a zero of $\Gamma_+^m 
-\Gamma_-^m$.

$a(z)=0$ if and only if $\binom{1}{0}$ is an eigenvector of $T_p(z)$, that is, exactly at the Dirichlet data points.

$\Gamma_+^m = \Gamma_-^m$ is equivalent to $\Gamma_+^{2m}=1$ since $\Gamma_- = 
\Gamma_+^{-1}$. This implies ${\lvert\Gamma_+\rvert} = {\lvert\Gamma_-\rvert}$, so can only happen on the bands. On the bands, by , $${\label}{2.37} 
\Gamma_+(x) = \exp (\pi i p\, k(x))$$ and $\Gamma_+^{2m}=1$ if and only if $${\label}{2.38} 
mp \, k(x) \in {{\mathbb{Z}}}$$ that is, if holds for some $q=0, \dots, m$. But at $q=0$ or $q=m$, $a(z)$ has a pole that cancels the zero of $\Gamma_+^m - \Gamma_-^m$, so the zeros of $p_{mp-1}$ are precisely given by (i) and (ii) of Theorem \[T2.1\].

We can use to analyze zeros of $p_{mp+b-1}$ for large $m$. We begin with the region away from the bands:

[T2.4]{} Let $z\in{{\mathbb{C}}}\backslash\text{bands}$ and let $b$ be fixed. Then $${\label}{2.39} 
\lim_{m\to\infty} \, \Gamma_+(z)^{-m} p_{mp+b-1}(z) = a(z) q_{b-2}(z) 
+ b(z) p_{b-1}(z)$$ In particular, if the right side of is called $j_b(z)$, then

If $j_b(z_0)\neq 0$, then $p_{mp+b-1}(z)$ is nonvanishing near $z_0$ for $m$ large.

If $j_b(z_0) =0$, then $p_{mp+b-1}(z)$ has a zero [[(]{}]{}$k$ zeros if $z$ has a $k$-th order zero at $z_0$[[)]{}]{} near $z_0$ for $m$ large.

