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Complex differentiable <-> real and imaginary parts satisfy C-R eqns and are cont.

  1. Oct 21, 2012 #1
    Complex differentiable <--> real and imaginary parts satisfy C-R eqns and are cont.

    Say we have a complex function f(z) we can break this in to real and imaginary parts:

    f(z)=u(x,y)+iv(x,y)


    In my book I am told the following:


    (1) f complex differentiable at z0 in ℂ --> the Cauchy Reimann equations are satisfied by the partials of u and v at z0


    I am also told seperately:


    (2) For some z0 in ℂ, if the partials of u and v exist and are continuous on an open set containing z0 and they satisfy the Cauchy Reimann equatiions, then f is complex differentiable at z0


    Why isn't this an iff thing? It seems like the only thing keeping it from being bidiriectional is the fact that in (2) I also have to check that the partials are continuous on an open set O containing z0. But this is where my doubt arises. I am also told in my book:


    (3) If u and v satisfy the Cauchy Reimann equations at z0 for some neighborhood O containingi z0, then u and v are both harmonic in O.


    Since being harmonic requires establishing an equality between the second partial derivatives, isn't such a thing only possible if the first partial derivatives are continuous (as continuity is a essential for differentiability)? So how if u and v are harmonic shouldn't their first partials be continuous? If (1) gaurantees the Cauchy Reimann equations are satisfied which allows me to invoke (3), then doesn't (1) also gaurantee that the partials are continuous?


    So why can't I simply say f is differentiable at z0 in ℂ ⇔ it's real and imaginary components have partial derivatives continuous on an open set O containing z0 and these partials satisfy the Cauchy Reimann equations. Why does this need to be broken in to two separate theorems?


    SIDE QUESTION: I know that if a complex function f is differentiable at z0 it is actually infinitely differentiable there. But are it's real and imaginary parts also infinitely differentiable? I am just getting so confused separating what I can infer about the complex function and what I can infer about its real and imaginary parts which are functions of real variables.
     
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  3. Oct 22, 2012 #2

    Erland

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    Re: Complex differentiable <--> real and imaginary parts satisfy C-R eqns and are con

    To conclude that f(z) is infinitely complex differentiable at a point z0 and that its real and imaginary parts have continuous partial derivatives of all orders in a neighborhood of z0, it is not sufficient to assume that f(z) is complex differentiable at z0, but we need to assume that this complex differentiability hold for all points in a neighborhood of z0, which is the same as saying that f is analytic (or holomorphic) at z0.

    In your proposed equivalence, you must therefore not talk only about a single point z0, but about all points in some region in C.

    For example, if we put f(z)=z^2 if |z| is rational and 0 otherwise, then f'(0) exists and is 0 at 0, and the first order partial derivatives of the real and imaginary parts of f exist and satisfy the Cauchy-Riemann equtaions at 0 (they are all 0 there), but your proposed theorem fails there, for the partial derivatives do not exist at any point other than 0, and furthermore, f is not complex differentiable or even continuous at any other point and f is only complex differentiable one time at 0.
    Your proposed theorem therefore fails for this f, but the original implication that complex differentiability at a point implies Cauchy-Riemann at that point is true also for this f.
     
    Last edited: Oct 22, 2012
  4. Oct 22, 2012 #3
    Re: Complex differentiable <--> real and imaginary parts satisfy C-R eqns and are con

    ok, I think I see where I was going wrong. So in (3) that I had posted originally, this is the condition where the partials satisfy the C-R equations throughout a neighborhood containing z0, which means they are continuous in that neighborhood and f is holomorphic at z0. So the equivalence holds if f is holomorphic at z0, not simply complex differentiable.


    FUrthermore, it seems you can have a function f that is complex differentiable at z0, so the C-R equations are satisfied at z0, but the partials are not necessarily harmonic. This happens when f is complex differentiable at z0 but not holomorphic at z0
     
  5. Oct 24, 2012 #4

    Bacle2

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    Re: Complex differentiable <--> real and imaginary parts satisfy C-R eqns and are con

    Take f(z)=|z|^2 = x^2+y^2 .

    Then u_x=2x ; u_y=2y ; v_x=0=v_y , u_x(0)=v_y ; u_y(0)=v_x , and

    u_xx=2 =u_yy.
     
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