Complex differentiability ($f^{\prime }(z)={\lim }_{h\to 0}(f(z+h)-f(z))/h$)

A complex-valued function is complex differentiable at $z_0$ ifexists. The same limit must hold for every approach, including curved paths. Equivalently,The error divided by $|h|$ tends to zero. Thus the derivative is a single complex multiplier, not an arbitrary real linear map of the plane. If the real coordinate map is differentiable, complex differentiability is equivalent to its satisfying the Cauchy-Riemann equations at that point.

The simplest failure is complex conjugation, $f(z)=\overline{z}$. At every $z_0$ its increment quotient is $\overline{h}/h$. For real $h=t\ne 0$ it is $1$, while for $h=it$ it is $-1$. More generally, $h=t{e}^{i\theta }$ gives $e^{-2i\theta }$. Although complex conjugation is a smooth real reflection and has every real directional derivative, these incompatible quotients rule out a complex derivative everywhere.

Agreement on two axes is necessary, but does not by itself establish a limit over the whole complex plane. The Cauchy-Riemann equations need the real differentiability condition above, or a sufficient regularity condition such as continuous partial derivatives near the point. There is a second distinction: $f(z)=|z{|}^2$ has complex derivative zero at zero because its quotient there is $\overline{h}\to 0$, but its Cauchy-Riemann equations fail at every nonzero point. It is complex differentiable at zero without being a holomorphic function on any neighbourhood of zero.

At zero, $[f(h)-f(0)]/h\to 1$, so the complex derivative exists. At a nonzero point, write $r=|z|$. The Cauchy-Riemann equations for $x/(1+r)$ and $y/(1+r)$ force $x^2=y^2$ and $xy=0$, impossible when $r>0$. Hence the function is complex differentiable only at zero and is holomorphic on no nonempty open subset. A complex derivative at one point gives no neighborhood of complex differentiability.