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Visually Representing Complex Derivatives

  1. Oct 8, 2012 #1
    I'm curious how the derivative of a complex function can be represented visually. It is defined as the limit of (f(z[itex]_{0}[/itex] + Δz) - f(z[itex]_{0}[/itex]) / Δz as Δz approaches 0. Is it right to say that f(z[itex]_{0}[/itex] + Δz) represents a neighborhood of radius Δz around z[itex]_{0}[/itex] in this case? Does the derivative still represent instantaneous change as in real functions?
     
  2. jcsd
  3. Oct 8, 2012 #2
    I do not recommend the neighbourhood 'picture', because you are moving on to a more general definition of the derivative.

    Although the picture idea just about works for complex numbers it is better to work with the fundamental definition of the derivative.

    That is because the formula you quote is pretty general and used for vectors, tensors and many other mathematical objects.

    Many objects, such as vectors, do not reside in the same space as the independent variables, but in their own space. This is reflected in complex transformations from the xy to vu planes.
     
  4. Oct 8, 2012 #3
    Then using a vector representation of the derivative definition, Δz is a vector that approaches zero to find the limit of the function. What if z=x+iy is viewed as a point rather than a vector?
     
  5. Oct 8, 2012 #4
    You might like to revies this thread, it is a bit off topic but post#13 explains better what I mean.

    https://www.physicsforums.com/showthread.php?t=640080&highlight=vector

    That is exactly the point (pun intended).

    z=x+iy is a point in one space but something else in another.

    Complex calculus is simpler than vector calculus because the spaces in the transformation are the same, as with real analysis. That is xy and uv are both complex planes.
     
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