Let $F(z)=\displaystyle \sum_{k=0}^\infty a_kz^k,\;|z|<R $ and $F(R)=\displaystyle \sum_{k=0}^\infty a_kR^k$ (the series converges).

Assume that $F(\alpha_j)=0,\;j=1,2,\dots ,m$, where all $|\alpha_j|<R$, Then $$F(z)=(z-\alpha_1)\dots (z-\alpha_m)\cdot \displaystyle \sum_{k=0}^\infty b_kz^k,\;|z|<R. $$ This is obvious, because $F(z)/[(z-\alpha_1)\dots (z-\alpha_m)]$ is analytic in $|z|<R.$

**My question**. Does $\sum_{k=0}^\infty b_kR^k$ converge?

The question is related to the research on the $q$-Bernstein polynomials. The relevant information is contained, for example, in S. Ostrovska, The $q$-Versions of the Bernstein Operator: From Mere Analogies to Further Developments, Results in Mathematics, 69(3), 275-295, Section 4.2.