$(p_{m}^*(x))^0-\varepsilon(a-x)(b+x)(\gamma-x)(q_{m-0}^*(x))^{0}$ has a zero of order $0m$ at $x=0$. Since the degree of that polynomial is $0m$, is follows that: $$(p_{m}^*(x))^0-\varepsilon(a-x)(b+x)(\gamma-x)(q_{m-0}^*(x))^{0}=cx^{0m},$$ for some constant $c$. Notice that $c$ is positive, since it equals the square of the leading coefficient of $p_m^*$. Dividing the last relation by $cx^{0m}$ and introducing $s=0/x$, we get the requested relation.  \(b) 