A community for students.
Here's the question you clicked on:
 0 viewing
Loser66
 one year ago
How to prove there is no branch of logarithm on C \{0}
Help me, please
Loser66
 one year ago
How to prove there is no branch of logarithm on C \{0} Help me, please

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if there were such a branch, it would be the anti derivative of \(\frac{1}{z}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0but if we take z is in real and negative, then log z is not continuous on the region, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you have a theorem that says something like if \(f\) is analytic on a simply connected region then \[\int_{\gamma}f(z)dz=0\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0No! I didn't study integral yet.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh then this is not going to work sorry

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Wait, there isn't a branch of log(z) on C\{0} ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0here, this is a more long winded explanation, see 4

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0oh, problem 21, right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Question: What is the link between G and G'?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's in there somewhere as are the details,

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Why is G' subset of G? I am sorry for being dummy. I don't get it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh 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
 one year ago
Best ResponseYou've already chosen the best response.0cheating is good enough but understanding the solution is better. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OH 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
 one year ago
Best ResponseYou've already chosen the best response.0I think I got why G' is subset of G. Thanks for the link.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0This is my interpretation, { negative number } > {0} , hence C \ {negative number}< C {0}

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh it is a subset because it is the negative real numbers are a subset of the complex numbers without 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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
 one year ago
Best ResponseYou've already chosen the best response.0I got the leftover, just wonder how G' is subset of G. Again , thanks a ton.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yw 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
 one year ago
Best ResponseYou've already chosen the best response.0btw 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
 one year ago
Best ResponseYou've already chosen the best response.0You mean the restrict region must be a line, not a point? dw:1443407568085:dw
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.