If $a,b\in \Bbb{C}$, with $|a|=|b|>1$ and $a^n-b^n$ is bounded, then $a=b$.

Question

Let $a,b\in \Bbb{C}$ with $|a|=|b|>1$. Show that, if $(a^n-b^n)_{n\geq 1}$ is a bounded sequence, then $a=b$.

What I've tried: Write $a=\rho (\cos \theta+i\sin \theta)$ e $b=\rho(\cos \varphi+i\sin \varphi)$, with $\rho>1$ and $\theta,\varphi\in [0,2\pi)$. Suppose that the sequence $(a^n-b^n)_{n\geq 1}$ is bounded and suppose, by contradiction, that $a\neq b$, i.e. $\theta \neq \varphi$. We have \begin{align*} |a^n-b^n|=&\rho^n|\cos n\theta-\cos n\varphi+i(\sin n\theta-\sin n\varphi)|\\ =&\rho^n(\cos^2n\theta-2\cos n\theta \cos n\varphi+\cos^2n\varphi\\ &+\sin^2n\theta-2\sin n\theta \sin n\varphi+\sin^2n\varphi)\\ =&\rho^n(2-2(\cos n\theta\cos n\varphi+\sin n\theta \sin n\varphi))\\ =&2\rho^n(1-\cos n(\theta-\varphi)), \end{align*} and I would like to show that there are natural numbers $n_k$, with $\lim_{k\to \infty}n_k=\infty$, such that $\lim_{k\to \infty}2\rho^{n_k}(1-\cos n_k(\theta-\varphi))= \infty$. I could not find such a sequence. I was able to show that, if $n(\theta-\varphi)$ has the form $2k\pi$, for some $k\in \Bbb{Z}$, then $(n+1)(\theta-\varphi)$ has not such a form. In this way, I could create a sequence $n_k\to \infty$ such that $2\rho^{n_k}(1-\cos n_k(\theta-\varphi))>0$, but not necessarily goes to infinity. I could conclude so if, for example, I could prove that \begin{align*} \inf \{1-\cos n_k(\theta-\varphi)\,:\, k\in \Bbb{N}\}>0. \end{align*}

Any ideas?

Answer 1

I think the sequence $n_k = ((2k+1)\pi/2)/\vert(\theta-\varphi)\vert$ will work $\forall\theta\neq\varphi$ as $\cos((2k+1)\pi/2) = 0 \ \forall k$.

Answer 2

Suppose $\{a^n-b^n\}$ is bounded sequence. Then, $\{a^n(1-(b/a)^n)\}$ must be bounded. Since $|a| >1$, $|a^n| \to \infty$ as $n\to \infty$. Therefore, $|(1-(b/a)^n)| \to 0$ as $n\to \infty$. Taking the limit we see that $b/a = 1$ yielding the result.