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Loser66

  • one year ago

How to prove there is no branch of logarithm on C \{0} Help me, please

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  1. Loser66
    • one year ago
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    @ikram002p @dan815

  2. anonymous
    • one year ago
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    if there were such a branch, it would be the anti derivative of \(\frac{1}{z}\)

  3. Loser66
    • one year ago
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    You mean log z?

  4. anonymous
    • one year ago
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    yes

  5. Loser66
    • one year ago
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    but if we take z is in real and negative, then log z is not continuous on the region, right?

  6. Loser66
    • one year ago
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    *undefined

  7. anonymous
    • one year ago
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    you have a theorem that says something like if \(f\) is analytic on a simply connected region then \[\int_{\gamma}f(z)dz=0\]

  8. Loser66
    • one year ago
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    No! I didn't study integral yet.

  9. anonymous
    • one year ago
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    oh then this is not going to work sorry

  10. anonymous
    • one year ago
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    dang

  11. Empty
    • one year ago
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    Wait, there isn't a branch of log(z) on C\{0} ?

  12. anonymous
    • one year ago
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    here, this is a more long winded explanation, see 4

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  13. Loser66
    • one year ago
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    where is 4 to see? :)

  14. Loser66
    • one year ago
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    oh, problem 21, right?

  15. Loser66
    • one year ago
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    Question: What is the link between G and G'?

  16. anonymous
    • one year ago
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    it's in there somewhere as are the details,

  17. Loser66
    • one year ago
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    Why is G' subset of G? I am sorry for being dummy. I don't get it

  18. anonymous
    • one year ago
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    oh now i have to explain it too ? i thought cheating was good enough' ok lets see, seems that \(G'\) .. actually that is a real good question

  19. Loser66
    • one year ago
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    cheating is good enough but understanding the solution is better. :)

  20. anonymous
    • one year ago
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    OH i didn't read carefully because \(z<0\) made no sense, but upon more careful reading it says \(z\in \mathbb{R}\)so it is negative real numbers

  21. Loser66
    • one year ago
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    I think I got why G' is subset of G. Thanks for the link.

  22. Loser66
    • one year ago
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    This is my interpretation, { negative number } > {0} , hence C \ {negative number}< C -{0}

  23. anonymous
    • one year ago
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    oh it is a subset because it is the negative real numbers are a subset of the complex numbers without 0

  24. anonymous
    • one year ago
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    the gimmick seems to be taking the limit as \(z=\to -1\) along the real line from the left and from the right, see that you get different answers

  25. Loser66
    • one year ago
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    I got the leftover, just wonder how G' is subset of G. Again , thanks a ton.

  26. anonymous
    • one year ago
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    yw i think the real idea, (if i remember correctly) is that in order to have a branch of the log, you have to cut the complex plane, not just delete one point

  27. anonymous
    • one year ago
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    btw short proof is that it would be the anti derivative of \(\frac{1}{z}\) making \[\int_{|z|=1 }\frac{dz}{z}=0\] but it isn't

  28. Loser66
    • one year ago
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    You mean the restrict region must be a line, not a point? |dw:1443407568085:dw|

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