 Conjugated Cauchy integral on a circle

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 giveswhere 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.