What is the meaning of a "Meromorphic Univalent Function"?

Question

I lecture this weak my teacher mentioned the term "Meromorphic Univalent Function" but he didn't explain it. I tried to find the meaning of this term but the only result I could find is the definition of "Meromorphic Function" and "Univalent Function". Does anyone know the meaning of a "Meromorphic Univalent Function"? Also, what is an example of Meromorphic Univalent Function?

Also, what is an example of a "Meromorphic univalent function of complex order"?

Thanks,

Answer 1

Univalent function means analytic (holomorphic) and injective, i.e. by correctly defining the two open sets $U,V = f(U)\subset \mathbb{C}$ you can see $f : U \to V$ as being bijective and analytic (biholomorphic).

Now look at $f(z) = 1/z$ it is bijective and analytic $\mathbb{C}^* \to \mathbb{C}$, or you can say it is bijective and meromorphic $\mathbb{C} \to \mathbb{C}$ it works too.

Note that if $g(z)$ has a pole at $z=a$ then $\frac{1}{g(z)}$ has a zero at $z=a$ so there is $r,R$ such that $\frac{1}{g(z)}$ takes all the values $|z| < r$ on $|z-a| < R$, and $g(z)$ takes all the values $|z| > 1/r$ on $|z-a| < R$. Also $h(z)$ analytic is locally injective (and bijective) iff $h'(z) \ne 0$.

Overall you see that a univalent meromorphic function can only have at most one pole of order $1$.

And there is also some theorems (that I can't find a reference) saying that in some cases (if $U$ is simply connected ?), a sufficient condition for $f(z)$ being biholomorphic is that $f'(z) \ne 0$

Answer 2

A meromorphic univalent function is a function which is meromorphic and univalent (univalent means injective). Example: $f(z)=1/z$.