## Loser66 one year ago What is the branch cut? Problem: Prove that there is no branch of the logarithm defined on C - {0} Please, help

1. anonymous

there is no branch defined on all of $$\mathbb{C}\setminus \{0\}$$ because of the discontinuity along the negative real axis, where the argument 'jumps' $$2\pi i$$

2. Loser66

We can't just say that on the proof, right? we need an algebraic proof for this, right?

3. Loser66

By contradiction, suppose there is a branch on S = C -{0}, then $$\forall z \in S$$ log z = log|z| + i arg z And we need prove it leads to something wrong. That is what I think but not know how to put it in need yet.

4. thomas5267
5. Loser66

Thanks for the link @thomas5267

6. thomas5267

I think the problem is that you have to choose a branch of $$arg(z)$$ or else it won't work.

7. Loser66

We cannot choose a case "it won't work" to prove the whole thing "won't work". I think.

8. anonymous

$$\log(z) = \log |z|+ i \arg(z)$$ is the principal branch; all other branches differ by a multiple of $$2\pi i$$. show that they are not continuous let alone analytic at a point the negative real axis (say, $$z=-1$$)

9. Loser66

So, we can prove the principal branch doesn't work on S, right?

10. thomas5267

Since a complex number can have $$\operatorname{Arg}(z)+2\pi k$$ as argument and it is still valid, in order to make the logarithm a function you have to choose a branch of argument. No branch of argument is continuous.

11. Loser66

like what oldrin says, let z =-1, then Log z = log|z|+iarg z= i(pi + 2kpi)

12. Loser66

The function log z = ln |z| + i arg z is continuous on $$\mathbb C\setminus \{re^{i\alpha}, r\geq 0\}$$

13. thomas5267

Put it in another way, $$e^{i\theta}$$ is surjective and not injective. $$e^{ix}=e^{iy}\not\implies x=y$$.