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[SOLVED] meromorphic functions

dwsmith

Well-known member
Feb 1, 2012
1,673
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

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
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.