Cauchy's integral Theorem proof clarify

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

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 $$\sup$$.