## Loser66 one year ago log z= log|z| + iarg z , $$\alpha\leq arg z\leq \alpha + 2\pi$$ Hence logz is analytic on $$\mathbb C -\{re^{i\alpha}\}, r\geq 0$$ My question: what is alpha? how to find it? Please, help

1. Loser66

@oldrin.bataku

2. anonymous

the point is that you can choose any $$\alpha$$, and you'll get a valid branch of $$\log z$$ with a branch cut along $$re^{i\alpha},r\ge 0$$

3. Loser66

Give me a concrete example, please.

4. anonymous

if you choose the range $$[\alpha,\alpha+2\pi)$$ for $$\arg z$$ then we get an analytic branch of $$\log z$$ aside from the branch cut along the ray $$z=re^{i\alpha}$$

5. anonymous

the standard example is with $$\alpha=-\pi$$, so we get arguments in $$[-\pi,\pi)$$ and a branch cut along the ray $$re^{-i\pi}=-r,r\ge 0$$, aka the negative real axis $$(-\infty,0)$$

6. Loser66

Now, let $$z =\sqrt2 /2 + i\sqrt 2/2$$ hence arg z = pi/4, How can I define alpha?

7. anonymous

but we could just as easily decide we want angles between $$0,2\pi$$ so $$\alpha=0$$, though this gives us instead a branch cut on the positive real axis $$(0,\infty)$$ since $$re^{i0}=r,r\ge 0$$ which is somewhat less convenient usually as we end up with no analyticity on the reals

8. Loser66

Should I take alpha = pi/4 also? hence the region is $$C - \{re^{i\pi/4}\} , r\geq 0$$

9. Loser66

|dw:1443754605557:dw|

10. anonymous

what do you mean take $$z=e^{i\pi/4}$$, you need to pick a range of angles to use for the plane

11. Loser66

I didn't say that. I said if $$z = \sqrt 2/2 + i\sqrt 2/2$$ then $$arg z = \pi/4$$

12. anonymous

that's what branch cuts have to do with; we have a Riemann surface that covers the plane however many times, and then we take a 'section' of it and deal with the 'defect' we get along the borders

13. anonymous

that doesn't narrow it down, @Loser66 -- that is true in any interval that contains $$\pi/4$$, including: $$(-\pi,\pi),\quad(0,2\pi),\quad(\pi/4,2\pi+\pi/4),\text{ etc.}$$

14. Loser66

Is there any video clip I can study ? I want to completely understand what it is. Please.

15. anonymous

the point is that if $$z=e^{i\theta}$$ then it follows $$z=e^{i(\theta+2\pi k)}$$ for all $$k\in\mathbb Z$$, so we need to fix some nonambiguous set of angles to use for $$\arg z$$

16. anonymous

the way to make $$\log z$$ close to analytic is to pick some contiguous interval like $$[\alpha,\alpha+2\pi)$$, which works well, except until we get close to an argument of $$\alpha+2\pi$$ and then it suddenly 'jumps' back down to $$\alpha$$, wrapping around

17. anonymous

this is the 'gap' you see between the different 'leaves' or branches of the complex logarithm here: https://upload.wikimedia.org/wikipedia/commons/thumb/a/ab/Riemann_surface_log.svg/791px-Riemann_surface_log.svg.png

18. anonymous

that's the reason analyticity fails along the ray $$\{re^{i\alpha}:r\ge 0\}$$ -- this is the ray over which the argument suddenly jumps by $$2\pi$$ in a discontinuous way: |dw:1443759197553:dw|

19. anonymous

on the left side of $$\alpha$$, the argument is very small, near $$\alpha$$; on the other side, the angle is just under $$\alpha+2\pi$$. at the ray, it suddenly jumps back down to $$\alpha$$

20. Loser66

Thank you so much. I think I got it. I was kicked out of the net while you were trying to help me. I apology for it.