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.
Meromorphic functions of non integral order
complex-analysismeromorphic-functions
Best Answer
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 $a$