Any complex function that is analytic on an open set is differentiable on that set.
But can a function fail to be analytic on an open set but still be differentiable?
For example, the function $f(z)=z|z|^2$ is not analytic on an open set. Is it differentiable? The Cauchy-Riemann equations are satisfied only at $z=0$, which isn't an open set.
How do you determine if a complex function is differentiable?
Best Answer
It is a little subtle. Even if a function satsifies Cauchy-Riemann's equations, it's not necessarily $\mathbb{C}$-differentiable. Take for example $$ f(z) = \begin{cases} \exp(-1/z^4), & z \neq 0 \\ 0 & z = 0 \end{cases}. $$ As an exercise, you can check that $f$ satsifies C-R everywhere (it's pretty much obvious outside the origin), but $f$ is certainly not an entire function. It isn't even continuous at $z=0$. The problem is that Cauchy-Riemann's equation only care about restrictions of $f$ to vertical and horizontal lines.
However, if you know that $f$ is $\mathbb{R}$-differentiable and satisfies C-R at a point, then $f$ is also $\mathbb{C}$-differentiable there. Your example has smooth ($C^\infty$) real and imaginary part, and is thus $\mathbb{R}$-differentiable. In other words, for your particular example, checking Cauchy-Riemann's equations is suffcient.