How to show alternate definition of holomorphic holds: there exists $a$ such that $f(z_{0}+h)-f(z_{0})-a h=h \psi(h)$? I am trying to justify a basic claim made about holomorphic functions which comes directly from the definition.

Definition. Let $\Omega$ be an open set in $\mathbb{C}$ and $f$ a complex-valued
function on $\Omega$. The function $f$ is holomorphic at the point
$z_{0} \in \Omega$ if the quotient
$$
 \frac{f\left(z_{0}+h\right)-f\left(z_{0}\right)}{h} \tag{1}$$ converges to a
limit when $h \rightarrow 0$. Here $h \in \mathbb{C}$ and $h \neq 0$
with $z_{0}+h \in \Omega$, so that the quotient is well defined. The
limit of the quotient, when it exists, is denoted by
$f^{\prime}\left(z_{0}\right)$, and is called the derivative of $f$ at
$z_{0}:$ $$ f^{\prime}\left(z_{0}\right)=\lim _{h \rightarrow 0}
 \frac{f\left(z_{0}+h\right)-f\left(z_{0}\right)}{h} $$

The author then goes on to say

It is clear from $(1)$ above that a function $f$ is holomorphic at
$z_{0} \in \Omega$ if and only if there exists a complex number $a$
such that  $$  f(z_{0}+h)-f(z_{0})-a h=h
 \psi(h), \tag{2} $$ where $\psi$ is a function defined for all small h and $\lim_{h\to0} \psi(h) = 0$

I would like to show that $(2)$ follows from $(1)$. Assume
$$
f'(z)=\lim _{h \rightarrow 0} \frac{f(z+h)-f(z)}{h}
$$
exists. Then by the definition of limit, there exists an $a\in \mathbb{C}$ with
\begin{align}
\forall \varepsilon>0, \exists \delta>0 \quad \text{ s.t. }\quad|z-h|<\delta 
&\implies\left|\frac{f(z+h)-f(z)}{h}-a\right|<\varepsilon
\\
&\iff \frac{|f(z+h)-f(z)-a h|}{\mid h \mid}<\varepsilon
\\
&\iff |f(z+h)-f(z)-a h|<\varepsilon{\mid h \mid}
\end{align}
However, I am not sure how to get the complex numbers out of the modulus, and to turn the inequality into a strict equality between the complex function and some function defined for small h. How would I continue from here?
 A: I don't think an $(\varepsilon,\delta)$ approach will be useful here, since proving the criterion provided by the author requires you to show that there is a function $\psi$ satisfying $\lim_{h\to 0}\psi(h)=0$ and
$$f(z_0+h)-f(z_0)-ah=h\psi(h)$$
for every $h$ whose modulus is sufficiently small, and the $(\varepsilon,\delta)$ definition does not immediately produce such a function. I think it's better to follow Kavi Rama Murthy's approach (in the comments) and start by defining $\psi$ with
$$\psi(h)=a-\frac{f(z_0+h)-f(z_0)}{h}$$
and observing that this immediately implies $\lim_{h\to 0}\psi(h)=0$ (this follows from the assumption that $f$ has derivative $a$ at $z_0$) and, after multiplying through by $h$ and rearranging,
$$f(z_0+h)-f(z_0)-ah=h\psi(h)$$
We can get the converse by starting with the assumption that there exists $a\in\mathbb{C}$ and a function $\psi$ satisfying $\lim_{h\to 0}\psi(h)=0$ and
$$f(z_0+h)-f(z_0)-ah=h\psi(h)$$
for every $h\neq 0$ sufficiently small, and working backwards to get
$$\lim_{h\to 0}\frac{f(z_0+h)-f(z_0)}{h}=a$$
