derivative and differentiability of complex function

Let $f(z)$ be given uniquely in a neighborhood of the point$z$ in $ℂ$.  If the difference quotient

$$\frac{\mathrm{\Delta }f}{\mathrm{\Delta }z}=\frac{f(z+\mathrm{\Delta }z)-f(z)}{\mathrm{\Delta }z}$$

tends to a finite limit $A$ as $\mathrm{\Delta }z\to 0$,then $A$ is the derivative of $f$ at the point $z$ and is denoted by

$f^{\prime }(z)=A=\underset{\mathrm{\Delta }z\to 0}{lim}{\displaystyle \frac{\mathrm{\Delta }f}{\mathrm{\Delta }z}}.$(1)

Thus the difference  $\lambda =\frac{\mathrm{\Delta }f}{\mathrm{\Delta }z}-A$  tends tozero simultaneously with $\mathrm{\Delta }z$, and $\mathrm{\Delta }f$ has the expansion

$$\mathrm{\Delta }f=A\mathrm{\Delta }z+\lambda \mathrm{\Delta }z.$$

If we denote  $|\mathrm{\Delta }z|=:\varrho$,  we have

$$\lambda \mathrm{\Delta }z=\frac{\lambda \mathrm{\Delta }z}{\varrho }\cdot \varrho =⟨\varrho ⟩\varrho$$

where $⟨\varrho ⟩$ means a complex number vanishing when$|\mathrm{\Delta }z|=\varrho \to 0$.  Consequently, (1) implies

$\mathrm{\Delta }f=A\mathrm{\Delta }z+⟨\varrho ⟩\varrho$(2)

in which  $A=f^{\prime }(z)$  and  $\varrho =|\mathrm{\Delta }z|$.  It’s easily seen thatthe conditions (1) and (2) are equivalent.  The latter expresses thedifferentiability of $f$ at $z$.  By it one can sayt that the increment of$f$ is “locally proportional” to the increment of $z$.  Cf. theconsideration of differential of real functions.