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