In our lecture notes for an undergraduate course on complex analysis, we have the following:
Let $D$ be a non-empty, open subset of $\mathbb{C}$.
A function $f:D \rightarrow \mathbb{C}$ is said to be complex differentiable at $z_0 \in D$ if the limit
$$f'(z_0) = \lim_{z \rightarrow z_0}\frac{f(z)-f(z_0)}{z-z_0}$$
exists. We note that $f:D \rightarrow \mathbb{C}$ is complex differentiable at $z_0 \in D$ with $f'(z_0) = a$ if and only if there exists a function $g:D \rightarrow \mathbb{C}$ such that
$$f(z) = f(z_0) + a(z-z_0) + |z-z_0|g(z)$$
and $\lim_{z \rightarrow z_0}g(z) = 0.$
I don't understand how the part following on from "We note that" follows. Can anyone provide an insight? Am I missing something easy?
The derivative exists iff $$ a = \lim_{z \to z_0} \frac{f(z)-f(z_0)}{z-z_0}, $$ or $$ 0 = \lim_{z \to z_0} \left( \frac{f(z)-f(z_0)}{z-z_0}-a \right), $$ or $$ 0 = \lim_{z \to z_0} \frac{f(z)-f(z_0)-a(z-z_0)}{z-z_0}, $$ which is equivalent to $$ 0 = \lim_{z \to z_0} \left| \frac{f(z)-f(z_0)-a(z-z_0)}{z-z_0} \right| $$ or $$ 0 = \lim_{z \to z_0} \frac{f(z)-f(z_0)-a(z-z_0)}{\lvert z-z_0 \rvert} \tag{1} $$ (where the absolute value functions go doesn't matter for complex limits that are zero). If we define $$ g(z) = \frac{f(z)-f(z_0)-a(z-z_0)}{\lvert z-z_0 \rvert} \quad (z \neq z_0), $$ we see that $(1)$ is equivalent to asking that $\lim_{z \to z_0} g(z) = 0$.
This is the notion that the derivative provides the best linear approximation to the function at a point. This form is useful for proving properties of derivatives that look like they require dividing by zero (the chain rule being a good example).