\section{Adding zeros near the circle}\label{sec:addzeros}

This section is about adding zeros of $B$ near the circle while $W$ is kept. We show that a critical point of $WB$ on the circle
can block this (Proposition~\ref{prop:BC}); for $W(w)\not\equiv w$ meromorphic past the circle we count exactly the zeros of $(WB)'$ in the disk when the new zeros lie on
fixed radii (Propositions~\ref{prop:extend} and~\ref{prop:tips}); and we show that in the polynomial family of~\cite[Lemma~6.2]{JatarNg2026b}
the number of zeros of $B$ is locally constant (Proposition~\ref{prop:KLC}).

We first recall the local effect of one new zero of $B$ near the circle, with $W$ kept. Let $(W,B)$ be a Blaschke model, $F=WB$,
$\xi_0$ a point of the circle and $a=(1-s)\xi_0$. Put $\chi=wF'/F$, and on the circle $u=\log|W|$, with derivatives in $\theta$. As
$|B|=1$ on the circle, $\frac d{d\theta}\log F(e^{i\theta})=i\chi$, so $\chi=\frac d{d\theta}\arg F-iu'$ there; in particular $\chi(\xi_0)$ is
real if $u'(\xi_0)=0$. By~\cite[Lemma~8.1, Rem.~8.3, Prop.~8.8]{JatarNg2026b}, if $u'=0$, $u''>0$ and $\chi<0$ at $\xi_0$, then $(W,BB_a)$ is a
Blaschke model for all small $s>0$, and its gain is larger by the factor $1/(1-s)$. By Proposition~\ref{prop:extend}(i) below, with $M=1$, $\zeta_1=\xi_0$ and $r=1-s$, the derivative
$(WBB_a)'$ has a zero in $\mathbb D$ for all small $s$ if $u'(\xi_0)\neq0$, or if $u'(\xi_0)=0$ and either $\chi(\xi_0)>0$ or $u''(\xi_0)<0$.

A critical point of $F$ on the circle can block this route. For $a\in\mathbb D\setminus\{0\}$ put $\widetilde B_a=-\frac{|a|}{a}B_a$, with $B_a$ as in Section~\ref{sec:prelim}, so that
$\widetilde B_a(0)=|a|$.

\begin{proposition}\label{prop:BC}
Let $(W,B)$ be admissible, and let $\zeta_0$ be a point of the circle such that $F=WB$ extends holomorphically to a neighbourhood of
$\zeta_0$, with $F'(w)=C(w-\zeta_0)^m(1+O(w-\zeta_0))$, $m\ge1$, $C\neq0$. Then $F(\zeta_0)\neq0$; put
$c_{\zeta_0}=(i\zeta_0)^m\zeta_0C/F(\zeta_0)$, so that $\chi(\zeta_0e^{it})=c_{\zeta_0}t^m+O(t^{m+1})$ for $\chi=wF'/F$, as in
Section~\ref{sec:count} (for $m=1$, $\operatorname{Im}c_{\zeta_0}=-u''(\theta_0)$ with $\zeta_0=e^{i\theta_0}$ and
$u=\log|W|$ on the circle).
\begin{enumerate}
\item[(a)] For all small $s>0$ the pair $(W,B\widetilde B_a)$ with $a=(1-s)\zeta_0$ is not admissible.
\item[(b)] Let $\eta_0>0$, and let $m\ge3$, or $m=2$ and $c_{\zeta_0}\notin(-\infty,0)$, or $m=1$ and $\operatorname{Im}c_{\zeta_0}>0$. There is
  $s_0>0$ such that $(W,B\widetilde B_a)$ with $a=(1-s)\xi_0$ is not admissible whenever $|\xi_0|=1$, $|\xi_0-\zeta_0|\ge\eta_0$ and
  $0<s<s_0$.
\end{enumerate}
\end{proposition}

\begin{proof}
Since $B\neq0$ near the circle, $W=F/B$ extends holomorphically near $\zeta_0$. If $W(\zeta_0)=0$, the points $w_n=(1-1/n)\zeta_0$
satisfy $W(w_n)\to0=W(0)$, an interior point of $W(\mathbb D)$, so $w_n\to W^{-1}(0)=0$ by continuity of $W^{-1}$, which is absurd;
hence $F(\zeta_0)\neq0$. On $\mathbb D\setminus\{a\}$ we have $(F\widetilde B_a)'=\widetilde B_a\,G_a$ with $G_a=F'+\Gamma_a$ and
$\Gamma_a(w)=(1-|a|^2)F(w)/((w-a)(1-\bar aw))$, so a zero of $G_a$ in $\mathbb D\setminus\{a\}$ is a zero of $(F\widetilde B_a)'$ in $\mathbb D$.

