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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

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  1. Loser66
    • one year ago
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    @oldrin.bataku

  2. anonymous
    • one year ago
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    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
    • one year ago
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    Give me a concrete example, please.

  4. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Now, let \(z =\sqrt2 /2 + i\sqrt 2/2 \) hence arg z = pi/4, How can I define alpha?

  7. anonymous
    • one year ago
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    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
    • one year ago
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    Should I take alpha = pi/4 also? hence the region is \(C - \{re^{i\pi/4}\} , r\geq 0\)

  9. Loser66
    • one year ago
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    |dw:1443754605557:dw|

  10. anonymous
    • one year ago
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    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
    • one year ago
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    I didn't say that. I said if \(z = \sqrt 2/2 + i\sqrt 2/2\) then \(arg z = \pi/4\)

  12. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Is there any video clip I can study ? I want to completely understand what it is. Please.

  15. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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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