Let $f(x+iy)=u(x,y)+i\,v(x,y)$
- Cauchy-Riemann Equations are satisfied at $z_0$
- $u, v, u_x, u_y, v_x, v_y$ are defined on some open neighbourhood of $z_0$
- $u, v, u_x, u_y, v_x, v_y$ are continuous at $z_0$
$\implies f'(z_0)$ exists.
Can you show me a complex-valued function $f$ such that $f'(z_0)$ exists but at least one of $(2)$ or $(3)$ is not satisfied? In other words, is there a counterexample to the converse?
The book went through a great deal of work to show that the converse is true if $f$ is analytic at $z_0$. So there is a possibility things fall apart in general.
Best Answer
Let $g:\mathbb{C}\to\mathbb{C}$ be an arbitrary bounded function. Then for $f(z)=z^2g(z)$, by definition, $f'(0)$ exists and is $0$. However, you may choose $g$ as pathological as possible, so that $u$ and $v$ have no continuity at all on $\mathbb{C}\backslash\{0\}$.