What is the branch cut? Problem: Prove that there is no branch of the logarithm defined on C - {0} Please, help

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What is the branch cut? Problem: Prove that there is no branch of the logarithm defined on C - {0} Please, help

Mathematics
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there is no branch defined on all of \(\mathbb{C}\setminus \{0\}\) because of the discontinuity along the negative real axis, where the argument 'jumps' \(2\pi i\)
We can't just say that on the proof, right? we need an algebraic proof for this, right?
By contradiction, suppose there is a branch on S = C -{0}, then \(\forall z \in S\) log z = log|z| + i arg z And we need prove it leads to something wrong. That is what I think but not know how to put it in need yet.

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https://en.m.wikipedia.org/wiki/Complex_logarithm#Branches_of_the_complex_logarithm Wikipedia!
Thanks for the link @thomas5267
I think the problem is that you have to choose a branch of \(arg(z)\) or else it won't work.
We cannot choose a case "it won't work" to prove the whole thing "won't work". I think.
\(\log(z) = \log |z|+ i \arg(z)\) is the principal branch; all other branches differ by a multiple of \(2\pi i\). show that they are not continuous let alone analytic at a point the negative real axis (say, \(z=-1\))
So, we can prove the principal branch doesn't work on S, right?
Since a complex number can have \(\operatorname{Arg}(z)+2\pi k\) as argument and it is still valid, in order to make the logarithm a function you have to choose a branch of argument. No branch of argument is continuous.
like what oldrin says, let z =-1, then Log z = log|z|+iarg z= i(pi + 2kpi)
The function log z = ln |z| + i arg z is continuous on \(\mathbb C\setminus \{re^{i\alpha}, r\geq 0\}\)
Put it in another way, \(e^{i\theta}\) is surjective and not injective. \(e^{ix}=e^{iy}\not\implies x=y\).

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