Suppose $f$ and $g$ are meromorphic functions on $\mathbb{C}$ such that $g(z)=f(1/z)$ for $z\neq 0$. Show that $f$ is a rational function.
Initially i thought that nonzero $a_i$'s are zeros(or poles) of $f$ then $\frac{1}{a_i}$'s are poles(or zeros) of $g(1/z)$ and i tried to express them in terms of factors in the quotient of the representation of $f$ and $g$ but could get nowhere. Any hint(s) will be highly appreciated. Thank you
Since you are given $f$ is meromorphic in $\mathbb{C}$, it suffices to show that $f$ has either a removable singularity or a pole at $\infty$. Notice $g(0)=f(\infty)$, and since $g$ is meromorphic on $\mathbb{C}$, this implies that either $g(0)$ is finite (which means $f$ has a removable singularity at $\infty$) or $0$ is a pole of $g$ (so $f$ has a pole at $\infty$). Thus $f$ is meromorphic on $\mathbb{\hat{C}}$ which implies that $f$ is rational.
A reference for the last line is here: Why is every meromorphic function on $\hat{\mathbb{C}}$ a rational function?
By the definition of meromorphic function, $f$ and $g$ don't have any singularity other than pole (neither essential singularity nor removable singularity). Let $A=\{a_i\in\mathbb{C}:g(a_i)=\infty\}$ and B=$\{b_j\in\mathbb{C}:g(b_j)=0\}$ be the sets of all the poles and all the zero's of $g$, respectively. Then, $f(1/a_i)=\infty$ and $f(1/b_j)=0$. So, there should exist entire functions $G(z)$ and $F(z)$, in both of which $F(z)\neq 0$ and $G(z)\neq 0$ for $\forall z\in\mathbb{C}$, s.t. $$g(z)=\frac{{\prod_j (z-b_j)^{h'_j}}}{\prod_i (z-a_i)^{h_i}}G(z)$$ $$f(z)=\frac{\prod_j (z-1/b_j)^{k'_j}}{\prod_i (z-1/a_i)^{k_i}}F(z)$$ where $h$ and $k$ are positive integers.
Case I: (and a trivial case $f(1/z)=g(z)=\infty$) If $f(z)<\infty$ at $z=0$, then $f(1/z)=g(z)$ for $z\in\mathbb{C}$ by identity theorem.
$$g(z)=f(1/z)=Cz^m\frac{\prod_j (z-1/b_j)^{k'_j}}{\prod_i (z-1/a_i)^{k_i}}F(1/z), m\in\mathbb{Z}, C\in\mathbb{C}...(1)$$
This equation holds for $z\in\mathbb{C}$, so $CF(1/z)=G(z)<\infty$. By Liouville's theorem, $f(z)$ is a rational function.
Case II: If $f(1/z)=\infty$ and $g(z)<\infty$ at $z=0$, then $f(1/z)$ has a removable singularity at $z=0$. Therefore, $f$ isn't a meromorphic function.
Case III: If $g(z)=\infty$ and $f(1/z)<\infty$ at $z=0$, then $g$ isn't a meromorphic function for the same reason.
Q.E.D.
