Proving differentiability of a complex function

Prove that if $f(z)$ is differentiable at $z_0$ then:

$f(z)=f(z_0)+f'(z_0)(z-z_0)+\lambda(z)(z-z_0)$, where $\lambda(z)\to 0$ as $z\to z_0$.

We begin by the definition:

\begin{equation} f'(z_0)=\lim_{\Delta z\to 0}\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z} \end{equation}

Then to:

\begin{equation} \lim_{\Delta z\to 0}f'(z_0)\Delta z=\lim_{\Delta z\to 0}f(z_0+\Delta z)-f(z_0) \end{equation}

set $\Delta z=(z-z_0)$ on the left side of =

\begin{equation} \lim_{(z-z_0)\to 0}f'(z_0)(z-z_0)=\lim_{\Delta z\to 0}[f(z_0+\Delta z)-f(z_0)] \end{equation}

But this is not the given results above.

Suppose that $f$ is defined in some neighbourhood $D$ of $z_0 \in \mathbb C$ and is differentiable at $z_0$. Let us define $\lambda : D \to \mathbb C$ by $$ \lambda (z): = \frac{{f(z) - f(z_0 )}}{{z - z_0 }} - f'(z_0 ) $$ if $D \ni z\neq z_0$ and $\lambda(z_0):=0$. Then by re-arranging the definition, $$f(z)=f(z_0)+f'(z_0)(z-z_0)+\lambda(z)(z-z_0).$$ It is clear from the definition of $f'(z_0)$ that $\lim_{z\to z_0}\lambda(z)=0$.