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 $\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
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.