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$$

|r(z)|\leq C|s(z)|,\quad\text{for all complex numbers} \ z.

$$

Prove that $r(z) = C_1s(z)$, for some complex number $C_1$.

This intuitively makes sense. Basically the isolated singularities i.e. poles of r and s are removable when we have r/s such that r/s is a constant. Would the use of Liouville's theorem be used here?