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Complex Analysis - Differentiability

  1. Apr 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that the function f defined by [itex]f(z) = 3\,{x}^{2}y+{y}^{3}-6\,{y}^{2}+i \left( 2\,{y}^{3}+6\,{y}^{2}+9\,x
    \right) [/itex] is nowhere differentiable.


    3. The attempt at a solution

    Computing the C.R equations for this, I am left with

    [itex]{y}^{2}+2\,y={\it xy}[/itex]

    and

    [itex]x^2+(y-2)^2 = 1[/itex]

    Now from the upper equation we can see that [itex]x={\frac {{y}^{2}+2\,y}{y}}[/itex]

    This means that there is a discontinuity at y = 0, and therefore this function is not differentiable?

    I somehow don't think is right as it would mean it's only not differentiable at that point, not necessarily everywhere, how would I go about this?
     
  2. jcsd
  3. Apr 24, 2012 #2
    x and y are real numbers right? So, is there ever a real solution to those two equations? That is can you substitute x=f(y) of one into the other, make a polynomial out of it, then solve for the roots? If those roots are only complex, then the CR equations are never satisfied.
     
  4. Apr 24, 2012 #3
    [tex]u_y = 3x^2 + 3y^2 - 12y = -9 = -v_x\Rightarrow x^2 + y(y -4) = -3[/tex]

    I think part of your CR has a mistake.
     
  5. Apr 24, 2012 #4
    I just completed the square, so y(y - 4) becomes (y-2)^2 and the addition of the 4 means the other side of equation becomes 1.
     
  6. Apr 24, 2012 #5
    Ah yes, I see now. Thanks for that!

    I got two sets of complex roots because I solved for two values of y, and when I substituted those, both results were complex. If only one were complex would that mean it's still differentiable but at that point?
     
    Last edited: Apr 24, 2012
  7. Apr 25, 2012 #6
    It would be differentialbe where ever x and y are real and satisfy the CR equations.
     
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