Cancellation of a sum of a product of analytic functions

Question

Consider a (finite, $k = 1, \ldots, K < \infty$) collection of analytic functions $p_k(z)$ and $q_k(z)$, where $p_k(0) = 0$ and $q_k(0) = 0$.

Suppose also that all of $p_k$ and $q_k$ have an infinite number of nonzero terms. That is to say

$$ p_k(z) = \sum_{n = 1}^\infty p_n^{(k)}z^n, $$

where $\forall N \in \mathbb{N}, \exists n > N$ s.t. $p_n^{(k)} \ne 0$.

Then define

\begin{align} T(z) &= \sum_{k = 1}^K p_k(z)q_k(1/z)\\ &= \sum_{n = -\infty}^\infty t_n z^n. \end{align}

Is it possible that $\exists n > 0$ such that $t_n \ne 0$ but $\forall n < 0$ we have $t_n = 0$? i.e. that $T(z)$ is one-sided?

It seems that it could only be possible that $T(z) = 0$ (i.e. if $p_1 = p_2$ and $q_1 = -q_2$), but that $T(z)$ can't be one-sided.

I'm not very well versed in this kind of function theory, but it seems there should be some theorems that can help to understand this situation.

Answer 1

Forgive me if I misunderstand, but I think modifying the idea of Mindlack for infinitely many nonzero terms, what about $p_k(z) = 2^{-k}z^2$ and $q_k(w)=2^{-k}w$. Then we have $$T(z)=\sum_{k \geq 0}4^{-k}z =z \sum_{k \geq 0} 4^{-k} = \frac{4}{3}z$$ which has a one-sided Laurent series.

Answer 2

I have synthesized the ideas of other posters into a definitively positive answer with $K = 2$.

Take $h(z) = \sum_{n = 2}^\infty h_n z^{n}$, with each $h_n \ne 0$. Then,

$p_1(z) = z^2 + h(z), p_2(z) = p_1(z)$

$q_1(z) = z + h(z), q_2(z) = z - h(z).$

This works out to

$p_1(z)q_1(1/z) + p_2(z)q_2(1/z) = 2z + 2z^{-1}h(z)$, which is one sided.