(f(z)^n - 1)/(g(z)^n - 1) is an integral function for each n

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(f(z)^n - 1)/(g(z)^n - 1) is an integral function for each n

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maxal

#1 h Mar 1, 2007, 1:19 AM • 3 Y

Y by Adventure10, Mango247, ngduchieu1903

A problem that very much resembles the "masterpiece".

Let $f(z)$ and $g(z)$ be integral functions (i.e., analytic at all finite points of the complex plane $\mathbb{C}$ ) such that for every fixed positive integer $n,$ the function $\frac{f(z)^{n}-1}{g(z)^{n}-1}$ is integral as well.
Prove that there exist positive integer $k$ such that $f(z) = g(z)^{k}.$

This post has been edited 1 time. Last edited by maxal, Dec 3, 2023, 4:21 PM
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#2 h Mar 19, 2007, 5:12 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

I have two corrections to the statement above. First: $g(z)$ must not be constant (consider $g(z)=0$ ). Second: the existence of POSITIVE integer $k$ does not generally hold ( $f(z)=e^{kz}$ , $g(z)=e^{z}$ give examples for ANY integer $k$ ).

So it needs to be proven that, for entire functions $f(z)$ and (nonconstant) $g(z)$ satisfying the condition in the post above, there is an integer $k$ such that $f(z)=g(z)^{k}$ . My proof is based on the following lemma:

Let $U(z)$ be a function analytic on some (open) domain $D$ having a common point with the circle $|z| = 1$ . Suppose that, for any $z \in D$ and any positive integer $n$ , $z^{n}= 1$ implies $U(z)^{n}= 1$ . Then $U(z) = C z^{k}$ for some integer $k$ and some constant $C$ .

(Proof.) Let's denote $E(z)=e^{2zi\pi}$ for brevity. Now take any $z=E(p/q)$ in $D$ (with $q>1$ and $p$ coprime to $q$ ); we must have $U(z)=E(r/q)$ for some integer $r$ . Moreover, for all sufficiently large integers $n$ there must exist integers $a_{n}$ such that $U(E(p/q+1/q^{n})) = E(r/q+a_{n}/q^{n})$ ; we may assume (using the continuity of $U$ ) that $\lim_{n\to\infty}(a_{n}/q^{n}) = 0$ . Now compute $U'(z)$ :
$U'(z)=\lim_{n\to\infty}\frac{E(r/q+a_{n}/q^{n})-E(r/q)}{E(p/q+1/q^{n})-E(p/q)}=E((r-p)/q)\lim_{n\to\infty}a_{n}\frac{(E(a_{n}/q^{n})-1)/(a_{n}/q^{n})}{(E(1/q^{n})-1)/(1/q^{n})}=E((r-p)/q)\lim_{n\to\infty}a_{n}.$
Thus $a_{n}$ has a limit $a$ , and therefore $a_{n}= a$ for all sufficiently large $n$ (recall that $a_{n}$ are integers). So the equation
$U(z) = C z^{a}\qquad(C = E((r-ap)/q))$
holds for $z=E(p/q)$ and for $z=E(p/q+1/q^{n})$ for all sufficiently large integers $n$ . The lemma follows by applying the principle of permanence.

Coming back to the solution; applying, for instance, Picard's theorem to $g(z)$ , we have a point $t$ such that $|g(t)|=1$ and $g'(t) \neq 0$ . By the inverse function theorem there is an open neighborhood $O$ of $t$ such that $g$ has an inverse function $h$ analytic on $g(O)$ . As for any positive integer $n$ and any $z \in O$ , $g(z)^{n}= 1$ implies $f(z)^{n}= 1$ , the function $U(z) = f(h(z))$ meets the conditions of the above lemma (with $D=g(O)$ ), so $f(z) = C g(z)^{k}$ for $C$ , $k$ from the lemma and for any $z \in O$ (hence for any complex $z$ ). Now $g(z)^{n}= 1$ implies $C^{n}= 1$ (for any positive integer $n$ and any complex $z$ ); applying Picard's theorem once again gives $C^{n}= 1$ for any positive integer $n$ except at most one. This means that $C = 1$ .

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