# 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.

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I don't see a question. Perhaps you intended to ask how to prove the statement in the first sentence? – joriki Sep 21 '12 at 9:17
Yes of course. My bad :) – Fuente Sep 21 '12 at 9:19
Probably you can differentiate the power series for $f$ centered at $a$ and estimate the remainder when $z$ is very close to $a$. – Siminore Sep 21 '12 at 9:20
I will try to do that. Thanks. – Fuente Sep 21 '12 at 9:22

$f'$ is continous as $f$ is holomorphic. So we can apply the definition of continuity with $\varepsilon:=\frac{|f'(a)|}2$.
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 $|z-a|<r$.
You don't need Cauchy's integral formula for this. A holomorphic function is infinitely differentiable, so its derivative is continuous. Your inequality is an instance of the $\delta$-$\epsilon$-definition of the continuity of $f'$, with $\delta=r$ and $\epsilon=|f'(a)|\gt0$.