(b) Let $|\xi_0-\zeta_0|\ge\eta_0$ and $s\le\eta_0/8$. On a closed disk $U'$ about $\zeta_0$ of radius $r_1\le\eta_0/4$ on which $F$ is
holomorphic, $|(w-a)(1-\bar aw)|\ge(\eta_0/2)^2$, so $\Gamma_a(\zeta_0+\hat h)=\lambda F(\zeta_0)/\zeta_0+O(s|\hat h|)$, where
$\lambda=(1-|a|^2)/|\zeta_0-a|^2\in[s/4,\,2s/(\eta_0-s)^2]$, and the
constant depends only on $F$ and $\eta_0$. Put $\rho=\lambda^{1/m}\rho_1$ with $\rho_1^m=-F(\zeta_0)/(\zeta_0C)$. Then
$G_a(\zeta_0+\rho t)=(\lambda F(\zeta_0)/\zeta_0)(1-t^m+E(t))$ with $E(t)=O(|\rho|)$ uniformly on $|t|\le2$. Choose $j$ and $q>0$ such that $\operatorname{Re}(\bar\zeta_0\rho e^{2\pi ij/m})\le-q|\rho|$. This is possible: for $m\ge3$ with
$q=1/2$, because some $m$-th root of unity lies within $\pi/m$ of any direction; for $m=2$ unless $\bar\zeta_0\rho_1$ is purely imaginary,
which happens exactly when $c_{\zeta_0}<0$; for $m=1$ because $\operatorname{Re}(\bar\zeta_0\rho)=-\lambda\operatorname{Im}c_{\zeta_0}/|c_{\zeta_0}|^2$.
The direction of $\rho$ depends neither on $\xi_0$ nor on $s$, so $j$ and $q$ do not either. Fix $r_0>0$ smaller than half the distance
between neighbouring $m$-th roots of unity and than $q/2$. By Rouch\'e's theorem, for $|\rho|$ small there is a zero $t'$ with
$|t'-e^{2\pi ij/m}|<r_0$. Since $|\zeta_0+\rho t'|^2=1+2\operatorname{Re}(\bar\zeta_0\rho t')+|\rho t'|^2$ and $r_0<q/2$, the point $\zeta_0+\rho t'$ lies in $\mathbb D$
when $|\rho|$ is small; it is then far from $a$, which lies near $\xi_0$. All constants depend only on $F$, $\eta_0$ and $m$.

(a) Now $a=(1-s)\zeta_0$. With $w=\zeta_0+\hat h$ one has exactly
$(w-a)(1-\bar aw)=-(1-s)\bar\zeta_0\hat h^2+s^2\hat h+s^2\zeta_0$, so
$N_a:=(w-a)(1-\bar aw)\,G_a=-C\bar\zeta_0\hat h^{m+2}(1+O(\hat h))+2s F(\zeta_0)+O(s|\hat h|+s^2)$. With $\hat h=\rho t$ and
$\rho^{m+2}=2s\zeta_0F(\zeta_0)/C$ this is $2s F(\zeta_0)(1-t^{m+2}+o(1))$ on $|t|\le2$. As in (b), since $m+2\ge3$, some $(m+2)$-th root of unity $\omega$ has
$\operatorname{Re}(\bar\zeta_0\rho\,\omega)\le-|\rho|/2$, and by Rouch\'e's theorem $N_a$ has a zero $\zeta_0+\rho t'$ with $t'$ near
$\omega$; as in (b) it lies in $\mathbb D$, at distance of order $s^{1/(m+2)}$ from $\zeta_0$, hence not at
$a$ or $1/\bar a$.
\end{proof}

The splitting of a multiple zero under a small perturbation is standard; the point of Proposition~\ref{prop:BC} is that the perturbation
has a direction that depends only on $F$ near $\zeta_0$. Case (b) is not covered by~\cite[Lemma~8.1, Prop.~8.8]{JatarNg2026b},
which treat regular pairs. For every boundary pair of Table~\ref{tab:onezero}, $(WB)'$ has a zero of order $5$ (pairs~\ref{pair:K2N18}--\ref{pair:O12}) or $3$
(pairs~\ref{pair:T6p} and~\ref{pair:T6}) on the circle (Section~\ref{sec:onezero}), so a new zero near any fixed boundary point, or
on the radius through that zero, destroys admissibility
when $W$ is kept. The proposition does not cover the case that $\xi_0$ tends to $\zeta_0$ together with $s$, the route
of~\cite[Lemma~2.1]{JatarNg2026b}, which first replaces $W$, or joint variations of $W$ and $B$.

For zeros added on fixed radii, Theorem~\ref{thm:count} gives an exact count. Let $W(w)\not\equiv w$, $P$, $u$ and $B$ (of degree $d$) be as in
Section~\ref{sec:count}, with $F=WB$, $\chi=wF'/F$, the sets $Z_\chi$ and the indices $\iota_\zeta$ there, and let $E$ be the set of
zeros of $u'$ on $\mathbb T\setminus P$, a finite set. A point $x\in E$ is \emph{$\chi$-negative}\label{term:chineg} if $\chi(x)<0$, with \emph{$\chi$-depth}
$\eta(x)=-\chi(x)$, and \emph{$\chi$-positive} if $\chi(x)>0$; the other points of $E$ are the zeros of $F'$ on $\mathbb T\setminus P$, and at each
of them $\iota\ge0$ by the definition of the index. A \emph{$\chi$-negative maximum} is a $\chi$-negative point at which $u$ has a strict local maximum, and
similarly for a \emph{$\chi$-negative minimum}. For distinct points $\zeta_1,\dots,\zeta_M$ of $\mathbb T\setminus P$ and $0<r<1$ put
$C_r=\prod_j\widetilde B_{r\zeta_j}$, so that $C_r(0)=r^M$, $F_r=FC_r$ and $\sigma_r=wC_r'/C_r$. On the circle $\sigma_r$ is real and
positive; since $|x-r\zeta|^2=(1-r)^2+r|x-\zeta|^2$ for $|x|=|\zeta|=1$,
\begin{equation}\label{eq:betar}
  \sigma_r(x)=\sum_{j=1}^M\frac{1-r^2}{(1-r)^2+r|x-\zeta_j|^2}\ \ge\ M\,\frac{1-r}{1+r},\qquad \sigma_r(\zeta_j)\ge\frac{1+r}{1-r},
\end{equation}
and $\sigma_r(x)\to0$ as $r\to1$ if $x$ is none of the $\zeta_j$. So $\chi_r:=wF_r'/F_r=\chi+\sigma_r$ has the imaginary part $-u'$ on
the circle, and $u$ does not depend on $r$.

