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Real/complex differentiable function

  1. Dec 2, 2011 #1
    1. The problem statement, all variables and given/known data

    Is the following function real and complex-differentiable everywhere?

    f(z) = Re(z)

    2. Relevant equations

    Cauchy-Riemann equations fy = ifx

    3. The attempt at a solution

    Let z = x + iy, z1 = x - iy

    Re(z) can be defined by Re(z) = (z + z1)/2

    A function is ℝ-differentiable if
    f(z+Δz) - f(z) = fx(z)Δx + fy(z)Δy + Ω(Δz)
    and
    Ω(Δz)/Δz -> 0 as Δz -> 0.

    Δz = Δx + iΔy


    So, f(z+Δz) = (z + Δz + (z + Δz)1)/2
    = x + Δx

    So f(z+Δz) - f(z) = Δx

    fx(z)Δx = ∂f/∂x (Δx) = ∂x/∂x (Δx) = (Δx)
    fy(z)Δy = 0

    f(z+Δz) - f(z) = fx(z)Δx + fy(z)Δx + Ω(Δz)
    So Δx = Δx + Ω(Δz)
    Thus Ω(Δz) = 0 and evidently 0/Δz -> 0 as Δz -> 0.

    So f(z) is ℝ-differentiable everywhere.


    I know from a theorem that if f is ℝ-differentiable and satisfies the Cauchy-Riemann equations then it is ℂ-differentiable.


    fy = ifx

    fy = 0
    ifx = i

    So f does not satisfy the equations, hence not ℂ-differentiable.


    Is this correct?
     
  2. jcsd
  3. Dec 2, 2011 #2
    You've shown that Re(z)y ≠ iRe(z)x. The Cauchy-Riemann condition is fy = ifx. What's the logical conclusion? ;o)
     
  4. Dec 2, 2011 #3
    I'm sorry? What do you mean?

    f(z) is Re(z).
     
  5. Dec 2, 2011 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    His point is that Re(z) does NOT satisfy the Cauchy-Riemann equations.
     
  6. Dec 2, 2011 #5
    So f is not complex differenctiable?
     
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