Cauchy's integral formula on a circle compactly contained in a disc of holomorphy

Define $g(\zeta )=(f(\zeta )-f(z))\mathrm{/}(\zeta -z)$ for $\zeta \mathrm{\ne }z$ and $g(z)=f^{\mathrm{\prime }}(z)$. By [L1], $g$ is continuous on $D(a,R)$ and holomorphic away from $z$.

If $z=a$, [L4] gives ${\int }_{\gamma }1\mathrm{/}(\zeta -z)d\zeta =2\pi i$. If $z\mathrm{\ne }a$, the finite identity $1\mathrm{/}(\zeta -z)={\sum }_{k=0}^N(z-a{)}^k\mathrm{/}(\zeta -a{)}^{k+1}+(z-a{)}^{N+1}\mathrm{/}((\zeta -a{)}^{N+1}(\zeta -z))$ holds on $\gamma$, and its remainder has modulus at most $(\mathrm{\mid }z-a\mathrm{\mid }\mathrm{/}r{)}^{N+1}\mathrm{/}(r-\mathrm{\mid }z-a\mathrm{\mid })$ there.

The disc is star-shaped with respect to $a$, since $\mathrm{\mid }(1-t)a+t\zeta -a\mathrm{\mid }=t\mathrm{\mid }\zeta -a\mathrm{\mid }<R$. The circle lies in the disc and is closed and rectifiable, so [L2] gives ${\int }_{\gamma }g(\zeta )d\zeta =0$.

In the second case, [L6] makes the remainder tend uniformly to zero; [L5] and [L4] then give ${\int }_{\gamma }1\mathrm{/}(\zeta -z)d\zeta =2\pi i$, because only the $k=0$ term has exponent $-1$. Thus the same kernel integral value holds also in the first case.

Since $\zeta \mathrm{\ne }z$ on $\gamma$, [L3] expands step 2.1 as ${\int }_{\gamma }f(\zeta )\mathrm{/}(\zeta -z)d\zeta =f(z){\int }_{\gamma }1\mathrm{/}(\zeta -z)d\zeta$.

Substitute step 2.2 into step 3.1 and divide by the nonzero number $2\pi i$ to obtain the formula.