Orthogonal Polynomial Solutions in $\frac{a}{z}+\frac{uz}{a}$ of Complex q-Difference Equations

Abstract

It is also possible to obtain real polynomial solutions of the (complex) q-difference equation (12.2.1)

$$\widehat{\varphi}(qz^*)\left(\mathcal{D}_q^2\widehat{y}_n\right)(z^*)+\widehat{\psi}(qz^*)\left(\mathcal{D}_q\widehat{y}_n\right)(z^*)=\widehat{\lambda_n}\widehat{\rho}(qz^*)\widehat{y}_n(qz^*),\quad n=0,1,2,\ldots$$

with argument $z^{*}:=\frac{a}{z}+\frac{uz}{a}$ where u∈ℝ∖{0} and a,z∈ℂ∖{0}. By using z=x+iy, a=α+i β with x,y,α,β∈ℝ, we find that the imaginary part of

$$\frac{a}{z}+\frac{uz}{a}=\frac{\alpha+i\beta}{x+iy}+\frac{u(x+iy)}{\alpha+i\beta}=\frac{(\alpha+i\beta)(x-iy)}{x^2+y^2}+\frac{u(x+iy)(\alpha-i\beta)}{\alpha^2+\beta^2}$$

equals

$$\frac{(\beta x-\alpha y)(\alpha^2+\beta^2)+u(\alpha y-\beta x)(x^2+y^2)}{(x^2+y^2)(\alpha^2+\beta^2)}=\frac{(\beta x-\alpha y)\left\{\alpha^2+\beta^2-u(x^2+y^2)\right\}}{(x^2+y^2)(\alpha^2+\beta^2)}.$$

This is equal to zero for all x∈ℝ and y∈ℝ if

$$x^2+y^2=r^2\quad\textrm{and}\quad ur^2=\alpha^2+\beta^2\,(=a\overline{a}).$$