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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.

Fuente
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3 Answers3

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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 $\epsilon=|f'(a)|\gt0$.

joriki
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  • Thank you all for the answers. I was so excited to have learned CIF that I couldn't see the forest for the trees. – Fuente Sep 21 '12 at 09:39
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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 $|z-a|<r$.

Siminore
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$f'$ is continous as $f$ is holomorphic. So we can apply the definition of continuity with $\varepsilon:=\frac{|f'(a)|}2$.

Davide Giraudo
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