Complex Analysis Problem

**K A D C Dilshan** · December 30th 2009, 05:10 PM

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

**tonio** · December 30th 2009, 07:08 PM

Originally Posted by K A D C Dilshan

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

**K A D C Dilshan** · December 31st 2009, 03:56 AM

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

**HallsofIvy** · December 31st 2009, 04:27 AM

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.

**K A D C Dilshan** · January 3rd 2010, 03:29 AM

ya ya it is true
thank you

**HallsofIvy** · January 3rd 2010, 04:05 AM

Originally Posted by K A D C Dilshan

ya ya it is true
thank you

Then how do you explain my example?

**mr fantastic** · January 3rd 2010, 04:09 AM

Originally Posted by HallsofIvy

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 ....

**K A D C Dilshan** · January 4th 2010, 03:09 PM

Originally Posted by mr fantastic

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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