= Conjugated Cauchy integral on a circle
{title2=$\frac1{2\pi i}\int_{|z|=1}\frac{\overline{f(z)}}{z-a}\,dz=\overline{f(0)}$}
If $f$ is <holomorphic> on a neighborhood of the closed <unit disc> and $|a|<1$, this boundary integral is independent of $a$. Parametrizing the <unit circle> by $z=e^{it}$ and taking a <complex conjugate> gives
$$
\overline{\frac1{2\pi i}\int_{|z|=1}\frac{\overline{f(z)}}{z-a}\,dz}
=\frac1{2\pi i}\int_{|z|=1}\frac{f(z)}{z(1-\overline a z)}\,dz=f(0),
$$
where the last equality is the <Cauchy integral formula> applied to the <holomorphic function> $f(z)/(1-\overline a z)$ at zero. The conjugation of the whole integral includes the factor $1/i$ and the differential; parametrization keeps their signs consistent. Equivalently, the conjugated boundary values have only nonpositive Fourier frequencies, so the Cauchy kernel detects their constant coefficient.
Back to article page
