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Loser66
 one year ago
What is the branch cut?
Problem: Prove that there is no branch of the logarithm defined on C  {0}
Please, help
Loser66
 one year ago
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there 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
 one year ago
Best ResponseYou've already chosen the best response.0We can't just say that on the proof, right? we need an algebraic proof for this, right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0By contradiction, suppose there is a branch on S = C {0}, then \(\forall z \in S\) log z = logz + 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.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0https://en.m.wikipedia.org/wiki/Complex_logarithm#Branches_of_the_complex_logarithm Wikipedia!

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for the link @thomas5267

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0I think the problem is that you have to choose a branch of \(arg(z)\) or else it won't work.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0We cannot choose a case "it won't work" to prove the whole thing "won't work". I think.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\log(z) = \log z+ i \arg(z)\) 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
 one year ago
Best ResponseYou've already chosen the best response.0So, we can prove the principal branch doesn't work on S, right?

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0Since 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
 one year ago
Best ResponseYou've already chosen the best response.0like what oldrin says, let z =1, then Log z = logz+iarg z= i(pi + 2kpi)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0The function log z = ln z + i arg z is continuous on \(\mathbb C\setminus \{re^{i\alpha}, r\geq 0\}\)

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0Put it in another way, \(e^{i\theta}\) is surjective and not injective. \(e^{ix}=e^{iy}\not\implies x=y\).
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