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

- Loser66

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

- schrodinger

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- Loser66

- anonymous

if there were such a branch, it would be the anti derivative of \(\frac{1}{z}\)

- Loser66

You mean log z?

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- anonymous

yes

- Loser66

but if we take z is in real and negative, then log z is not continuous on the region, right?

- Loser66

*undefined

- 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

No! I didn't study integral yet.

- anonymous

oh then this is not going to work sorry

- anonymous

dang

- Empty

Wait, there isn't a branch of log(z) on C\{0} ?

- anonymous

here, this is a more long winded explanation, see 4

##### 1 Attachment

- Loser66

where is 4 to see? :)

- Loser66

oh, problem 21, right?

- Loser66

Question: What is the link between G and G'?

- anonymous

it's in there somewhere as are the details,

- Loser66

Why is G' subset of G? I am sorry for being dummy. I don't get it

- 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

cheating is good enough but understanding the solution is better. :)

- 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

I think I got why G' is subset of G. Thanks for the link.

- Loser66

This is my interpretation, { negative number } > {0} , hence C \ {negative number}< C -{0}

- anonymous

oh it is a subset because it is
the negative real numbers are a subset of the complex numbers without 0

- 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

I got the leftover, just wonder how G' is subset of G. Again , thanks a ton.

- 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

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

You mean the restrict region must be a line, not a point? |dw:1443407568085:dw|

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