in between (hence values not of the form $2 k \pi$.)2012-02-01
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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 .

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    -1, not nearly "understandable to a nonexpert"2012-02-01
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    Thanks, but I'm afraid this is over my head at this point.2012-02-01
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    Dear @Ricky, yours is a perfectly legitimate opinion, but I consider that the elementary theory of coverings is part of the basic training of a mathematician. Many institutions provide such a course at an undergraduate level. Anyway the point of view that you have to be an expert (in what?) to know about coverings strikes me as unusual.2012-02-01
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    Dear @Hallyu, I have added another answer based on complex functions that might be more to your liking.2012-02-01
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    I don't think it's bad to have lots of answers at different levels.2012-02-01
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    I don't agree with the downvoting, largely because I think that the "Accept" feature is a good enough way to identify which answer is most helpful to the OP, as distinguished from the answer with the widest appeal. But if it led to a third good answer to this question, it can't be all bad.2012-02-01
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    Thanks for your point of view, @Dylan...2012-02-01
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    ...and @Jonas (I split the sentence because the software prevents me from notifying two users in one comment)2012-02-01
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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 $\int _\gamma\frac{1}{z} dz=2i\pi$.

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    I am giving this alternative proof, bypassing coverings, for users more happy with complex analysis.2012-02-01