= Solution
{c}
A <holomorphic function> on the <open set> $\Omega$ is <complex differentiable> at every point of $\Omega$: the <limit>
$$
f'(z)=\lim_{h\to0,\ h\in\mathbb C}\frac{f(z+h)-f(z)}h
$$
exists independently of the direction of approach. Write $f(x+iy)=u(x,y)+iv(x,y)$. The <Cauchy-Riemann equations> are
$$
\boxed{u_x=v_y,\qquad u_y=-v_x.}
$$
To prove necessity, approach first with real $h$, obtaining $f'(z)=u_x+iv_x$. Approach instead with $h=it$ for real $t$, obtaining
$$
f'(z)=\frac1i(u_y+iv_y)=v_y-iu_y.
$$
Equality 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 $h=a+ib$,
$$
f(z+h)-f(z)=f_xa+f_yb+o(|h|).
$$
The <Cauchy-Riemann equations> imply $f_y=if_x$, so this is $f_xh+o(|h|)$. Dividing by $h$ proves <complex differentiability> and $f'(z)=f_x(z)$ at every point. Merely having first <partial derivatives> and the equations at a point, without a suitable differentiability hypothesis, would not justify this argument.
Back to article page
