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This question is from Brown and Churchill's Complex Variables and Applications, 8ed., Section 52, Question 6.

Let $f(s)$ denote a continuous function taken along a simple contour, $C$ enclosing a region $D$, prove that the function:

$$ g(z) = \frac{1}{2 \pi\mathrm{i}}\int \frac{f(s)}{s-z} ds$$

is analytic $\forall z \in D$ and that

$$ g'(z) = \frac{1}{2 \pi\mathrm{i}}\int \frac{f(s)}{(s-z)^2} ds$$

I know to prove Cauchy's Integral formula I'm to assume that some ball with radius $R$ creates a region $D$ and $f(s)$ is holomorphic on $D$, except at the point $z$. I can create an open ball with radius $r$ centered at $z$. Then using a limit, I reduce the radius $r$ until it converges on the singular point $z$ and show that this limit (or residue?) is zero. i.e. Show:

$$\lim_{r\to 0}\, \left\lvert \int_{C_R} \frac{f(s)}{s-z}\ ds - 2\pi\mathrm{i} f(z) \right\rvert = 0. $$

The problem is that in the above proof I've assumed that $f(s)$ is holomorphic, therefore analytic everywhere therefore continuous everywhere. If I'm only allowed to assume that $f(s)$ is continuous how can I show that $f(s)$ is also analytic? Does this have something to do with existence of a derivative in an open ball around the singularity? How can I make the jump between continuity and analyticity?

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