Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

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

share|cite|improve this answer
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 9:39

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 $|z-a|<r$.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.