\begin{proposition}\label{prop:extend}
With this notation:
\begin{enumerate}
\item[(i)] For all $r<1$ close enough to $1$, $F_r'$ has no zero on $\mathbb T\setminus P$, and
\[
  N_{\mathbb D}(F_r')=N_{\mathbb D}(F')+\sum_{j=1}^M\bigl(1+\iota^*_j\bigr)+\sum_\zeta\iota_\zeta ,
\]
where $\iota^*_j=\iota_{\zeta_j}$ if $\zeta_j\in Z_\chi$ and $\iota^*_j=0$ otherwise, and the last sum is over the zeros $\zeta$ of
$F'$ on $\mathbb T\setminus P$ that are none of the $\zeta_j$. Every term is non-negative. Hence $(W,BC_r)$ is admissible for all $r$
close to $1$ if and only if $(W,B)$ is admissible, every $\zeta_j$ is a $\chi$-negative minimum, and every zero of $F'$ on $\mathbb T\setminus P$
has index $0$; its gain is then $r^{-M}/B(0)$.
\item[(ii)] For every $r\in(0,1)$ such that $\chi(x)+\sigma_r(x)\neq0$ for all $x\in E$,
\begin{equation}\label{eq:Nr}
  N_{\mathbb D}(F_r')=d+M-\sum\bigl\{\varepsilon_x:\ x\in E,\ \chi(x)+\sigma_r(x)<0\bigr\}.
\end{equation}
\item[(iii)] Let $M\ge1$, suppose that the conditions of (i) hold, and let $r^*$ be the largest $r\in(0,1)$ at which
$\chi(x)+\sigma_r(x)=0$ for a $\chi$-negative maximum $x$ or for $x=\zeta_j$. Then $r^*$ exists, and $(W,BC_r)$ is admissible for every
$r\in(r^*,1)$. If exactly one point of $E$ has $\chi+\sigma_{r^*}=0$, with a change of sign, then $N_{\mathbb D}(F_r')=1$ for $r<r^*$
close to $r^*$.
\item[(iv)] In the setting of (iii), at least $d+M$ $\chi$-negative maxima are none of the $\zeta_j$, and $1-r^*<2\eta(x)/M$ for each
of them with $2\eta(x)/M<1$. If $M=1$ and $\eta(\zeta_1)\le1$, then $1-r^*$ is the least root in $(0,1)$, over the $\chi$-negative maxima $x$, of
\[
  f_x(Y)=(1+\eta)Y^2-(2+\eta D^2)Y+\eta D^2,\qquad \eta=\eta(x),\ D=|x-\zeta_1| ;
\]
here $f_x(1-r)=\bigl((1-r)^2+rD^2\bigr)\bigl(\eta(x)-\sigma_r(x)\bigr)$, which has the sign of $\eta(x)-\sigma_r(x)$. If $\eta(x)<1$, then $f_x$ has exactly one root in $(0,1)$, and it is $\eta D^2(1+O(\eta))/2$.
\end{enumerate}
\end{proposition}

\begin{proof}
(i) By Theorem~\ref{thm:count}(i), which holds for every finite Blaschke product, $N_{\mathbb D}(F')=d-\sum_{Z_\chi}\iota$ and
$N_{\mathbb D}(F_r')=d+M-\sum_{Z_{\chi_r}}\iota$. As $E$ is finite, by~\eqref{eq:betar} we may take $r$ so close to $1$ that
$\sigma_r(x)<\eta(x)$ at every $\chi$-negative point $x$ that is none of the $\zeta_j$, and $\sigma_r(\zeta_j)>|\chi(\zeta_j)|$ for every $j$. Outside
$E$, $\operatorname{Im}\chi_r=-u'\neq0$. At a point $x\in E$ that is none of the $\zeta_j$: if $\chi(x)\ge0$, then
$\chi_r(x)\ge\sigma_r(x)>0$; if $\chi(x)<0$, then $\chi_r(x)<0$, so $x\in Z_{\chi_r}$ with $m_x=0$ and index $\varepsilon_x$, its index
for $\chi$ as well. At the $\zeta_j$ in $E$, $\chi_r>0$. So $\chi_r$ has no zero on $\mathbb T\setminus P$, nor has $F_r'$, and $Z_{\chi_r}$
is the set of $\chi$-negative points other than the $\zeta_j$, with the same indices. Subtracting, $N_{\mathbb D}(F_r')-N_{\mathbb D}(F')$
is $M$ plus the sum of $\iota$ over the points of $Z_\chi$ outside $Z_{\chi_r}$, that is, over the $\zeta_j$ in $Z_\chi$ and over the
zeros of $F'$ on the circle that are none of the $\zeta_j$: this is the formula. The index is at least $0$ at a zero of $F'$ and is
$\varepsilon\ge-1$ at a $\chi$-negative point, so every term is non-negative, and $N_{\mathbb D}(F_r')=0$ if and only if $N_{\mathbb D}(F')=0$
and every term vanishes. The term $1+\iota^*_j$ vanishes if and only if $\iota^*_j=-1$, that is, $\zeta_j$ is a $\chi$-negative point with
$\varepsilon=-1$; then no $\zeta_j$ is a zero of $F'$, and the last sum vanishes if and only if all zeros of $F'$ on the circle have
index $0$. Finally $W$ is unchanged and $BC_r(0)=r^MB(0)>0$.

