A specific consequence of Cauchy's integral formula If f is holomorphic in an open subset $G \subset \mathbb{C}$, and if $f'(a)\neq0$ for some $a \in G$, then there exists $r>0$ such that \begin{eqnarray}|f'(z)-f'(a)|<|f'(a)|,\end{eqnarray} for $z \in D(a,r)$ ($D$ for 'disk' with centre $a$, radius $r$).
The above is what I intend to prove.
I've tried to use Cauchy's integral formulae, i.e \begin{eqnarray}2\pi i f'(z)=\int_{\partial D(a,r)}\frac{f(w)}{(w-z)^2} \, dw,\end{eqnarray} or \begin{eqnarray}2\pi i f'(z)=\int_{\partial D(a,r)}\frac{f'(w)}{w-z} \, dw,\end{eqnarray}
but I don't get anywhere? If someone would give a hint I'd appreciate it.

Differentiate
$$
f(z)=f(a)+f'(a)(z-a)+\frac{f''(a)}{2} (z-a)^2 + \ldots
$$
and deduce
$$
f'(z)=f'(a)+f''(a)(z-a)+\ldots
$$
Since the power series $f''(a)(z-a)+\ldots$ vanishes at $a$ and $f'(a) \neq 0$, there exists a radius $r>0$ such that $|f''(a)(z-a)+\ldots| < |f'(a)|$ whenever $|z-a|<r$.
