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Holomorphic function help with checking the solution

  1. Jan 29, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm sorry in advance but this will contain a lot of words to describe the soluton as I"m not good with Latex.

    z_=conjugate
    Show that f(z) = z^2 * z_ is not holomorphic in C. At which points is it
    complex-differentiable?
    I think I solved the problem I"m just looking for a second view.(or multiples).


    2. Relevant equations
    Cauchy–Riemann equations


    3. The attempt at a solution
    Ok first I wrote f(z)=z*z*z_=z*|z^2|=(x+yi)(x^2+y^2)
    Using the first Cauchy–Riemann equation and calculating the partial diff with respect to x of the real part then calculating the partial diff with respect to y for the imaginary part I get:
    2x^2+y^2=3yi+ix^2,this is not holomorphic but is complex diff at x=y=0.
    Is this correct?If not what should I do.(if you can provide answer with examples,or at least detailed, that with be great)
    Thank you.
     
  2. jcsd
  3. Jan 29, 2012 #2

    Dick

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    Your conclusion is right, but the partial derivatives are wrong and there is no 'i' in the Cauchy-Riemann equations. What are u(x,y) and v(x,y)? Try and clean that up.
     
  4. Jan 29, 2012 #3
    ...f(z)=x^3 +x*y^2+i(y*x^2+y^3).
    u(x,y)=x^3 +x*y^2
    v(x,y)=i(y*x^2+y^3)

    C-R=>2x^2+y^2=3y^2+x^2....this is equal for x=y=0...
    Is there anything else I should do?
     
  5. Jan 29, 2012 #4

    Dick

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    v(x,y)=y*x^2+u^3. There is no 'i' in v(x,y). There are two C-R equations. I assume you are doing u_x=v_y. If so your x partial derivative of u(x,y) is still wrong.
     
  6. Jan 29, 2012 #5
    sorry about that, typo mistake...3x^2+y^2=3y^2+x^2.....for any x=y I assume
     
  7. Jan 29, 2012 #6

    Dick

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    Yes, any x=y works. x=(-y) also works. But you aren't using the other CR equation.
     
  8. Jan 29, 2012 #7
    I can but if one of the C-R eq is not satisfied then the holomorphic part is done....
     
  9. Jan 29, 2012 #8

    Dick

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    True. But there are lots of solutions of the CR equation you just wrote down. That doesn't mean there are lots of points where the function has a complex derivative.
     
    Last edited: Jan 29, 2012
  10. Jan 29, 2012 #9
    Understood.Thank you
     
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