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Complex differentiation

  1. Oct 28, 2006 #1
    I can't follow a vertain step in my book in complex analysis (Kwok):

    [tex]\frac{d \bar{z}}{dz} = lim _{\Delta z \rightarrow 0} \frac{\Delta \bar{z}}{\Delta z} = lim _{\Delta z \rightarrow 0} e^{-2i Arg \Delta z}[/tex]

    I can't see how you could arrive at this last expression, could anyone give me a hint as how to do this?
  2. jcsd
  3. Oct 28, 2006 #2

    matt grime

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    I guess you're thinking about this in the form z=x+iy. If you can't get that to work, think of z as r*exp{iArg(z)} instead, because then it is immediate.
  4. Oct 28, 2006 #3
    How is it then immediate? I can't unambiguously see what happens when I vary z as in real analysis, do you vary r and argz independently in some way?
  5. Oct 28, 2006 #4

    matt grime

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    For any complex number w*/w (* for conjugate), is precisely exp{-2iArg(w)}. That is why I said it was immediate.

    The point is that that limit is not well defined. It depends on how delta(z) tends to zero. Just as in real analysis you need limits to be independent of the way you let things tend to zero (or some point). This is multivariable calc with some extra structure.
    Last edited: Oct 28, 2006
  6. Oct 28, 2006 #5
    Aha, now I see... and because this is so for any complex number it also holds in the variation of this number. Great, thx!
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