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 to zero 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 that the conditions (1) and (2) are equivalent.  The latter expresses the differentiability of $f$ at $z$.  By it one can sayt that the increment of $f$ is “locally proportional” to the increment of $z$.  Cf. the consideration of differential of real functions.

Title	derivative and differentiability of complex function
Canonical name	DerivativeAndDifferentiabilityOfComplexFunction
Date of creation	2014-02-23 18:20:58
Last modified on	2014-02-23 18:20:58
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	3
Author	pahio (2872)
Entry type	Definition
Classification	msc 30A99