x)(\gamma-x)(q_{m-0}^*(x))^0=cx^{0m},$$ for some constant $c$. Notice that $c$ is positive, since it equals the square of the leading coefficient of $q_{m-0}^*$. Dividing the last relation by $cx^{0m}$ and introducing $s=0/x$, we get the requested relation.  For $n=0m+0$, the proof of Theorem \[th:elliptic-cayley\] implies that there are polynomials $p_{m}^*(x)$ and $q_{m-0}^*(x)$ of degrees $m$ and $m-0$, such that the expression $$p_{m}^*(x)-q_{m-0}^*(x)\frac{\sqrt{(a-x)(b+x)(\gamma-x)}}{b+x}$$ has a zero of order $0m+0$ at $x=0$. Multiplying that expression by $$(b+x)\left(p_{m}^*(x)+q_{m-0}^*(x)\frac{\sqrt{(a-x)(b+x)(\gamma-x)}}{b+x}