Jensen's formula on a disc

Assume first that $f$ has no zero on $\mathrm{\mid }z\mathrm{\mid }=R$. Because the closed disc is compact, $f$ has only finitely many zeros in $\mathrm{\mid }z\mathrm{\mid }<R$; applying [F2] repeatedly gives $f(z)={\prod }_{k=1}^N(z-a_k)g(z)$ on a neighbourhood of the closed disc, where $g$ is holomorphic and zero-free there.

By [F3], choose a holomorphic logarithm $L$ of $g$ on $\mathrm{\mid }z\mathrm{\mid }<R$. Applying [F1] to $L$ on the circle $\mathrm{\mid }z\mathrm{\mid }=R$ and taking real parts yields $\log \mathrm{\mid }g(0)\mathrm{\mid }=\frac{1}{2\pi }{\int }_0^{2\pi }\log \mathrm{\mid }g(R{e}^{it})\mathrm{\mid }dt$.

For each zero $a_k$ with $\mathrm{\mid }{a}_k\mathrm{\mid }<R$, write $R{e}^{it}-a_k=R{e}^{it}(1-a_k{R}^{-1}{e}^{-it})$. The factor $1-a_k{R}^{-1}z$ is zero-free on the closed unit disc, so the same argument as in step 2.1 shows $\frac{1}{2\pi }{\int }_0^{2\pi }\log \mathrm{\mid }1-a_k{R}^{-1}{e}^{-it}\mathrm{\mid }dt=0$; hence $\frac{1}{2\pi }{\int }_0^{2\pi }\log \mathrm{\mid }R{e}^{it}-a_k\mathrm{\mid }dt=\log R$.

Taking logarithms of the factorization in step 1.1 on the boundary circle and averaging, step 2.1 gives the mean for $g$ and step 3.1 contributes one $\log R$ for each zero. Rearranging yields $\log \mathrm{\mid }f(0)\mathrm{\mid }=\frac{1}{2\pi }{\int }_0^{2\pi }\log \mathrm{\mid }f(R{e}^{it})\mathrm{\mid }dt-{\sum }_{k=1}^N\log (R\mathrm{/}\mathrm{\mid }{a}_k\mathrm{\mid })$.

If $f$ has zeros on $\mathrm{\mid }z\mathrm{\mid }=R$, apply step 4.1 to radii $r<R$ with no zero on $\mathrm{\mid }z\mathrm{\mid }=r$; as $r\uparrow R$, the zero list inside $\mathrm{\mid }z\mathrm{\mid }<r$ stabilizes except when $r$ crosses one of finitely many zero moduli, and the boundary integral converges to the stated radial limit.