1. ## 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 ?

2. 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 ?

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

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

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

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

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

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

Tonio

3. 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

4. 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 $\displaystyle w(x,y)= x^2+ y^2$. Now, $\displaystyle \frac{\partial^2 w}{\partial x^2}+ \frac{\partial^2 w}{\partial y^2}= 2+ 2= 4$ but $\displaystyle 4|x+ iy|^2= 4(x^2+ y^2)$. They are not equal.

5. ya ya it is true
thank you

6. Originally Posted by K A D C Dilshan
ya ya it is true
thank you
Then how do you explain my example?

7. 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 ....

8. 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)