Prove that $f(z)=|z|^2$ is nowhere analytic.

Question

Prove that $f(z)=|z|^2$ is nowhere analytic in z=0.

My attempt:

Let $z=x+iy$ then $f(z)=x^2+y^2=u(x,y)+iv(x,y)$ with $u,v:\mathbb{R}^2\rightarrow\mathbb{R}$

Moreover, we know that

$u_x=2x\quad u_y=2y\quad v_y=0\quad v_x=0$

The Cauchy-Riemann equation suggest that

$2x=0$ and $2y=0$ and this happen iff $x=y=0$ and this implies $z=0$

Here i'm stuck. Can someone help me?

Answer 1

It follows from what you did (together with the continuity of the partial derivatives) that $f$ is differentiable at $0$ and only at $0$. But when an analytic function is differentiable at a point $z_0$, it is also differentiable at every point of some neighborhood of $z_0$.