(ii) If $\chi+\sigma_r\neq0$ on $E$, then $\chi_r$ has no zero on $\mathbb T\setminus P$, the points of $Z_{\chi_r}$ have $m=0$ and
index $\varepsilon$, and Theorem~\ref{thm:count}(i) gives~\eqref{eq:Nr}.

(iii) Let $O$ be the set of $\chi$-negative points other than the
$\zeta_j$. By (i) and~\eqref{eq:Nr} for $r$ near $1$, $\sum_{x\in O}\varepsilon_x=d+M$; so at least $d+M$ points of $O$ are $\chi$-negative
maxima. The set of $r$ in the definition of $r^*$ is closed in $(0,1)$ and, by~\eqref{eq:betar}, avoids a neighbourhood of $1$.
Let $\tilde r<1$ be its largest element, or $\tilde r=0$ if it is empty. As $r$ decreases from $1$ to $\tilde r$, the points with
$\chi\ge0$ stay $\chi_r$-positive, as $\chi+\sigma_r\ge\sigma_r>0$ there; by the choice of $\tilde r$, the $\zeta_j$ stay $\chi_r$-positive and the $\chi$-negative maxima in $O$
stay $\chi_r$-negative; so only $\chi$-negative minima and points with $\varepsilon=0$ can leave the set of $\chi_r$-negative points. No $\chi$-negative minimum leaves:
otherwise let $r_y\in(\tilde r,1)$ be the largest $r$ at which a $\chi$-negative minimum $y\in O$ has $\chi(y)+\sigma_r(y)=0$. At $r_y$ the $\chi_r$-negative
points are those of $O$ less $y$ and less some points with $\varepsilon\le0$, so their $\varepsilon$ add up to at least
$d+M+1$, and the zeros of $F_{r_y}'$ on the circle have index at least $0$; Theorem~\ref{thm:count}(i) gives
$N_{\mathbb D}(F_{r_y}')\le d+M-(d+M+1)<0$, which is absurd. So on $(\tilde r,1)$ the $\varepsilon$ of the $\chi_r$-negative points add up
to $d+M$: $N_{\mathbb D}(F_r')=0$ where $\chi+\sigma_r\neq0$ on $E$, by~\eqref{eq:Nr}, and at the other $r$ the zeros of $\chi_r$ on $E$
have $\varepsilon=0$, so Theorem~\ref{thm:count}(i) gives $N_{\mathbb D}(F_r')=-\sum\iota\le0$, hence $0$ again. So $(W,BC_r)$ is
admissible on $(\tilde r,1)$. As $r\to0$, $C_r$ tends locally uniformly to a unimodular constant times $w^M$, and $(w^MF)'$ has a zero of
order $M$ at $0$, since $F(0)=0\neq F'(0)$; by Hurwitz's theorem $N_{\mathbb D}(F_r')\ge M\ge1$ for small $r$. So $\tilde r>0$, and
$r^*=\tilde r$ exists. If exactly one point $x$ has $\chi+\sigma_{r^*}=0$ at $r^*$, with a change of sign, then just below $r^*$ the
set of $\chi_r$-negative points loses the $\chi$-negative maximum $x$ or gains the point $x=\zeta_j$, of $\varepsilon=-1$, and~\eqref{eq:Nr} gives $N_{\mathbb D}=1$.

(iv) The proof of (iii) shows that at least $d+M$ points of $O$ are $\chi$-negative maxima. If $x\in O$ is a $\chi$-negative maximum
with $2\eta(x)/M<1$, then at $1-r=2\eta(x)/M$, by~\eqref{eq:betar},
$\sigma_r(x)\ge M(1-r)/(1+r)>M(1-r)/2=\eta(x)$, while $\sigma_r(x)<\eta(x)$ for $r$ near $1$; so $\chi(x)+\sigma_r(x)=0$ for some $r$
with $1-r<2\eta(x)/M$, and $r^*>1-2\eta(x)/M$. For $M=1$ put $Y=1-r$; then $1-r^2=Y(2-Y)$ and $(1-r)^2+rD^2=Y^2+(1-Y)D^2$, so
$f_x(Y)=\bigl(Y^2+(1-Y)D^2\bigr)\bigl(\eta(x)-\sigma_r(x)\bigr)$ has the sign of $\eta(x)-\sigma_r(x)$. As
$\sigma_r(\zeta_1)\ge(1+r)/(1-r)>1\ge\eta(\zeta_1)$, the point $\zeta_1$ gives no $r$ in the definition of $r^*$, and $1-r^*$ is the least
root in $(0,1)$ of the $f_x$. If $\eta<1$, then $f_x(0)=\eta D^2>0>\eta-1=f_x(1)$, and $f_x$ is a quadratic with positive leading
coefficient, so it has exactly one root in $(0,1)$; the expansion of the smaller root is $\eta D^2(1+O(\eta))/2$.
\end{proof}

