There are at most $2p+2b-3$ points in ${{\mathbb{C}}}\backslash\text{bands}$ where $j_b (z_0)$ is zero.

is immediate from and ${\lvert\Gamma_-/\Gamma_+\rvert}<1$. (1) and (2) then follow by Hurwitz’s theorem if we show that $j_b(z)$ is not identically zero.

By and near $z=\infty$, $$T_p(z) = \biggl( \, \prod_{j=1}^p a_j \biggr)^{-1} z^p 
\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} +O(z^{p-1})$$ which implies $\Gamma_+ = (\prod_{j=1}^p a_j)^{-1} z^p + O(z^{p-1})$ and $\Gamma_-(z) 
=O(z^{-p})$. It follows that $a(z)\to 0$ as $z\to\infty$ and $b(z)\to 1$. Thus, since $p_{b-1}$ has degree $b-1$, shows that as $z\to\infty$ on the main sheet, $f(z_0)$ has a pole of order $b-1$.

On the other sheet, $P_+$ changes to $P_-$, so $a(z)\to 0$ and $b(z)\to 0$ on the other sheet. It follows that $j(z)$ has a pole at $\infty$ of degree at most $b-2$. $j$ also has poles of degree at most $1$ at each branch point. Thus, $j_b(z)$ as a function on ${{\mathcal S}}$ has total degree at most $2p+ (b-1) + (b-2) =2p + 2b-3$ which bounds the number of zeros.

Finally, we turn to zeros on the bands. A major role will be played by the function on the right side of ($j$ is for “Jost" since this acts 