MHB Proving Meromorphic Function Equality w/ Liouville's Theorem

  • Thread starter Thread starter Dustinsfl
  • Start date Start date
  • Tags Tags
    Functions
Dustinsfl
Messages
2,217
Reaction score
5
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?
 
Physics news on Phys.org
dwsmith said:
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.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
Back
Top