\emph{An example: one added zero.} Let $(W,B)$ be admissible with $\deg B=d$, let $F'$ have no zero on $\mathbb T\setminus P$, and
add one zero ($M=1$) on the radius of a $\chi$-negative minimum $\zeta_1$ with $\eta(\zeta_1)\le1$. In~(i), $\iota^*_1=\iota_{\zeta_1}=
\varepsilon_{\zeta_1}=-1$ and the last sum is empty, so $N_{\mathbb D}(F_r')=0+(1-1)=0$: for $r$ close to $1$ the pair $(W,BC_r)$ is
admissible, with gain $r^{-1}/B(0)$. (At a $\chi$-negative maximum the term would be $1+1=2$, and $F_r'$ would have two zeros in the
disk.) By the proof of~(iii), the $\varepsilon_x$ of the $\chi$-negative points $x\neq\zeta_1$ add up to $d+1$, so at least $d+1$ of
them are $\chi$-negative maxima. At such a point $\chi(x)+\sigma_r(x)=0$ means $\sigma_r(x)=\eta(x)$, and $\zeta_1$ gives no such $r$,
as $\sigma_r(\zeta_1)>1\ge\eta(\zeta_1)$. So $r^*$ is the largest $r<1$ at which $\sigma_r(x)=\eta(x)$ for one of these maxima, and
by~(iv) $1-r^*$ is the least root in $(0,1)$ of their quadratics $f_x$.

Proposition~\ref{prop:extend}(i) is the exact form of Proposition~\ref{prop:BC} for zeros added on fixed radii: with $W$ kept, a zero
$\zeta_0$ of $F'$ on the circle that is none of the $\zeta_j$ blocks the construction if and only if $\iota_{\zeta_0}>0$, and it adds exactly $\iota_{\zeta_0}$ zeros
of $(WBC_r)'$ in the disk; if $\zeta_0$ is one of the $\zeta_j$, its term in (i) is $1+\iota_{\zeta_0}\ge1$, so it always blocks, as in
Proposition~\ref{prop:BC}(a). Among the cases left open there, $m=1$
with $\operatorname{Im}c_{\zeta_0}<0$ (a minimum of $u$) has index $0$, and $m=2$ with $c_{\zeta_0}<0$ has index $1+\varepsilon_{\zeta_0}$, so it
blocks unless $u$ has a minimum at $\zeta_0$. Proposition~\ref{prop:BC}(b) is uniform in the added point, which (i) does not give.
For the two pairs of Remark~\ref{rem:countrecords}, whose zero of order $5$ has index $3$ ($m=5$ by (C) of Lemma~\ref{lem:certify}; $u$ has a strict maximum with
$k_\zeta=5$ by Section~\ref{sec:flatmax}), every choice
of zeros added on fixed radii near the circle, with $W$ kept, leaves at least $3$ zeros of $(WBC_r)'$ in the disk. For regular pairs the ``if'' of (i) with $M=1$ at a nondegenerate minimum ($u''(\zeta_1)>0$)
is~\cite[Prop.~8.8]{JatarNg2026b}; (i) allows minima of any odd order. The lower bound
in~\eqref{eq:betar} is the mechanism of~\cite[Thm.~8.7(a)]{JatarNg2026b}. By (iv), a pair whose $\chi$-negative maxima are shallow has little
room: for one added zero, $1-r^*$ is about the least of the numbers $\eta(x)D^2/2$.

Let
$\mathcal Q_n$ be the monic polynomials of degree $n$, with the norm $\|Q\|$ given by the largest modulus of a coefficient. For
$Q\in\mathcal Q_n$ and a zero $b\in\mathbb D$ of $Q$ let $A(Q,b)$ be the multiset of zeros of $Q$ in $\mathbb D$ with one copy of $b$
removed, and $H_{Q,b}=Q\prod_{a\in A(Q,b)}(1-\bar az)/(z-a)$, a polynomial with $H_{Q,b}(b)=0$. Let $U_n$ be the set of such $(Q,b)$
with $H_{Q,b}$ univalent in $\mathbb D$. The pairs of~\cite[Lemma~6.2]{JatarNg2026b}, and their analogues with $k$ zeros of $B$, are built
from points of $U_n$ with $|A(Q,b)|=k$. In this family the number of zeros of $B$ cannot change along a continuous path in $U_n$ (degree $n$ fixed).

