Cauchy-kernel contour integrals may be differentiated by a direct difference-quotient estimate

For every $z\in V$, the function $\zeta \mapsto \phi (\zeta )\mathrm{/}(\zeta -z{)}^n$ is continuous on the trace, so $F_n(z)$ exists by [L1]. Moreover, [L3] applied to $\phi \circ \gamma$ gives a finite $M\ge 0$ with $\mathrm{\mid }\phi (\zeta )\mathrm{\mid }\le M$ on the trace.

Fix $z_0\in V$. Choose $\rho >0$ with $B(z_0,\rho )\subseteq V$; because the trace is disjoint from $V$, $w=\zeta -z_0$ satisfies $\mathrm{\mid }w\mathrm{\mid }\ge \rho$, and if $0<\mathrm{\mid }h\mathrm{\mid }<\rho \mathrm{/}2$ then $\mathrm{\mid }w-h\mathrm{\mid }\ge \rho \mathrm{/}2$.

The finite power identity gives $((w-h{)}^{-n}-w^{-n})\mathrm{/}h={\sum }_{j=0}^{n-1}1\mathrm{/}(w^{j+1}(w-h{)}^{n-j})$, and after subtracting $n\mathrm{/}{w}^{n+1}$ the remainder is $h{\sum }_{j=0}^{n-1}{\sum }_{k=0}^{n-j-1}1\mathrm{/}(w^{j+k+2}(w-h{)}^{n-j-k})$; by step 1.2 and [L4], its modulus is at most $\mathrm{\mid }h\mathrm{\mid }n(n+1)(2\mathrm{/}\rho {)}^{n+2}\mathrm{/}2$, uniformly on the trace.

By [L2], the difference between $(F_n(z_0+h)-F_n(z_0))\mathrm{/}h$ and $n{F}_{n+1}(z_0)$ has modulus at most a fixed finite constant times $ML(\gamma )\mathrm{\mid }h\mathrm{\mid }$, which tends to zero. Hence $F_n^{\mathrm{\prime }}(z_0)=n{F}_{n+1}(z_0)$; since $z_0$ was arbitrary, $F_n$ is holomorphic on $V$. The estimate also covers $n=1$ and a constant contour, while $h=0$ is only the excluded difference-quotient value.