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Math Help - Complex Analysis Problem

  1. #1
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    Smile Complex Analysis Problem

    I spent half an hour to solve this problem with the aid of cauchy-rieman equations, but it did not work for me.
    Can somebody help me to solve this problem descriptively ?



    Thanks in advance
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  2. #2
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    Quote Originally Posted by K A D C Dilshan View Post
    I spent half an hour to solve this problem with the aid of cauchy-rieman equations, but it did not work for me.
    Can somebody help me to solve this problem descriptively ?



    Thanks in advance

    I'm not sure what you mean by 4\times|f(x+iy)|^2, , but if we write f(x+iy)=u(x,y)+iv(x,y) , then Re(f)=u\,,\,Im(f)=v , and then w=|f|=u^2+v^2 , and since f is analytic then

    ** u_x=v_y\,,\,\,u_y=-v_x

    ** w(x,y)=u^2+v^2 , so

    ** \frac{\partial w}{\partial x}=2uu_x+2vv_x=2uu_x-2vu_y

    ** \frac{\partial^2 w}{\partial x^2}=2u_x^2+2uu_{xx}+2v_x^2+2vv_{xx}=2u_x^2+2uv_{y  x}+2u_y^2-2vx_{yx}

    Now get, from symmetry, a similar expression for \frac{\partial^2w}{\partial y^2} and adding both and after cancelling a lot of stuff (remember! by continuity of the 2nd partial derivative we have that u_{xy}=u_{yx} and then same for the mixed der. of v) we get... 4(u_x^2+u_y^2)\neq 4|f(x+iy)|^2 , unless that "X" you wrote there has another meaning different from multiplication.

    Tonio
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  3. #3
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    Unhappy

    no that is simply the multiplication
    I did the same thing which you have you too done here
    but it did not give me an answer
    please
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  4. #4
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    What tonio is telling you is that you can't prove- it simply isn't true.

    A simple counter-example: suppose f(z)= z= x+ iy. Then w(x,y)= x^2+ y^2. Now, \frac{\partial^2 w}{\partial x^2}+ \frac{\partial^2 w}{\partial y^2}= 2+ 2= 4 but 4|x+ iy|^2= 4(x^2+ y^2). They are not equal.
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  5. #5
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    ya ya it is true
    thank you
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  6. #6
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    Quote Originally Posted by K A D C Dilshan View Post
    ya ya it is true
    thank you
    Then how do you explain my example?
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    Then how do you explain my example?
    I assumed the "It is true" remark refered to what you and tonio said .... If it does, all is well. If it doesn't, then there's nothing more can be said so all is well ....
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  8. #8
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    Smile

    Quote Originally Posted by mr fantastic View Post
    I assumed the "It is true" remark refered to what you and tonio said .... If it does, all is well. If it doesn't, then there's nothing more can be said so all is well ....
    It is not the same f(Z) it is the first derivative of f(Z) to be shown
    sorry for the mistake
    what we have to prove is 4 times the square of modulus of first derivative of f(z)
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