I have to show that $|z|^2$ is nowhere analytic, where $z=x+iy$.
I took $f(z) =|z|^2 = x^2+y^2$
$\implies u(x,y)=x^2+y^2$ and $v(x,y)=0$
Then I started off by checking whether the Cauchy-Riemann equations were satisfied, and got,
$u_x=2x$ , $u_y=2y$
$v_x=0$ , $v_y=0$
I concluded that the given function $f(z)$ can only be analytic at the origin. But since I had to prove that it is nowhere analytic I decided to check whether or not it is differentiable at the origin.
Using the definition of differentiability I found out that the $f(z)$ is differentiable at zero.
But since it is not differentiable in a neighbourhood of zero therefore it cannot be analytic at zero and hence is nowhere analytic.
Have I done this correctly?
Best Answer
Yes.
But since $u$ and $v$ are $C^1$, it's enough to check Cauchy-Riemann's equation. They are satsified only at $z=0$, hence there is no open set where the function is $\mathbb{C}$-differentiable, and the function is therefore nowhere analytic.