\begin{proposition}\label{prop:KLC}
\begin{enumerate}
\item[(a)] For $(Q,b)\in U_n$, $\min_{|z|=1}|Q(z)|\ge(1-|b|^2)|H_{Q,b}'(b)|/4$, and this is at least $(1-|b|^2)|Q'(b)|/4$ if $b$ is a
  simple zero of $Q$.
\item[(b)] $|A(Q,b)|$ is locally constant on $U_n$, and $U_n$ and each of the sets $\{|A(Q,b)|=k\}$ are closed in
  $\mathcal Q_n\times\mathbb D$.
\item[(c)] Let $b_0\in\mathbb D$ be a simple zero of $Q_0\in\mathcal Q_n$, $m_0=(1-|b_0|)|Q_0'(b_0)|$ and $\|Q_0\|_1$ the sum of the moduli of
  the coefficients of $Q_0$. If $0<\delta_1\le(1-|b_0|)/2$, $n(n+1)\delta/2+2n(n+1)\|Q_0\|_1\delta_1\le m_0/2$ and
  $(n+1)\delta<(1-|b_0|)m_0/16$, then no $(Q,b)\in U_{n+1}$ has $|b-b_0|<\delta_1$ and $\|Q-Q_0\,(z-\xi)\|<\delta$ for some $|\xi|=1$.
\end{enumerate}
\end{proposition}

\begin{proof}
A univalent $H$ that is continuous on the closed disk takes no value of $H(\mathbb D)$ on the circle (if $H(\xi)=H(z_0)$ with
$|\xi|=1$, then $H((1-1/m)\xi)\to H(z_0)\in H(\mathbb D)$, so $(1-1/m)\xi\to z_0$ by continuity of $H^{-1}$, although
$(1-1/m)\xi\to\xi\neq z_0$). (a) Each factor $(1-\bar az)/(z-a)$ has modulus $1$ on the circle, so $|H_{Q,b}|=|Q|$
there. With $\varphi(w)=(w+b)/(1+\bar bw)$, Koebe's one-quarter theorem~\cite[Ch.~2]{Duren1983} applied to $H_{Q,b}\circ\varphi$ shows
that $H_{Q,b}(\mathbb D)$ contains the disk of radius $(1-|b|^2)|H_{Q,b}'(b)|/4$ about $0$, which the values on the circle avoid. If $b$ is simple,
$H_{Q,b}'(b)=Q'(b)\prod_a(1-\bar ab)/(b-a)$ and $|1-\bar ab|^2-|b-a|^2=(1-|a|^2)(1-|b|^2)>0$. (b) By (a), $Q$ has no zero on the
circle, so Rouch\'e's theorem keeps the number of zeros in $\mathbb D$ fixed on the ball $\|\widetilde Q-Q\|<\min_{|z|=1}|Q|/n$. For closedness, let
$(Q_m,b_m)\in U_n$ converge to $(Q_\infty,b_\infty)\in\mathcal Q_n\times\mathbb D$, pass to a subsequence along which
$A(Q_m,b_m)=\{a_{1,m},\dots,a_{k,m}\}$ (with $k$ fixed) converges to $\{\alpha_1,\dots,\alpha_k\}$, and write
$Q_m=\prod_j(z-a_{j,m})\,P_m$. Monic division is continuous and the remainders vanish, so $P_m\to P_\infty=Q_\infty/\prod_j(z-\alpha_j)$,
a monic polynomial of degree $n-k\ge1$. Then
$H_{Q_m,b_m}=P_m\prod_j(1-\bar a_{j,m}z)\to H_\infty=P_\infty\prod_j(1-\bar\alpha_jz)$ uniformly on the closed disk; $H_\infty$ is not
constant ($P_\infty$ is monic of degree at least $1$ and the product has constant term $1$), so it is univalent by Hurwitz's
theorem (if $H_\infty$ took some value at two points, Rouch\'e's theorem on two small disjoint circles would make $H_{Q_m,b_m}$
take it twice), and $H_\infty(b_\infty)=0$. Hence $H_\infty$ has no zero on the circle. So no $\alpha_j$ lies on the circle, as
$H_\infty(\alpha_j)$ would vanish through the factor $1-|\alpha_j|^2$, and $P_\infty$ has no zero there; $P_\infty$ has exactly one zero in $\mathbb D$, namely $b_\infty$, so
$A(Q_\infty,b_\infty)=\{\alpha_j\}$ and $(Q_\infty,b_\infty)\in U_n$. (c) Put $Q_\xi=Q_0\,(z-\xi)$; then $|Q_\xi'(b_0)|\ge m_0$,
$|Q_\xi''|\le2n(n+1)\|Q_0\|_1$, $|Q-Q_\xi|<(n+1)\delta$ and $|Q'-Q_\xi'|<n(n+1)\delta/2$ on the closed disk. So $|Q'(b)|\ge m_0/2$, $b$ is a simple
zero of $Q$, $1-|b|^2\ge(1-|b_0|)/2$, and (a) gives $\min_{|z|=1}|Q|\ge(1-|b_0|)m_0/16>(n+1)\delta>|Q(\xi)-Q_\xi(\xi)|=|Q(\xi)|$, which is
absurd.
\end{proof}

So a pair of degree $n+1$ with two zeros cannot be reached from a one-zero pair of degree $n$ by adding a zero of $Q$ near the circle
while $b$ stays near its old place, and pairs with $k$ zeros of fixed degree form a union of components of $U_n$. This is in line
with~\cite[Rem.~8.11]{JatarNg2026b}, where a zero of $B$ may escape to the circle in a limit: inside the polynomial family this cannot
happen while $b$ stays in a compact subset of $\mathbb D$. The restriction to $b$ in a compact subset of $\mathbb D$ is needed: $(z^n(z-b),b)\in U_{n+1}$ for every
$b\in\mathbb D$, with $H=z-b$.

