Meromorphic functions of non integral order

Question

We know that an entire function of non integral order has infinitely many zeros. What can be said about the zeros and poles of a meromorphic function of non integral order? In fact, is there any representation of meromorphic functions of non integral order? The order of a meromorphic function is defined as $\sigma=\displaystyle{\limsup\limits_{r\rightarrow\infty}\frac{\log T(r,f)}{\log r}}$, where $T(r,f)$ is the Nevanlinna characteristic function.

Answer 1

Sketch for the answer (the lecture notes of Hayman has proofs)- a good intuitive definition of the order of meromorphic functions is that if we write $f=g/h$, $g,h$ entire and "minimal" (no common factors), the order of $f$ is the maximum of the orders of $g,h$; rigorously we need to use the Nevalinna characteristic of course.

In particular, this immediately implies that a meromorphic function of non-integral order takes all but at most one values (including infinity) infinitely many times as either $g$ or $h$ must have non-integral order hence so does $g-ah$ for at most one finite $a$ and same for the poles of $f$ which are the zeroes of $h$.

We cannot in general do better as taking $(\cos \sqrt z)/z$ which has precisely one pole and non-integral order $1/2$ shows. Same for $(a\cos \sqrt z-1)/(\cos \sqrt z)$ and finite $a$