Can $\frac{zR(z)-aR(a)}{z-a}$ have an inside root if $R$ does not?

Question

Let $R:\mathbb C \to \mathbb C$ be a rational function, and for some $a\in \mathbb C$ let $$S(z)\equiv\frac{zR(z)-aR(a)}{z-a}.$$ Can $S$ have a root inside the unit circle if $R$ does not?

Answer 1

Assuming that $S$ is assigned the limiting value at $z=a$ the answer is yes. Take $R(z)=z-2a$. If $0<|a| <1$ then $S(a)=0$ so there is a root at $a$ but $R(a)=-a \neq 0$.