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Do functions with complex parts differentiate/integrate like usual?

  1. Aug 14, 2013 #1
    To get straight to the point: If I have a function f(
    x)= 3x2+2ix
    Would f'(x)= 6x+2i?

    More generally, does complex differentiation follow the same rules as normal differentiation? (Ie; power rule, quotient rule, product rule, chain rule etc.)
    Thanks!
     
  2. jcsd
  3. Aug 14, 2013 #2
    There are analogs of the power/product/quotient rules in complex analysis, but, in general, differentiation and integration of complex functions is a bit more subtle (see, for instance, the Cauchy-Reimann equations). The answer to your question is yes with a "but", but the "but" is large enough that the answer is effectively "no".
     
  4. Aug 14, 2013 #3
    Could you give me a quick example? Also is my example in the first post correct? Thanks :)
     
  5. Aug 14, 2013 #4

    D H

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    ##f(z) = \bar z## is the canonical example of a complex function that does not have a complex derivative.

    For a real function f(x), a limit exists at some point a if both ##\lim_{x\to a_-}\, f(x)## and ##\lim_{x\to a_+} \, f(x)## exist and if these two one-sided limits are equal. With real functions, you have to worry about are two paths, one from below, the other from above. There are an uncountable number of paths in complex space along which z can approach some point a. This means that the mere existence of the complex derivative at some point places very strong conditions on the nature of the function.
     
  6. Aug 14, 2013 #5

    Office_Shredder

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    cm, the properties of differentiation depend as much on the domain as the range of the function. For your example f(x) is x supposed to be a real number, or a complex number?

    Because if you have a function f from reals to the complex numbers, then to differentiate it you can differentiate the complex and real parts as separate real differentiable functions, and everything in the theory works exactly as it does in the real to real case (this requires proving of course, but it's not particularly hard). If it's from the complex numbers to the complex numbers then things are significantly different even though superficially they appear the same
     
  7. Aug 14, 2013 #6
    Okay I think I'm getting the jist of it, But could you, for example; differentiate the Gamma function Γ(z) on some continuous interval?
     
  8. Aug 14, 2013 #7

    SteamKing

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    See this article: http://en.wikipedia.org/wiki/Gamma_function

    For complex z, the derivative of [itex]\Gamma[/itex](z) is defined in terms of the polygamma function [itex]\psi[/itex](z), and exists for z > 0.
     
  9. Aug 15, 2013 #8

    mathwonk

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    it helps to actually define differentiation. for a function like the one you gave first, the derivative means the best real linear approximation. this is computed exactly as you said.

    but for a complex valued function of a complex variable, the complex derivative means the best complex linear approximation (in a precise sense of best).

    since "zbar" is real linear but not complex linear, it has a real derivative but not a complex derivative.

    in Riemann's treatment he expressed the real derivative in terms of dz and dzbar, and called a function complex differentiable if the dzbar part was zero.

    I have explained this in excruciating detail in a long post somewhere here.

    maybe this thread:

    https://www.physicsforums.com/showthread.php?t=563301&highlight=dzbar
     
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