## Loser66 one year ago How to prove there is no branch of logarithm on C \{0} Help me, please

1. Loser66

@ikram002p @dan815

2. anonymous

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

3. Loser66

You mean log z?

4. anonymous

yes

5. Loser66

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

6. Loser66

*undefined

7. anonymous

you have a theorem that says something like if $$f$$ is analytic on a simply connected region then $\int_{\gamma}f(z)dz=0$

8. Loser66

No! I didn't study integral yet.

9. anonymous

oh then this is not going to work sorry

10. anonymous

dang

11. Empty

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

12. anonymous

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

13. Loser66

where is 4 to see? :)

14. Loser66

oh, problem 21, right?

15. Loser66

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

16. anonymous

it's in there somewhere as are the details,

17. Loser66

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

18. 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

19. Loser66

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

20. 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

21. Loser66

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

22. Loser66

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

23. anonymous

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

24. 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

25. Loser66

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

26. 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

27. 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

28. Loser66

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

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