Consider $f$ and $g$ entire function s.t. $\lim_{|z|\to\infty} |f(z)/g(z)|=1$. Show that $f=cg$.

Question

Consider $f$ and $g$ entire function s.t. $\lim_{|z|\to\infty} |f(z)/g(z)|=1$.

(a) Suppose that $f$ and $g$ have no zeroes. Show that $f=cg$.

(b) Does (a) also hold without the assumptions that $f$ and $g$ have no zeroes?

My attempt:

(a) Let $h(z)=f(z)/g(z)$. The given limit means: there is some $M\in\mathbb{R}$ s.t. $$|z|>M\Rightarrow |h(z)-1|\le\varepsilon \text{ for all } \varepsilon>0.$$ Then, $|h(z)|\le |h(z)-1|+1 \le \varepsilon+1$. So, $h$ is an entire function that is bounded on $\mathbb{C}$ and therefore constant. So, $f=cg$ for some constant $c\in\mathbb{C}$.

(b) Suppose that $f$ has zeroes $a_1,\dots,a_n$ and $g$ has zeroes $b_1,\dots,b_m$. Then $$h(z):=\frac{f(z)(z-b_1)\cdots (z-b_m)}{g(z)} \text{ is entire}.$$ So, $\lim_{|z|\to\infty}|h(z)|= \lim_{|z|\to\infty} |z-b_1|\cdots |z-b_m|=\infty.$ Meaning, that as $|z|> N$, then $|h(z)|> K$, i.e. $|f(z)/g(z)|>K|p(z)|$, $p\in\mathbb{C}[z]$. But we're given that $|f/g|\to 1$ as $|z|>N$. Does this imply that $f=cg$?

Thanks.

Answer 1

$f(z) = z$ and $g(z) = z-1$ (or any polynomials with the same degree and same leading coefficient) shows that (b) does not hold.

What you can conclude is that the meromorphic function $h = f/g$ has only finitely many poles and only finitely many zeros, and therefore is (the restriction of) a rational function.

Answer 2

(a) You did not prove that $h$ is bounded; all that you have proved was that, for some $M>0$, the restriction of $h$ to $\{z\in\mathbb C\mid\lvert z\rvert>M\}$ is bounded.

There is some $M>0$ such that $\lvert z\rvert>M\implies\left\lvert\frac{f(z)}{g(z)}\right\rvert<2\implies\bigl\lvert h(z)\bigr\rvert<2$. Let $N$ be the maximum of the restriction of $\lvert h\rvert$ to $\overline{D(0,M)}. Then$$(\forall z\in\mathbb C):\bigl\lvert h(z)\bigr\rvert<\max\{2,N\}.$$So, $H$ is bounded and therefore constant.