 Solution

Use the following boundary version of Cauchy's integral theorem: if a function is holomorphic on a neighborhood of a compact region bounded by finitely many piecewise smooth simple closed curves, its integral around the positively oriented boundary is zero. On an annulus this means that its integrals around the two circles, both taken counterclockwise, are equal.

For $0<\rho <r$, the function $f(z)/(z-w)$ is holomorphic on a neighborhood of $\rho \le |z-w|\le r$. The theorem therefore givesAs $\rho \downarrow 0$, continuity of $f$ makes the integrand converge uniformly to $f(w)$. The right side tends to $f(w)$, proving Cauchy integral formula without assuming that formula in an intermediate step.

Let $I$ denote the requested boundary integral. Parametrize the unit circle by $z=e^{it}$. Since $dz=i{e}^{it}dt$,Taking complex conjugates and using $\overline{e^{it}}=e^{-it}$ givesThis is the conjugated Cauchy integral on a circle identity. The function $h(z)=f(z)/(1-\overline{a}z)$ is holomorphic on a neighborhood of the closed unit disc, because $|a|<1$. Applying Cauchy integral formula at zero gives $\overline{I}=h(0)=f(0)$. ThusThe complex conjugates in these formulas are essential; the boundary integrand contains $\overline{f(z)}$.