Given $f(z) = f(x+iy)=u(x,y)+iv(x,y)$ where $i$ is imaginary, we have the Cauchy-Riemann equations $u_x = v_y$ and $u_y = -v_x$
If the C-R equations are not satisfied, then $f(z)$ is differentiable nowhere.
However, if they are satisfied, then does this mean that $f(z)$ differentiable everywhere or is further testing required?
Thanks