= Solution
Let $m$ be the <multiplicity> at zero, and write $f(z)=z^m h(z)$ with $h(0)\ne0$. The proof in <2/a/solution> gives $|f|\le1$. For radii without boundary zeros, <Jensen's formula> says
$$
\int\log|f(re^{i\theta})|\,dm
=m\log r+\log|h(0)|+\sum_{0<|a_n|<r}\log\frac r{|a_n|}.
$$
The absolute logarithmic hypothesis makes the left side tend to zero. The nonnegative zero sum increases to $\sum_n\log(1/|a_n|)$ by the <monotone convergence theorem>, so this series is finite. Hence $\sum_n(1-|a_n|)<\infty$, and the <Blaschke condition> supplies a <Blaschke product> $B$ with exactly these zeros and their <multiplicities>.
The quotient $g=f/B$ extends <holomorphically> across every zero because the <multiplicities> agree, and the extensions have nonzero values. Thus \b[$f=Bg$ with $g$ zero-free and holomorphic]. We will also need the stronger <Blaschke factorization of a bounded holomorphic function>, namely $|g|\le1$. Let $B_N$ contain the factor $z^m$ and the first $N$ other zero factors. The quotient $f/B_N$ is <holomorphic>. On $|z|=R$ enclosing those finitely many zeros, its modulus is at most $1/\min_{|z|=R}|B_N(z)|$. The <maximum modulus principle> transfers this bound to the interior. Since the finite <Blaschke product> has boundary modulus one continuously, letting $R\uparrow1$ gives $|f/B_N|\le1$. Now let $N\to\infty$ using <locally uniform convergence> of the <Blaschke product>, first off the zeros and then by <continuity> at them. This proves $|g|\le1$.
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