Loser66
  • Loser66
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Loser66
  • Loser66
@oldrin.bataku
anonymous
  • 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 \(re^{i\alpha},r\ge 0\)
Loser66
  • Loser66
Give me a concrete example, please.

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anonymous
  • 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 \(z=re^{i\alpha}\)
anonymous
  • 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 \((-\infty,0)\)
Loser66
  • Loser66
Now, let \(z =\sqrt2 /2 + i\sqrt 2/2 \) hence arg z = pi/4, How can I define alpha?
anonymous
  • 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
Loser66
  • Loser66
Should I take alpha = pi/4 also? hence the region is \(C - \{re^{i\pi/4}\} , r\geq 0\)
Loser66
  • Loser66
|dw:1443754605557:dw|
anonymous
  • anonymous
what do you mean take \(z=e^{i\pi/4}\), you need to pick a range of angles to use for the plane
Loser66
  • Loser66
I didn't say that. I said if \(z = \sqrt 2/2 + i\sqrt 2/2\) then \(arg z = \pi/4\)
anonymous
  • 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
anonymous
  • anonymous
that doesn't narrow it down, @Loser66 -- that is true in any interval that contains \(\pi/4\), including: $$(-\pi,\pi),\quad(0,2\pi),\quad(\pi/4,2\pi+\pi/4),\text{ etc.}$$
Loser66
  • Loser66
Is there any video clip I can study ? I want to completely understand what it is. Please.
anonymous
  • 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 \(\arg z\)
anonymous
  • 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
anonymous
  • 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
anonymous
  • 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|
anonymous
  • 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 \(\alpha\)
Loser66
  • 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.

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