 Solution

A holomorphic function on the open set $\Omega$ is complex differentiable at every point of $\Omega$: the limitexists independently of the direction of approach. Write $f(x+iy)=u(x,y)+iv(x,y)$. The Cauchy-Riemann equations areTo prove necessity, approach first with real $h$, obtaining $f^{\prime }(z)=u_x+i{v}_x$. Approach instead with $h=it$ for real $t$, obtainingEquality of real and imaginary parts yields the displayed Cauchy-Riemann equations.

A sufficient condition is that $u$ and $v$ have continuous first partial derivatives throughout $\Omega$ and satisfy the Cauchy-Riemann equations there. Indeed, real differentiability gives, for $h=a+ib$,The Cauchy-Riemann equations imply $f_y=i{f}_x$, so this is $f_{x}h+o(|h|)$. Dividing by $h$ proves complex differentiability and $f^{\prime }(z)=f_x(z)$ at every point. Merely having first partial derivatives and the equations at a point, without a suitable differentiability hypothesis, would not justify this argument.