For pairs of the type of~\cite[Lemma~8.1]{JatarNg2026b}, with all zeros of $B$
near the circle, there is an exact count of the zeros of $(WB)'$. Let $W$, $P$, $u$, $g$ and $Z_-$ be as in
Section~\ref{sec:ratext}, with $W(w)\not\equiv w$, so that $u$ is constant on no arc; let $\mathrm{Crit}$ be the set of zeros of $W'$ on
$\mathbb T\setminus P$. By Lemma~\ref{lem:D}(b), which holds on $\mathbb T\setminus P$ (proof of Lemma~\ref{lem:Dplus}), these zeros
are simple, so $g$ has a simple zero at each $c\in\mathrm{Crit}$. A zero of $u'$ is a \emph{maximum}, a \emph{minimum} or an
\emph{inflection} of $u$ if $u$ has a strict local maximum there, a strict local minimum, or neither. A point of $Z_-$ that is a
minimum of $u$ is a \emph{downward crossing}, and one that is a maximum is an \emph{upward crossing}. For distinct
$\zeta_1,\dots,\zeta_M\in\mathbb T\setminus(P\cup\mathrm{Crit})$ and $0<r<1$ let $C_r$ be the Blaschke product with zeros $r\zeta_j$
and $C_r(0)=r^M$.

\begin{proposition}\label{prop:tips}
Suppose that no point of $\mathrm{Crit}$ is an inflection of $u$. Then for all $r<1$ close enough to $1$, $(WC_r)'$ has no zero on
$\mathbb T\setminus P$, and the number of its zeros in $\mathbb D$ is
\[
  \#\{c\in\mathrm{Crit}:\ c\text{ is a maximum of }u\}+\sum_{j=1}^M\hat\nu_j,\qquad
  {\hat\nu}_j=\begin{cases}0&\text{if $\zeta_j$ is a downward crossing,}\\ 2&\text{if $\zeta_j$ is an upward crossing,}\\ 1&\text{otherwise.}\end{cases}
\]
In particular $(W,C_r)$ is admissible for all $r$ close to $1$ if and only if no point of $\mathrm{Crit}$ is a maximum of $u$ and every
$\zeta_j$ is a downward crossing.
\end{proposition}

\begin{proof}
Let $F_r=WC_r$, $\sigma_r=wC_r'/C_r$, real and positive on $\mathbb T$, and $\chi_r=g+\sigma_r=wF_r'/F_r$. In the closed disk minus $P$,
$F_r'$ and $\chi_r$ have the same zeros, since $F_r$ vanishes only at $0$ and at the $r\zeta_j$, where $F_r'\neq0$. Put
$s=1-r$. Fix $\rho_0>0$ such that the closed disks $\Delta_x$ of radius $\rho_0$ about the points $x$ of
$P\cup\mathrm{Crit}\cup\{\zeta_j\}$ are disjoint and lie in $U$, $g$ is injective on each $\Delta_c$, $c\in\mathrm{Crit}$, and on
$\mathbb T\cap\Delta_x$ the function $u'$ vanishes at most at $x$ for $x\in\mathrm{Crit}\cup\{\zeta_j\}$; moreover
$\operatorname{Im}g\neq0$ on $\mathbb T\cap\Delta_j$ if $g(\zeta_j)\notin\mathbb R$, and $\operatorname{Re}g>0$ there if $g(\zeta_j)>0$.
On the closed disk outside $\Delta_j$, $|w-r\zeta_j|\ge\rho_0/2$ and $|1-r\bar\zeta_jw|\ge|w-r\zeta_j|$, so $|\sigma_r|\le Cs$ there.

\emph{Away from the disks.} On the closed disk minus the open disks $\Delta_x$, $g$ is continuous and zero-free, so $|g|$ is bounded
below, and $\chi_r\neq0$ for small $s$. \emph{Poles.} On $\Delta_p$ minus $p$, $g=-mp/(w-p)+O(1)$, so $|g|\ge1$ for $\rho_0$
small, and $\chi_r\neq0$.

\emph{Critical points.} Let $c\in\mathrm{Crit}$. On $\partial\Delta_c$, $|\sigma_r|\le Cs<\min|g|$, so by Rouch\'e's theorem
$\chi_r$ has exactly one zero $w_r$ in $\Delta_c$. On the arc $A=\mathbb T\cap\Delta_c$, $\operatorname{Im}\chi_r=-u'$ vanishes only at
$c$, where $\chi_r(c)=\sigma_r(c)>0$; so $w_r\notin\mathbb T$, and $w_r\in\mathbb D$ if and only if $\chi_r$ winds once around $0$ along the
boundary of $\Delta_c\cap\mathbb D$, traversed as $A$ followed by the inner arc. On the inner arc $\chi_r=g+O(s)$, and $g$ is
conformal at $c$ with $g(c)=0$, so there the argument of $\chi_r$ increases by $\pi+O(\rho_0)+o(1)$. If $c$ is a maximum of $u$, then
$u'>0$ before $c$ and $u'<0$ after it, so on $A$ the curve $\chi_r$ runs from the lower half-plane to the upper one through a positive
point; its argument increases by an amount in $(0,2\pi)$ which, as the end points are $\mp g'(c)ic\rho_0+O(\rho_0^2)+o(1)$, is
$\pi+O(\rho_0)+o(1)$. The total is $2\pi$, and $w_r\in\mathbb D$. If $c$ is a minimum, the half-planes are exchanged, the change on $A$
is $-\pi+O(\rho_0)+o(1)$, the total is $0$, and $w_r$ lies outside the closed disk.

