$f(z)$ and $g(z)$ are Meromorphic functions such $|f(z)|\le|g(z)|$ for all $z\in\mathbb{C} $ then $ f=ag$

We know that if $f(z)$ and $g(z)$ are entire functions such that $g(z)\ne0$ and $|f(z)|\le|g(z)|$ for all $z\in\mathbb{C} $ then by Liouville's theorem $$ f=ag$$ for some constant $a\in \mathbb{C} $ .

Now my question is this that similar to above argument, if $f(z)$ and $g(z)$ are Meromorphic functions such $|f(z)|\le|g(z)|$ for all $z\in\mathbb{C} $ then I want to show $$ f=ag$$ for some constant $a\in \mathbb{C} $ .

I am thinking in this way that because because poles and zeros of Meromorphic functions are isolated , by the Riemann's theorem on removable singularities and By using analytic continuation to eliminate removable singularities also we can have $$ f=ag$$ for some constant $a\in \mathbb{C} $ .

Is this true way?

Solution 1:

Yes, exactly. You know that $\frac{f}{g}$ is holomorphic on $\mathbb{C}\setminus (P_f \cup Z_g)$, where $P_f$ is the set of poles of $f$ and $Z_g$ the set of zeros of $g$, and since both of these are closed discrete subsets of $\mathbb{C}$, so is $P_f \cup Z_g$.

Since $\frac{f}{g}$ is bounded, each of the points in $P_f \cup Z_g$ is a removable singularity, so, after removing them, $\frac{f}{g}$ is a bounded entire function. Hence constant.