The Cauchy–Pompeiu formula with fixed signs

For $0<r<\mathrm{dist}(z,\mathrm{\partial }D)$ set $D_r=D\setminus \overline{B_r(z)}$ and $g(\zeta )=f(\zeta )\mathrm{/}(\zeta -z)$ on its closure. Since $1\mathrm{/}(\zeta -z)$ is holomorphic there, [F3] gives ${\mathrm{\partial }}_{\bar{\zeta }}g=({\mathrm{\partial }}_{\bar{\zeta }}f)\mathrm{/}(\zeta -z)$. Writing $dg=({\mathrm{\partial }}_{\zeta }g)d\zeta +({\mathrm{\partial }}_{\bar{\zeta }}g)d\bar{\zeta }$ and using $d(gd\zeta )=dg\wedge d\zeta$, the repeated $d\zeta$ term vanishes, so $d(gd\zeta )=-({\mathrm{\partial }}_{\bar{\zeta }}f)\mathrm{/}(\zeta -z)d\zeta \wedge d\bar{\zeta }$. This is the needed type component of $d$ from [F4].

The continuous derivative ${\mathrm{\partial }}_{\bar{\zeta }}f$ is bounded on the compact $\bar{D}$. Near $z$ the absolute area density is at most $C\mathrm{\mid }\zeta -z{\mathrm{\mid }}^{-1}dA$, whose integral over $B_{ϵ}(z)$ is at most $2\pi Cϵ$; away from $z$ the integrand is bounded on the bounded domain. Thus the area term is absolutely integrable and its integral over the region $D_r$ defined in step 1.1 converges to that over $D$ as $r\downarrow 0$. Parametrizing the positively oriented circle by $\zeta =z+r{e}^{it}$ gives ${\int }_{\mathrm{\partial }{B}_r(z)}f(\zeta )(\zeta -z{)}^{-1}d\zeta =i{\int }_0^{2\pi }f(z+r{e}^{it})dt\to 2\pi if(z)$ by continuity of $f$.

The boundary of $D_r$ from step 1.1 is the disjoint union $\mathrm{\partial }D$ and the negatively oriented circle $-\mathrm{\partial }{B}_r(z)$. The given full AC is the premise in [F1], so [F2] applies to $gd\zeta$ on $D_r$; if $D_r$ is disconnected, each component has C¹ boundary and there are finitely many components because the compact C¹ boundary has a finite graph-chart cover, each chart meeting only one local interior component. Apply [F5] to the components and add. Using step 1.1 gives ${\int }_{\mathrm{\partial }D}f(\zeta )(\zeta -z{)}^{-1}d\zeta -{\int }_{\mathrm{\partial }{B}_r(z)}f(\zeta )(\zeta -z{)}^{-1}d\zeta =-{\int }_{D_r}({\mathrm{\partial }}_{\bar{\zeta }}f)(\zeta -z{)}^{-1}d\zeta \wedge d\bar{\zeta }$.

Letting $r\downarrow 0$ in step 2.2 and using step 2.1 yields ${\int }_{\mathrm{\partial }D}f(\zeta )(\zeta -z{)}^{-1}d\zeta +{\int }_D({\mathrm{\partial }}_{\bar{\zeta }}f)(\zeta -z{)}^{-1}d\zeta \wedge d\bar{\zeta }=2\pi if(z)$. Division by $2\pi i$ proves the first formula, with the plus sign fixed by the inner boundary orientation and the wedge swap in step 1.1.

Since $d\zeta \wedge d\bar{\zeta }=(dx+idy)\wedge (dx-idy)=-2idA$, the area term in step 3.1 equals $-{\pi }^{-1}{\int }_D({\mathrm{\partial }}_{\bar{\zeta }}f)(\zeta -z{)}^{-1}dA$. This proves the equivalent area form.