Testing $q$-binomial theorem at $z=1/a$ The $q$-binomial theorem states that when $|q|<1$ $$ \frac{(az;q)_\infty}{(z;q)_\infty}={}_1 \phi_0 (a;\;;q;z)\equiv \sum_{n=0}^\infty \frac{(a;q)_n}{(q;q)_n}z^n \quad \text{for } |z|<1$$ with _no restriction_ on $a$. The proof can be found in Gasper & Rahman - _Basic hypergeometric series_ or in Andrews, Askey, Roy - _Special functions_. Since there is no restriction on $a \in \mathbb{C}$, we can choose $a$ such that $|a|>1$. My question is the following: the left-hand side of this equation has a _zero_ at $z=a^{-1}$ since $(1,q)_\infty=(1-1)(1-q)\dots=0$, while the right-hand side is not obviously zero at $z=a^{-1}$. How are these two facts reconciled?

You made a good observation and it is not obvious, but the following may be of interest to you.

If $z=1/a=q^k$ for some positive integer $k$, then $\sum_{n=0}^k \frac{(a;q)_n}{(q;q)_n}z^n = 0$ and also the terms of the summation for $n>k$ are zero because then $(a;q)_n=0$.

Perhaps this makes the identity more plausible for you.