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Loser66
 one year ago
log z= logz + 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
Loser66
 one year ago
log z= logz + 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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the point is that you can choose any \(\alpha\), and you'll get a valid branch of \(\log z\) with a branch cut along \(re^{i\alpha},r\ge 0\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Give me a concrete example, please.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if 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 \(z=re^{i\alpha}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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 \((\infty,0)\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Now, let \(z =\sqrt2 /2 + i\sqrt 2/2 \) hence arg z = pi/4, How can I define alpha?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but 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

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Should I take alpha = pi/4 also? hence the region is \(C  \{re^{i\pi/4}\} , r\geq 0\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what do you mean take \(z=e^{i\pi/4}\), you need to pick a range of angles to use for the plane

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I didn't say that. I said if \(z = \sqrt 2/2 + i\sqrt 2/2\) then \(arg z = \pi/4\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that'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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that doesn't narrow it down, @Loser66  that is true in any interval that contains \(\pi/4\), including: $$(\pi,\pi),\quad(0,2\pi),\quad(\pi/4,2\pi+\pi/4),\text{ etc.}$$

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Is there any video clip I can study ? I want to completely understand what it is. Please.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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 \(\arg z\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this 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/791pxRiemann_surface_log.svg.png

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that'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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0on 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 \(\alpha\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Thank 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.
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