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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

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  1. anonymous
    • one year ago
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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' \(2\pi i\)

  2. Loser66
    • one year ago
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    We can't just say that on the proof, right? we need an algebraic proof for this, right?

  3. Loser66
    • one year ago
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    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.

  4. thomas5267
    • one year ago
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    https://en.m.wikipedia.org/wiki/Complex_logarithm#Branches_of_the_complex_logarithm Wikipedia!

  5. Loser66
    • one year ago
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    Thanks for the link @thomas5267

  6. thomas5267
    • one year ago
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    I think the problem is that you have to choose a branch of \(arg(z)\) or else it won't work.

  7. Loser66
    • one year ago
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    We cannot choose a case "it won't work" to prove the whole thing "won't work". I think.

  8. anonymous
    • one year ago
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    \(\log(z) = \log |z|+ i \arg(z)\) 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, \(z=-1\))

  9. Loser66
    • one year ago
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    So, we can prove the principal branch doesn't work on S, right?

  10. thomas5267
    • one year ago
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    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.

  11. Loser66
    • one year ago
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    like what oldrin says, let z =-1, then Log z = log|z|+iarg z= i(pi + 2kpi)

  12. Loser66
    • one year ago
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    The function log z = ln |z| + i arg z is continuous on \(\mathbb C\setminus \{re^{i\alpha}, r\geq 0\}\)

  13. thomas5267
    • one year ago
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    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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