Loser66
  • Loser66
How to prove there is no branch of logarithm on C \{0} Help me, please
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Loser66
  • Loser66
@ikram002p @dan815
anonymous
  • anonymous
if there were such a branch, it would be the anti derivative of \(\frac{1}{z}\)
Loser66
  • Loser66
You mean log z?

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anonymous
  • anonymous
yes
Loser66
  • Loser66
but if we take z is in real and negative, then log z is not continuous on the region, right?
Loser66
  • Loser66
*undefined
anonymous
  • anonymous
you have a theorem that says something like if \(f\) is analytic on a simply connected region then \[\int_{\gamma}f(z)dz=0\]
Loser66
  • Loser66
No! I didn't study integral yet.
anonymous
  • anonymous
oh then this is not going to work sorry
anonymous
  • anonymous
dang
Empty
  • Empty
Wait, there isn't a branch of log(z) on C\{0} ?
anonymous
  • anonymous
here, this is a more long winded explanation, see 4
1 Attachment
Loser66
  • Loser66
where is 4 to see? :)
Loser66
  • Loser66
oh, problem 21, right?
Loser66
  • Loser66
Question: What is the link between G and G'?
anonymous
  • anonymous
it's in there somewhere as are the details,
Loser66
  • Loser66
Why is G' subset of G? I am sorry for being dummy. I don't get it
anonymous
  • anonymous
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
Loser66
  • Loser66
cheating is good enough but understanding the solution is better. :)
anonymous
  • anonymous
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
Loser66
  • Loser66
I think I got why G' is subset of G. Thanks for the link.
Loser66
  • Loser66
This is my interpretation, { negative number } > {0} , hence C \ {negative number}< C -{0}
anonymous
  • anonymous
oh it is a subset because it is the negative real numbers are a subset of the complex numbers without 0
anonymous
  • anonymous
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
Loser66
  • Loser66
I got the leftover, just wonder how G' is subset of G. Again , thanks a ton.
anonymous
  • anonymous
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
anonymous
  • anonymous
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
Loser66
  • Loser66
You mean the restrict region must be a line, not a point? |dw:1443407568085:dw|

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