Understanding Jänich´s proof of Cauchy's integral formula for a circular disk

Question

I want to understand a particular proof. For the proof we need this lemma.

Lemma: Let $f: U \rightarrow \mathbb{C}$ be holomorphic and $Q$ be a rectangle in $\mathbb{C}$ as above, with boundary curve $\gamma$, but instead of the condition $Q \subset U$ let now be given a $C^{1}$-mapping $\varphi: Q \rightarrow U$. Then $$ \int_{\varphi \circ \gamma} f(z) d z=0 . $$

Now I formulate the Theorem:

Cauchy's integral formula for a circular disk: Let $f$ be holomorphic in an open set containing the circular disk $\{z\mid |z-z_0| \leq r\}$. Then for any $a$ in the interior ($|a-z_0| < r$) of the circular disk $$ f(a)=\frac{1}{2 \pi i} \int_{\left|z-z_0\right|=r} \frac{f(z)}{z-a} d z. $$

And now the proof. It is from the Complex Analysis Book by Klaus Jänich:

Proof: From Cauchy's integral theorem for rectangular images it follows that $$ \int_{\left|z-z_0\right|=r} \frac{f(z)}{z-a} d z=\int_{|z-a|=\varepsilon} \frac{f(z)}{z-a} d z $$ holds for sufficiently small $\varepsilon\leq r-|a-z_0|$. So this integral is independent of $\varepsilon$, in particular it is equal to $$ \lim_{\varepsilon \rightarrow 0}\int_{|z-a|=\varepsilon}\frac{f(z)}{z-a} d z=\lim_{\varepsilon \rightarrow 0}\int_{|z-a|=\varepsilon}\frac{f(z)-f(a)}{z-a}dz+\lim_{\varepsilon \rightarrow 0}\int_{|z-a|=\varepsilon}\frac{f(a)}{z-a}dz. $$ The first of the two limits on the right-hand side is zero because the integrand remains bounded; we must calculate the second: For $\gamma(t):=a+\varepsilon e^{i t}$ we have to calculate $\int_\gamma \frac{f(a)}{z-a} d z$, which is $$ \int_0^{2 \pi} \frac{f(a)}{\varepsilon e^{i t}} \cdot \varepsilon i e^{i t} d t=2 \pi i f(a), $$ and the Cauchy formula is proved. $\square$

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My questions:

I do not understand why the conclusion from Cauchy's integral theorem for rectangular images he made is true and what it means and how he derived it. Why does the integral not depent on $\varepsilon$ and why does he take the limit to zero? Then he splits the integral in two parts and I get the second one but why is the first one zero. He says, because ,,the integrand remains bounded" but what does this mean? If anyone could provide a more detailed version of this particular proof this would be a great help. Thank you.

Answer 1

1. Using polar coordinates, the annular region between two circles can be tiled by many smaller regions (polar cells) whose oriented boundaries are $C^1$ images of true rectangles. The Lemma tells us that the integral $\oint_C \frac{f(z}{z-a}$ must vanish on each oriented cellular boundary $C$. By pairwise cancellation of such terms at shared edges, we deduce that the outer circular integral agrees with the inner circle (once orientations for the two circles are chosen to match). That justifies deforming the circular to a small one, and shows that the answer must be independent of the small radius.

(2) The first integral $\frac{ f(z)- f(a)}{z-a}$ stays bounded because this difference quotient can be estimated well by $f'(a) $ (the correction being a a term that is also bounded since it tends to zero as $\epsilon \to 0$.