\subsection{Removing a disk around each zero}\label{sec:FH}

Lemma~\ref{lem:P} uses only that the image of the inverse branch of Lemma~\ref{lem:branch} at $0$, pulled back to the disk by $W^{-1}$, misses
the points of $A$. In fact it misses the whole pulled-back image of the branch at each zero, and that image contains a disk of explicit radius.

\begin{lemma}\label{lem:FH}
For every zero $a$ of $B$, with $r=|a|$ and
\[
  s(r):=r_{\rm in}\,\frac{(1-r^2)(1-r)^2}{4r},
\]
one has $s(r)<r$ and
\[
  r_{\rm in}K_0\ \le\ \frac1{\lambda_{\mathbb D_a}(0)},\qquad \mathbb D_a:=\mathbb D\setminus\overline{D(a,s(r))}.
\]
\end{lemma}

\begin{proof}
For $b\in\{0,W(a)\}$ let $g_b:D(0,r_{\rm in})\to E_b$ be the inverse of Lemma~\ref{lem:branch}, and put
$h_b(u)=W^{-1}(g_b(r_{\rm in}u))$. The maps $h_0$ and $h_{W(a)}$ are univalent on $\mathbb D$, and their images are disjoint open
subsets of $\mathbb D$, since $E_0\cap E_{W(a)}=\emptyset$. As in the proof of Lemma~\ref{lem:Z}, $h_{W(a)}(0)=a$ and
$|h_{W(a)}'(0)|=r_{\rm in}/|W(a)B'(a)|$. Since $B(a)=0$, the Schwarz--Pick lemma gives $|B'(a)|\le1/(1-r^2)$, and the growth
theorem for $W\in\mathcal S$ gives $|W(a)|\le r/(1-r)^2$. Hence $|h_{W(a)}'(0)|\ge4s(r)$, and by Koebe's one-quarter theorem
$h_{W(a)}(\mathbb D)$ contains the open disk $D(a,s(r))$. The open set $h_0(\mathbb D)$ misses $h_{W(a)}(\mathbb D)$, so it misses
$D(a,s(r))$ and therefore also its closure. So $h_0$ maps $\mathbb D$ into $\mathbb D_a$; in particular $0=h_0(0)$ is not in the closed disk,
that is $s(r)<r$. Finally $h_0(0)=0$ and $|h_0'(0)|=r_{\rm in}/|B(0)|=r_{\rm in}K_0$ (proof of Lemma~\ref{lem:P}), and the
Schwarz--Pick lemma for the hyperbolic metrics gives $\lambda_{\mathbb D_a}(0)\,r_{\rm in}K_0\le\lambda_{\mathbb D}(0)=1$.
\end{proof}

The density $\lambda_{\mathbb D_a}(0)$ is explicit. Rotate so that $a=r>0$, put $x_1=r-s$, $x_2=r+s$ with $s=s(r)$, and
\[
  p=\frac{1+x_1x_2-\sqrt{(1+x_1x_2)^2-(x_1+x_2)^2}}{x_1+x_2},\qquad q=\frac{x_2-p}{1-px_2}.
\]
The M\"obius map $w\mapsto(w-p)/(1-pw)$ sends $\partial\mathbb D$ to itself and the circle $|w-r|=s$ to $|z|=q$, so
$\lambda_{\mathbb D_a}(0)=(1-p^2)\,\lambda_{A_q}(p)$, where $A_q=\{q<|z|<1\}$ has the density
\[
  \lambda_{A_q}(t)=\frac{\pi}{2t\log(1/q)\sin\bigl(\pi\log(1/t)/\log(1/q)\bigr)} .
\]
Since $s(r)$ is proportional to $r_{\rm in}$ and a larger disk removed gives a larger density, the bound $1/(r_{\rm in}\lambda_{\mathbb D_a}(0))$ of
Lemma~\ref{lem:FH} is largest at $r_{\rm in}=1/4$; we write $\operatorname{bd}(r)$ for its value there. Every Blaschke model satisfies
$K_0\le\operatorname{bd}(|a|)$ for each of its zeros $a$.

\begin{corollary}\label{cor:FH}
Every zero $a$ of $B$ satisfies $|a|>1/2.7051=0.369672\ldots$.
\end{corollary}

\begin{proof}
$K_0\ge1/|a|$, since every factor $1/|a_j|$ of $K_0$ is at least $1$, and $|a|\ge\rho_*=0.31088\ldots$ by Corollary~\ref{cor:P}(b). On the
interval $0.3108\le r\le1/2.7051$ the function $\operatorname{bd}$ is at most $2.7050891<2.7051\le1/r$, so no zero has its modulus
there. The bound on $\operatorname{bd}$ is a ball-arithmetic computation~\cite{Johansson2017} at $200$ bits: the interval is cut into
$11$ pieces, and on a piece $[r_-,r_+]$ the closed disk about the midpoint of radius $s(r_+)-(r_+-r_-)/2$ lies inside
$\overline{D(r,s(r))}$ for every $r$ in the piece, since $s$ is decreasing ($(1-r^2)(1-r)^2$ decreases and $4r$ increases); a smaller
disk removed gives a smaller density, so the
value of $1/(r_{\rm in}\lambda)$ at that one disk bounds $\operatorname{bd}$ on the piece.
\end{proof}

Corollary~\ref{cor:FH} improves the bound $|a|\ge1/3$ that~\cite[Thm.~7.1]{JatarNg2026b} gives for every zero. For $\deg B=1$,
Theorem~\ref{thm:B1upper} gives $|a|>0.84904$.