Proving Meromorphic Function Equality w/ Liouville's Theorem

[Dustinsfl]

Let $r(z)$ and $s(z)$ be meromorphic functions on the complex plane. Assume that there is a real number $C$ such that
$$
|r(z)|\leq C|s(z)|,\quad\text{for all complex numbers} \ z.
$$
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?

 

[Opalg]

Let $r(z)$ and $s(z)$ be meromorphic functions on the complex plane. Assume that there is a real number $C$ such that
$$
|r(z)|\leq C|s(z)|,\quad\text{for all complex numbers} \ z.
$$
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?

Yes, that is correct. The function r/s is meromorphic, and bounded by C. Therefore it has no poles, and Liouville's theorem shows that it must be constant.