Differentiability complex functions If u, v : Ω → C are C^1 functions in an open set Ω ⊂ C, and f = u + iv. Then if the Cauchy–Riemann equations is not satisfied at any points of Ω, does it mean that f is not differentiable?
 A: As I understand (I am not very experienced in complex-analysis), the Cauchy–Riemann equations must be satisfied in that point (if not, $f$ is not differentiable at that point). Even if these equations are satisfied, you need  $u$ and $v$  to be differentiable in the point $(x,y)$ to be sure that $f$ is diferenciable. (The famous Pollard example: $f(z) = \frac{(1+i)x^3 - (1-i)y^3}{x^2 + y^2}$. In the point $z=0$ , the C-R equations are satisfied, but $ f'(0) $ doesn't exist)
We speak about differentiability in a point of domain. If $f$ is differentiable at all points of domain, it is holomorphic.   You can also have the situation that a complex function is differentiable in one point (for example, $f(z) = |z|^2$ in $z=0$) but isn't in the other.
So, a function can stil be differentiable at some points of your domain if it is not in some of its points, but if C-R aren't satisfied in any point the domain, the function surely isn't analythic (holomorphic), and also isn't differetiable at that point.
