Loser66
  • Loser66
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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katieb
  • katieb
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anonymous
  • anonymous
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' \(2\pi i\)
Loser66
  • Loser66
We can't just say that on the proof, right? we need an algebraic proof for this, right?
Loser66
  • Loser66
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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thomas5267
  • thomas5267
https://en.m.wikipedia.org/wiki/Complex_logarithm#Branches_of_the_complex_logarithm Wikipedia!
Loser66
  • Loser66
Thanks for the link @thomas5267
thomas5267
  • thomas5267
I think the problem is that you have to choose a branch of \(arg(z)\) or else it won't work.
Loser66
  • Loser66
We cannot choose a case "it won't work" to prove the whole thing "won't work". I think.
anonymous
  • anonymous
\(\log(z) = \log |z|+ i \arg(z)\) 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, \(z=-1\))
Loser66
  • Loser66
So, we can prove the principal branch doesn't work on S, right?
thomas5267
  • thomas5267
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.
Loser66
  • Loser66
like what oldrin says, let z =-1, then Log z = log|z|+iarg z= i(pi + 2kpi)
Loser66
  • Loser66
The function log z = ln |z| + i arg z is continuous on \(\mathbb C\setminus \{re^{i\alpha}, r\geq 0\}\)
thomas5267
  • thomas5267
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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