= 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 gives
$$
\frac1{2\pi i}\int_{|z-w|=r}\frac{f(z)}{z-w}\,dz
=\frac1{2\pi}\int_0^{2\pi}f(w+\rho e^{it})\,dt.
$$
As $\rho\downarrow0$, 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 $dz=ie^{it}dt$,
$$
I=\frac1{2\pi}\int_0^{2\pi}
\frac{\overline{f(e^{it})}}{1-ae^{-it}}\,dt.
$$
Taking <complex conjugates> and using $\overline{e^{it}}=e^{-it}$ gives
$$
\overline I=\frac1{2\pi}\int_0^{2\pi}
\frac{f(e^{it})}{1-\overline a e^{it}}\,dt
=\frac1{2\pi i}\int_{|z|=1}\frac{f(z)}{z(1-\overline a z)}\,dz.
$$
This 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 $\overline I=h(0)=f(0)$. Thus
$$
\boxed{I=\overline{f(0)}.}
$$
The <complex conjugates> in these formulas are essential; the boundary integrand contains $\overline{f(z)}$.
Back to article page
