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Complex Limit

  1. Jan 25, 2012 #1
    Trying to remember how to use the definition of a complex limit.

    [tex]\lim_{\Delta z\to 0}\frac{f(z+h)-f(z)}{\Delta z}[/tex]

    [tex]f(z) = |z| = \sqrt{x^2+y^2}[/tex]

    [tex]\Delta z = \Delta x + i\Delta y[/tex]

    [tex]\lim_{\Delta x\to 0}\frac{\sqrt{(x+\Delta x)^2+(y+\Delta y)^2}- \sqrt{x^2+y^2}}{\Delta x}[/tex]

    Is that correct? Or do I just have the delta x with the x? Or is there a x + delta x and y + delta y?

    Thanks.
     
  2. jcsd
  3. Jan 25, 2012 #2

    HallsofIvy

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    Well, that is almost the correct formula. The only things missing are that you need to take the limit as both [itex]\Delta x[/itex], [itex]\Delta y[/itex] go to 0 and the denominator must be [itex]\sqrt{(\Delta x)^2+ (/Delta y)^2}[/itex], the distance from [itex](\Delta x, \Delta y)[/itex] to (0, 0), and you have to take the limit without assuming a relation between [itex]\Delta x[/itex] and [itex]\Delta y[/itex]. Strictly spealing, we take the limit as the point [itex](\Delta x, \Delta y)[/itex] goes to (0, 0) along any possible path in two dimensions.
     
  4. Jan 25, 2012 #3
    So the denominator is [tex]\Delta x + \Delta y[/tex] then?
     
  5. Jan 25, 2012 #4

    HallsofIvy

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    Sorry, you posted while I was editing my post- the denominator should be the distance from the point [itex](\Delta x, \Delta y)[/itex] to (0, 0):
    [tex]\sqrt{(\Delta x)^2+ (\Delta y)^2}[/tex]

    That is difficult to calculate. What is often done is to asssume that the function is differentable so that the limits along, say, parallel to the real axis or parallel to the imaginary axis are sufficient.
     
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