Cauchy integral formula

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.

The clockwise definitions in quarter-plane oblique boundary spectral functions require the minus sign in this reconstruction. Indeed integrating first in $k$ gives $i/(z-w)$, and clockwise boundary traversal has Cauchy integral $-2\pi i{q}_z(z)$. Reversing both boundary spectra or both reconstruction-ray orientations reverses this sign; mixing the two conventions gives the negative of the desired holomorphic function.

// Target: analysis.bigb

For a function holomorphic on a disk containing the contour, its Taylor series converges uniformly on the integration circle. Integrating termwise leaves only the term with exponent $-1$, giving the displayed coefficient extraction. The power is $-n-1$, not $-n$; the latter extracts the coefficient at level $n-1$.

For a continuously differentiable function on a bounded planar domain, the Cauchy-Pompeiu formula supplements the Cauchy integral formula with an area integral of its Wirtinger derivative. Apply the Generalized Stokes theorem to $f(\zeta )d\zeta /(\zeta -z)$ on a punctured domain; the small-circle integral supplies $2\pi if(z)$. If the boundary term vanishes at infinity, it gives the Cauchy-Green operator representation $f(z)={\pi }^{-1}\int f_{\bar{\zeta }}(\zeta )/(z-\zeta )dA$.

Let $g$ be a meromorphic function with simple poles ${\alpha }_{+}$ above $c$ and ${\alpha }_{-}$ below $-c$, and $zg(z)\to 0$ uniformly at infinity in $\mathrm{Im}z\ge -c$. The left-to-right line integrals $H(\lambda )=(2\pi i{)}^{-1}{\int }_{\mathbb{R}-ic}g(z)/(z-\lambda )dz$ and $J(\lambda )=-(2\pi i{)}^{-1}{\int }_{\mathbb{R}+ic}g(z)/(z-\lambda )dz$ satisfy $H=g-r_{+}/(\lambda -{\alpha }_{+})$ above $-c$ and $J=r_{+}/(\lambda -{\alpha }_{+})$ below $c$, where $r_{+}$ is the upper residue. Upper semicircle closure proves both formulas by the residue theorem; subtracting the principal part makes $H$ holomorphic at ${\alpha }_{+}$ too. Consequently $H+J=g$ on the intervening strip. No lower-half-plane decay is needed.

If $f$ is holomorphic on a neighbourhood of the closed disc $|z-a|\le R$ and $|f(z)|\le M$ on its boundary, differentiating the Cauchy integral formula gives

For an entire function $f(z)={\sum }_j{c}_j{z}^j$, equality $|c_n|={\max }_{|z|=r}|f(z)|/r^n$ at a fixed $r>0$ holds exactly when $f(z)=c{z}^n$. The coefficient is the average of $r^{-n}f(r{e}^{i\theta })e^{-in\theta }$. Equality in its modulus bound forces the continuous integrand to have constant modulus and argument, hence to be constant. The identity theorem then extends the monomial identity from the circle to the whole plane. The zero function and the constant case $n=0$ are included.