Let \(\gamma\) be a smooth curve joining two distinct points \(a, b\in \CC\).
Prove that the function \[ f(z) \definedas \int_\gamma {g(w) \over w-z} \,dw \] is analytic in \(\CC\setminus\gamma\).
Solution
Goal: For a smooth curve \(\gamma\) joining distinct points \(a, b\) and continuous \(g\) on \(\gamma\), prove \(f(z) = \int_\gamma \frac{g(w)}{w-z}\, dw\) is analytic on \(\CC \setminus \gamma\).
1. Fix \(z_0 \notin \gamma\); let \(d = \operatorname{dist}(z_0, \gamma) > 0\) and \(r = d/2\).
\(\gamma\) is compact and \(z_0\) is not on it, so the distance \(d\) between them is strictly positive.
2. For \(\abs{z - z_0} < r\) and \(w \in \gamma\), the geometric series \(\frac{1}{w - z} = \sum_{n=0}^\infty \frac{(z - z_0)^n}{(w - z_0)^{n+1}}\) converges uniformly in \(w\).
\(\abs{w - z_0} \geq d = 2r\) while \(\abs{z - z_0} < r\), so the ratio \(\abs{\frac{z - z_0}{w - z_0}} \leq \frac{r}{2r} = \frac12 < 1\) uniformly, and the geometric series converges uniformly on the compact set \(\gamma\).
3. \(f(z) = \sum_{n=0}^\infty (z - z_0)^n \int_\gamma \frac{g(w)}{(w - z_0)^{n+1}}\, dw\) for \(\abs{z - z_0} < r\).
Substitute the uniformly convergent expansion of 2 into the definition of \(f\) and interchange sum and integral (justified by uniform convergence on \(\gamma\); \(g\) is continuous, hence bounded on \(\gamma\)).
4. \(f\) is analytic at \(z_0\).
3 represents \(f\) near \(z_0\) as a convergent power series in \(z - z_0\), which is analytic in its disk of convergence; hence \(f\) is analytic at \(z_0\).
Since \(z_0 \notin \gamma\) was arbitrary, \(f\) is analytic on \(\CC \setminus \gamma\).
5. Q.E.D.
4 proves the claim.