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Complex Differentiation?

  1. Jun 8, 2010 #1

    I was wondering if you have a complex function [tex]f: A \subset C \to C[/tex] everywhere differentiable on A, can you say anything about the existence of [tex]\frac{d Re(f)}{d z}[/tex] or [tex]\frac{d Im(f)}{d z}[/tex]?

    I was thinking that it's easy to proof that if [tex]f : A \subset R \to C[/tex], then Re(f) and Im(f) are differentiable, because

    [tex]\frac{Re(f(x)) - Re(f(a))}{x-a} = Re \left( \frac{f(x) - f(a)}{x-a} \right)[/tex]

    So [tex]\frac{d Re(f)}{d x} = Re \left( \frac{df}{dx} \right)[/tex]

    Is that correct?

    But generally, when [tex]A \subset C[/tex], I'm not sure how to find a counter-example or proof that it is always differentiable...

    We do have:

    [tex]\frac{f(x) - f(a)}{x-a} = \frac{Re(f(x)) - Re(f(a))}{x-a} + \frac{Im(f(x)) - Im(f(a))}{x-a}i[/tex], but all I can get out of that is that if either Re(f) or Im(f) is differentiable, then the other is also.

    Any ideas?

    EDIT: weird, I thought I had posted this in Analysis (which would seem more logical...), I hope this isn't a problem
    Last edited: Jun 9, 2010
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  3. Jun 8, 2010 #2


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  4. Jun 8, 2010 #3
    Well we haven't seen those yet, I'm a first year undergraduate. I'm sure they are general equations that will surely give me the result I'm looking for, but is there anyone who can respond to my questions directly? It's in an effort to get a bit more feeling with complex differentiation. Thanks for your post.
  5. Jun 8, 2010 #4


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    It is true, in the complex numbers as in the real numbers, that d(f+ g)/dz= df/dz+ dg/dz.

    In particular, if [itex]f(z)= Re(z)+ i Im(z)[/itex], then [itex]\frac{df}{dz}= \frac{dRe(z)}{dz}+ \frac{dIm(z)}{dz}[/itex]. Those two derivatives must exist.
  6. Jun 8, 2010 #5
    Our professor of Analysis stressed the fact slogans like "d(f+ g)/dz= df/dz+ dg/dz" must be used with caution, because they can only be used when you already know f and g are differentiable, so you can't use that to proof they are, right?

    For example, if f: R -> R: x -> |x| - |x|, then f is the zero function and obviously differentiable, but clearly the terms seperately aren't
  7. Jun 8, 2010 #6


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    It follows directly from the Cauchy-Riemann equations.
  8. Jun 8, 2010 #7
    I'd say those derivatives exist since for differentiable function [itex]f(z)=u(x,y)+iv(x,y)[/itex]:



    and those derivatives exists since [itex]f(z)[/itex] is differentiable by assumption.
  9. Jun 8, 2010 #8
    Hm... I'm sorry if I'm being high-maintenance here and I thank you all for trying to help, but I was wondering if there's a way to see they're differentiable without the theory of the Cauchy-Riemann equations or the use of partial derivatives (but for example with the definition of derivative or close to that)
  10. Jun 8, 2010 #9
    I'm not a mathematics major, but I don't think there is. I mean, if you've been exposed to multivariable calculus, then you should have learned that a limit can only be a limit if it's the limit coming from all directions. It's the same thing in complex differentiation. The limit has to be defined from all directions on the real and imaginary axes, otherwise differentiating doesn't make sense. The formal way of expressing this condition (to check) is the cauchy-riemann equations.
  11. Jun 8, 2010 #10
    But is there a reason why the principle of a limit being taken from infinite directions should demand the use of partial derivates? In my course of Analysis I we never saw one partial derivative (material of Analysis II), but we did see a little bit of complex differentiation, just straight with the definition of derivative.

    For example, in my OP I proved (I think, if it is correct) that in the case of f: R -> C, the Re(f) and Im(f) are differentiable if f is, I'm looking for a similar proof for f: C -> C.
  12. Jun 9, 2010 #11
    This is the definition of a partial derivative - the derivative of a function when you're approaching a point from different directions. So when you are talking about function of two variables (like in the case of complex functions) you must talk about partial derivatives. (However, if a complex function is analytic, then the derivative wrt z is not called a partial derivative)

    As to your question, taking a derivative wrt a complex variable z, is much different than taking a derivative wrt a real variable x, and even different than taking partial derivatives. The constraint is that if you want to define this derivative, then it must be independent of the direction you approach (not only that it has to be defined from any direction, it has to be independent). Some would describe it as "the function is respecting the complex structure of z". This constraint leads directly to the Cauchy-Riemann equation.

    For example, your expression


    Is undefined unless Re(f) is constant.

    The approach here is entirely different.
  13. Jun 9, 2010 #12
    Hm... But in Analysis I we never had to worry about partial derivatives. We simply had a function f: C -> C and then its derivative would be complex and thus hold all the information. Is this only possible with analytic functions? That seems odd (but possible...) But I don't see why [tex]\frac{d Re(f)}{dz}[/tex] should be ill-defined? It can approach it in infinite independent ways, no? (Look at my proof below to see what I actually mean with Re(f))

    I think I have found a way to show that (elibj123, maybe you could tell me if this is only valid for analytic functions?) if f is differentiable, then so is Re(f) and Im(f):

    So we can write out that:

    [tex]\frac{f(z) - f(a)}{z-a} = \frac{Re(f(z)) - Re(f(a))}{z-a} + \frac{Im(f(z)) - Im(f(a))}{z-a}i[/tex]

    And now the trick is to notice that if a certain complex sequence [tex]z_n = a_n + b_n i[/tex] converges to [tex]z_n \to z = a + bi[/tex], then (and only then) [tex]a_n \to a[/tex] and [tex]b_n \to b[/tex]. So if [tex]\frac{Re(f(z_n)) - Re(f(a))}{z_n-a}[/tex] and [tex]\frac{Im(f(z_n)) - Im(f(a))}{z_n-a}[/tex] did not converge for all sequences [tex]z_n \to a[/tex], then f couldn't be differentiable in a and we'd have a contradiction.

    Does that seem correct?
    Last edited: Jun 9, 2010
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