Can a complex function be complex-differentiable at a point and not in a neighborhood?

Question

Is it possible for a function $f:\mathbb{C} \to \mathbb{C}$ to be complex-differentiable at a point $z_0\in \mathbb{C}$ without being analytic in a neighborhood of $z_0$? How can we prove this?

Answer 1

Yes; try $f(z)=|z|^2$ ; then Cauchy-Riemann:

$u_x=2x; u_y=2y; v_x=v_y=0$, is satisfied only at $(x,y)=(0,0)$ , tho maybe that is not

exactly what you wanted?