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Consider two entire functions with no zeroes and having a ratio equal to unity at infinity. Use Liouville’s Theorem to show that they are in fact the same function.

My attempt Consider $h(z) = f(z)/g(z)$. First of all, $h$ is entire, since $f$ and $g$ are entire, and $g(z)$ is nonzero for all $z$ in $\mathbb{C}$.

The fact that $\lim_{z→\infty} h(z) = \lim_{z→\infty} f(z)/g(z) = 1$ suggests that $h(z)$ is bounded as well.

Why: Since $\lim_{z→\infty} f(z)/g(z) = 1$, there exists $N > 0$ such that $|f(z)/g(z) - 1| < 1$ for all $|z| > N$. (Note that I am taking $\epsilon = 1$ for concreteness.)

Then, $|f(z)/g(z) - 1| \geq ||f(z)/g(z)| - 1|$ $\implies ||f(z)/g(z)| - 1| < 1$ $\implies |f(z)/g(z)| - 1 < 1$ or $1 - |f(z)/g(z)| < 1$ $\implies 0 < |f(z)/g(z)| < 2$.

That is, $|f(z)/g(z)|$ is bounded above by $2$ for $|z| > N$. Moreover, we know that $|f(z)/g(z)|$ has a maximum $M$ on $|z| \leq N$ by maximum modulus principle (or simply from $|z| \leq N$ being compact). Hence, $|f(z)/g(z)|$ is bounded above by $\max \{2, M\}$ for all z in $\mathbb{C}$.

Hence, $h(z)$ is constant by Liouville's Theorem, i.e. $h(z) = f(z)/g(z) = c$ for some constant $c$.

Since this is true for all $z$ in $\mathbb{C}$, taking the limit as $z\to \infty$ yields $c = 1$. Hence $f(z) = g(z)$ for all $z$ in $\mathbb{C}$.

Is correct my work?

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    The invocation of the maximum modulus principle is incorrect, but continuity and the compactness you mentioned give boundedness on $|z|\leq N$.2012-06-07

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Since both are entire functions without zeros, $h(z) = \frac{f(z)}{g(z)}$ is an entire function. $\lim_{z \rightarrow \infty} \frac{f(z)}{g(z)} \rightarrow 1$ where $\infty$ can be interpreted as the point at infinity, implies that $h$ is a bounded. A bounded entire function is constant $c$. $h(z) = c$ implies that $f(z) = c g(z)$. However $\lim_{z \rightarrow \infty} h(z) = 1$ implies that $c = 1$.

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    @Gerry: Which is fine by me if it suits her. The question was answered only in the comments I suppose, or perhaps implicitly by William's endorsement by repetition, and comments can't be accepted, so it is at least a way to signal that no further answer is requested.2012-06-13