On page 152 H.A. Priestly introduction to complex analysis.

The theorem is stated as: Let $f$ be holomorphic inside and on a positively oriented contour $\gamma$. Then if $a$ is inside $\gamma$ then

\begin{aligned} f(a) = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(\omega)}{\omega - a} d \omega \end{aligned}

There is a step which I do not understand the book states that this is due to the Estimation theorem, but I can really not see how.. So it goes as follows:

\begin{aligned} &|\frac{1}{2 \pi i} \int_{0}^{2 \pi} \frac{f(a+re^{i \theta}) - f(a)}{re^{i \theta}} ire^{i \theta} d \theta|\\ &\leq \frac{1}{ 2 \pi} \cdot 2 \pi \cdot sup _{\theta \in [0, 2 \pi]} |f(a+re^{i \theta}) - f(a)| \end{aligned}

Any clarification would be much appreciated


1 Answer 1


Clearly, those $re^{i\theta}$ cancel each other, as well as those $i$'s. And, for each $\theta$, $|f(a+re^{i\theta})-f(a)|$ is smaller than or equal to the $\sup$ that appears in that inequality. So, yes, the estimation theorem is what you need to prove what you want to prove, since it says that if $f\colon[a,b]\longrightarrow\mathbb C$ is integrable and if there is a $M$ such that you always have $|f(x)|\leqslant M$, then $\bigl|\int_a^bf(x)\,\mathrm dx\bigr|\leqslant M(b-a)$. So, take $a=0$, $b=2\pi$, and $M$ equal to that $\sup$.

  • $\begingroup$ Yes I see that cancellation, but I am not sure how estimation theorem is applied here could you be more specific $\endgroup$
    – Chengdu
    May 25, 2019 at 19:00
  • $\begingroup$ I've edited my answer. What do you think now? $\endgroup$ May 25, 2019 at 19:13
  • $\begingroup$ ahh yes in that way it follows immediately thank you $\endgroup$
    – Chengdu
    May 25, 2019 at 19:21
  • 1
    $\begingroup$ I'm glad I could help. $\endgroup$ May 25, 2019 at 19:23

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