# Awkward sum of complex numbers

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.

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..]

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?

$$B$$ 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, $$|B| < N^k$$.