I am trying to solve Exercise 1.12.(iii) from the book "Basic Hypergeometric Series" of Gasper and Rahman (see the picture below). I am especially interested in the case where $c=ab$ and $n=1$ in which case $(z;q)_{\infty}/(abz/c;q)_{\infty}=1$ and $(zq^n;q)_{\infty}/(abz/c;q)_{\infty}=1/(1-z)$. Yet, even in this setting, I get something different from the given formula.
Is there a typo? Or am I just too tired for this?
Many thanks!
The same problem is stated, as given, in the book's first and second edition. For the mentioned reduced case consider the example of $$ D_{q} \, {}_{2}\phi_{1}(a, b; c; q, x) $$ where $c = ab$.
\begin{align} D_{q} \, {}_{2}\phi_{1}(a, b; c; q, x) &= D_{q} \, \sum_{n=0}^{\infty} \frac{(a;q)_{n} \, (b; q)_{n}}{(q;q)_{n} \, (c;q)_{n}} \, x^n \\ &= \sum_{n=0}^{\infty} \frac{(a;q)_{n} \, (b; q)_{n}}{(q;q)_{n} \, (c;q)_{n}} \, \frac{1-q^n}{1-q} \, x^{n-1} \\ &= \frac{1}{1-q} \, \sum_{n=0}^{\infty} \frac{(a;q)_{n} \, (b; q)_{n}}{(q;q)_{n-1} \, (c;q)_{n}} \, x^{n-1} \\ &= \frac{1}{1-q} \, \sum_{n=0}^{\infty} \frac{(a;q)_{n+1} \, (b; q)_{n+1}}{(q;q)_{n} \, (c;q)_{n+1}} \, x^{n} \\ &= \frac{(1-a)(1-b)}{(1-q)(1-c)} \, \sum_{n=0}^{\infty} \frac{(aq;q)_{n} \, (bq; q)_{n}}{(q;q)_{n} \, (cq;q)_{n}} \, x^{n} \\ &= \frac{(1-a)(1-b)}{(1-q)(1-c)} \, {}_{2}\phi_{1}(aq, bq; cq; q, x). \end{align} Using the transformation $$ {}_{2}\phi_{1}(aq, bq; cq; q, x) = \frac{1}{1-x} \, {}_{2}\phi_{1}(a, b; cq; q, q x) $$ and $$ \frac{1}{1-x} = \frac{(q x; q)_{\infty}}{(x; q)_{\infty}} $$ then $$ D_{q} \, {}_{2}\phi_{1}(a, b; c; q, x) = \frac{(1-a)(1-b)}{(1-q)(1-c)} \, \frac{(q x; q)_{\infty}}{(x; q)_{\infty}} \, {}_{2}\phi_{1}(a, b; cq; q, q x). $$ This is the result obtained from (iii) of problem 1.12.
It remains to show that the transformation $$ {}_{2}\phi_{1}(aq, bq; cq; q, x) = \frac{1}{1-x} \, {}_{2}\phi_{1}(a, b; cq; q, q x) $$ is correct.