I was trying to prove this using the method of contradiction, so,
I assumed on the contrary that their is such an analytic function $f=u+iv$ with $u(x,y)=x^2+y^2$.
Now since $f$ is analytic therefore it must be differentiable and must satisfy the Cauchy Riemann equations $u_x$=$v_y$ and $u_y$=$-v_x$.
So i started off by calculating $u_x=2x$ and using the first CR equation concluded that $v_y=2x$
=>$v=2xy+h(x)$
Similarly using the second CR equation i found out that $v_x =-2y$
=>$v=-2xy+g(y)$
=>$2xy+h(x)=-2xy+g(y)$
=>$4xy+h(x)=g(y)$
I am stuck at this point and can't figure out how to proceed further to get a contradiction.Any help will be highly appreciated.
Best Answer
An analytic function has real (and imaginary) part a harmonic function. That is, sum of second partials is zero. But for $x^2+y^2$ this sum is $4$.
Note I mean the two pure partials, not the mixed partial, in the sum. So $u_{xx}+u_{yy}=0.$