turn to the analysis of zeros of $\varphi_{mp+b}(z)$, $b=0,1,\dots, p-1$; $m=0,1,2,\dots$. The analog of is, by , $${\label}{3.22} 
\varphi_{mp+b} =\bigg\langle \binom{1}{0}, T_b (T_p)^m \binom{1}{1}\bigg\rangle$$ As in Section \[s2\], we write, for $z\in{{\mathbb{C}}}\backslash\text{bands}$: $${\label}{3.23} 
z^{-p/2} T_p(z) = \Gamma_+(z) P_+(z) + \Gamma_-(z) P_-(z)$$ where $P_\pm$ are $2\times 2$ matrices which are complementary projections, that is, / hold. / are replaced by $$\begin{aligned}
P_+ &= {\frac}{z^{-p/2} T_p(z) - \Gamma_-(z)\boldsymbol{1}}{\Gamma_+ - \Gamma_-} {\label}{3.24} \\
P_- &= {\frac}{z^{-p/2} T_p(z) - \Gamma_+(z)\boldsymbol{1}}{\Gamma_- - \Gamma_+} {\label}{3.25} \end{aligned}$$ So, in particular, $P_\pm$ have meromorphic continuations to ${{\mathcal S}}$, and $P_+$ continued to the other sheet is $P_-$.

Define $$\begin{aligned}
a(z) &= {\frac}12\, \bigg\langle\binom{1}{1}, P_+ \binom{1}{1}\bigg\rangle {\label}{3.26} \\ 
b(z) &= {\frac}12\, \bigg\langle\binom{1}{-1}, P_+ \binom{1}{1}\bigg\rangle {\label}{3.27} \end{aligned}$$ so that, by , $$\begin{aligned}
{\frac}12\, \bigg\langle\binom{1}{1}, P_-\binom{1}{1}\bigg\rangle &= 1-a(z) {\label}{3.28} \\ 
{\frac}12\, \bigg\langle\binom{1}{-1}, P_-\binom{1}{1}\bigg\rangle &= -b(z) {\label}{3.29}\end{aligned}$$ Thus, since ${\frac}{1}{\sqrt2}\binom{1}{1}, {\frac}{1}{\sqrt2}\binom{1}{-1}$ are an orthonormal basis, $${\label}{3.30} 
\begin{split}
z^{-mp/2} T_{mp}\binom{1}{1} &= \Gamma_+^m \biggl[ a(z) \binom{1}{1} + b(z) \binom{1}{-1}\biggr] \\
& \qquad \qquad + \Gamma_-^m \biggl[ (1-a(z))\binom{1}{1} - b(z)\binom{1}{-1}\biggr] 
\end{split}$$ Therefore, by , , and , $${\label}{3.31} 
\begin{split} 
\varphi_{mp+b}(z) &= \varphi_b(z) [a(z)z^{mp/2}\Gamma_+^m + (1-a) z^{mp/2}\Gamma_-^m] \\
& \qquad \qquad + \psi_b(z) [b(z) z^{mp/2} \Gamma_+^m -b(z) z^{mp/2}\Gamma_-^m] 
\end{split}$$ We thus define $${\label}{3.32} 
j_b(z) =a(z) \varphi_b(z) 