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Complex derivative

  1. Aug 21, 2011 #1
    I'm solving complex derivatives of holomorphic functions, my question is if I take the derivative of a holomorphic function will the result always be holomorphic too?

    The examples I made always have a holomorphic result but I'm trying to find a counter example.
  2. jcsd
  3. Aug 21, 2011 #2
    Welcome to PF!

    How long have you been studying complex analysis? There is actually a major theorem on this subject whose content you can find on Wikipedia. A key part of the argument is the Cauchy integral formula, so you'll need some background in complex analysis to read the article.
  4. Aug 21, 2011 #3
    Thanks for being the first person to help me :)
    My experience with complex analysis is limited, but I know the general theorems.
    So if f'(z)=[itex]\partial[/itex]x f(z) and f'(z)= -i[itex]\partial[/itex]y f(z). Then [itex]\partial[/itex]x f(z) + i[itex]\partial[/itex]y f(z) = 0. So the derivative of z which is holomorphic will always result in a holomporphic function. Is this correct?
  5. Aug 21, 2011 #4
    Hm, I didn't understand your reply. Does [itex] \partial_x[/itex] mean [itex] \frac{\partial}{\partial x} [/itex]? It looks like you're doing something like the proof of the Cauchy-Riemann equations.

    Could you state your homework problem as it appears? Are you really to show that the derivative of a holomorphic function is holomorphic or to provide a counterexample?
  6. Aug 22, 2011 #5
    The actual question is: If f(z) is holomorphic and continuously differentiable in an open area [itex]\Omega[/itex], then it's complex derivative f'(z) is holomorphic. True or False?

    In my previous attempt ∂x means ∂/∂x.
  7. Aug 22, 2011 #6


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    Have you looked at the definition of "holomorphic"? A function is holomorphic (on a set) if and only if it is analytic for all x in that set. When a set is not specified it is typically taken to be all complex numbers.

    And, of course, one definition of "analytic" at a point is that its Taylor series exist at that point and converge to the value of the function in some neighborhood of the point. If a function is "holomorphic", it follows that its Taylor series exists and converge to the value of the function for all complex numbers.
  8. Aug 22, 2011 #7
    I disagree. That is not the definition of holomorphic, but rather an important theorem in complex analysis. This is the definition that I'm familiar with: http://planetmath.org/encyclopedia/Holomorphic.html [Broken].

    Are you supposed to provide a proof, Andz? If not, HallsofIvy's post contains the answer.
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