Awkward sum of complex numbers

Question

Let $k \geq 1$ be an integer and $u_{1}, \ldots, u_{k}$ and $z_{1}, \ldots, z_{k}$ be distinct nonzero complex numbers. If the set $$ \mathscr{A}= \left\{a_{n} : a_{n}= \sum_{i =1}^k u_i \, z_{i}^n \right\} $$ for $n \in \mathbb{N}$ is finite, then prove that there exists a positive integer $d$ such that $a_{n}=a_{n+d}$ for all $n \in \mathbb{N}$.

As per the model solutions provided, this question is from José Luis Díaz-Barrero. I was given this question in a practice sheet.

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The solutions goes as below:

If the set $\mathscr{A}$ is finite then the set $$\mathscr{B} = \{(a_n, a_{n+1}, \dots, a_{n+k-1}\}$$ for $n \in \mathbb{N}$ is finite as well. Therefore there exists $p < q$ such that $$(a_p, a_{p+1}, \dots, a_{p+k-1}) = (a_q, a_{q+1}, \dots , a_{q + k -1})$$ that is to say $$(a_p = a_q) , (a_{p+1} = a_{q+1}), \dots ,( a_{p+k-1} = a_{q+k-1})$$ [Rest of the solution..]

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My question is: Why should $p,q$ exist and what guarantees that the terms will be equal if $p < q$ ? Isn't it sort of what we are trying to prove?

Answer 1

$B$ is the set of ordered k-tuples $\{(a_n, ..., a_{n+k-1})\}$ for all $n\in \mathbb N$. Since there are infinitely many $n$ but only finitely many distinct elements of $B$, there must be repeated elements. Thus there exist some $p, q$ with $(a_p, ..., a_{p+k-1}) = (a_q, ..., a_{q+k-1})$, and without loss of generality we can assume $p < q$.

As for why $B$ is finite, if $A$ has $N$ elements, then there are only $N^k$ distinct ordered $k$-tuples that can be made from elements of $A$. Thus, $|B| < N^k$.