\emph{The zeros of $C_r$.} In $\Delta_j\cap\mathbb D$, $\chi_r$ has exactly one pole, at $r\zeta_j$, as $1/(r\bar\zeta_j)$ lies outside.
On $A_j=\mathbb T\cap\Delta_j$, $\chi_r$ has no zero: if $g(\zeta_j)\notin\mathbb R$, then $\operatorname{Im}\chi_r=\operatorname{Im}g\neq0$;
if $g(\zeta_j)>0$, then $\operatorname{Re}\chi_r>\operatorname{Re}g>0$; and if $g(\zeta_j)<0$, then $\operatorname{Im}\chi_r=-u'$ vanishes
only at $\zeta_j$, where $\chi_r=g(\zeta_j)+(1+r)/(1-r)+O(s)>0$. On the inner arc $\chi_r=g+O(s)$, and $g$ has no zero
or pole in $\Delta_j\cap\overline{\mathbb D}$, so the number of zeros of $\chi_r$ in $\Delta_j\cap\mathbb D$ is
\[
  1+\frac1{2\pi}\bigl(\Delta_{A_j}\arg\chi_r-\Delta_{A_j}\arg g\bigr)+o(1),
\]
where both curves have the same end points up to $o(1)$; we show that it is $\hat\nu_j$. If $g(\zeta_j)\notin\mathbb R$ or
$g(\zeta_j)>0$, both curves stay in one open half-plane, and the number is $1$. At a downward crossing $\operatorname{Im}g=\operatorname{Im}\chi_r$ passes from positive to negative; $g$
crosses $\mathbb R$ once, at $g(\zeta_j)<0$, so its argument increases by an amount in $(0,2\pi)$, while $\chi_r$ crosses once at a positive
point, so its argument changes by an amount in $(-2\pi,0)$; the difference is $-2\pi$ and the number is $0$. At an upward crossing the
difference is $2\pi$, and the number is $2$. At an inflection with $g(\zeta_j)<0$, $\operatorname{Im}\chi_r=\operatorname{Im}g$ keeps one
sign on both sides of $\zeta_j$, so both curves stay in one closed half-plane minus $0$, where a continuous argument gives both changes as
the same difference of end values, and the number is $1$. Summing over the disks proves the count, and the last statement follows.
\end{proof}

If $c\in\mathrm{Crit}$ is the end of a slit, that is, a punctured arc of $\mathbb T$ about $c$ lies in $\mathrm{Cont}$, then the partner
map $\beta$ of Lemma~\ref{lem:D}(e) fixes $c$ and reverses the circle near $c$, and $u\circ\beta=u$. So $u-u(c)$ has one sign
on both sides of $c$: $c$ is never an inflection, and it is a maximum of $u$ exactly when the distance to $0$ decreases as one leaves the end along
the slit. The count of the $\hat\nu_j$ is the local form of the proof of~\cite[Lemma~8.1]{JatarNg2026b}, and for Blaschke models it is
Proposition~\ref{prop:extend}(i) with $B\equiv1$; at a point of $\mathrm{Crit}$ with $u''\neq0$, the maximum case is Proposition~\ref{prop:BC}(b) with
$m=1$ and $B$ constant.

\begin{corollary}\label{cor:hook}
Let $W$ be as above, and let it map $\mathbb D$ onto a slit plane. There are $M\ge1$ and points $\zeta_1,\dots,\zeta_M$ for which
$(W,C_r)$ is admissible for all $r<1$ close to $1$ if and only if the distance to $0$ increases as one leaves each end of a slit and
some slit has an interior strict local minimum of this distance.
\end{corollary}

\begin{proof}
Each point of a slit is the image of two points of the circle, one from each edge. At an interior strict local minimum of the distance
to $0$ along the slit, the slit is tangent to a circle about $0$ and $\arg W$ moves in opposite directions on the two edges: one edge
gives a downward crossing, the other a point with $g>0$. An interior strict local maximum gives an upward crossing and a point with
$g>0$. With the paragraph before the corollary, which treats the ends of the slits, the corollary follows from the last statement of
Proposition~\ref{prop:tips}.
\end{proof}

If, for instance, the slit is a single arc with one finite end, then the distance to $0$ must increase from
the end, decrease to an interior minimum and increase again to $\infty$: the slit turns back toward $0$, a \emph{hook}\label{term:hook}. Straight slits
never qualify (Corollary~\ref{cor:straight} shows more), and by Proposition~\ref{thm:ratslit} no rational map onto a slit plane has a hook.
The gains $r^{-M}$ of these pairs tend to $1$ as $r\to1$.