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

Question

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

loser66 · Answer

@ikram002p  @dan815

anonymous · Answer

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

loser66 · Answer

You mean log z?

anonymous · Answer

yes

loser66 · Answer

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

loser66 · Answer

*undefined

anonymous · Answer

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 · Answer

No! I didn't study integral yet.

anonymous · Answer

oh then this is not going to work sorry

anonymous · Answer

dang

empty · Answer

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

anonymous · Answer

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

loser66 · Answer

where is 4 to see?  :)

loser66 · Answer

oh, problem 21, right?

loser66 · Answer

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

anonymous · Answer

it's in there somewhere as are the details,

loser66 · Answer

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

anonymous · Answer

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 · Answer

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

anonymous · Answer

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 · Answer

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

loser66 · Answer

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

anonymous · Answer

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

anonymous · Answer

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 · Answer

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

anonymous · Answer

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 · Answer

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 · Answer

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