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  "body": "\n## Statement\n\nLet $a, b, c, q, z \\in \\C$ with $0 \u003c \\abs{q} \u003c 1$, $\\abs{z} \u003c 1$, $0 \u003c \\abs{b} \u003c 1$, and with $c$ and $az$ not of the form $q^{-m}$ for an integer $m \\ge 0$.\nThen\n$${}_2\\phi_1 \\left( \\begin{matrix} a, b \\\\ c \\end{matrix} ; q, z \\right) = \\frac{(b; q)_\\infty\\, (az; q)_\\infty}{(c; q)_\\infty\\, (z; q)_\\infty}\\; {}_2\\phi_1 \\left( \\begin{matrix} c/b, z \\\\ az \\end{matrix} ; q, b \\right),$$\nwhere ${}_2\\phi_1$ is the [[basic hypergeometric series|EB4N9RQR51]] and $(\\cdot\\,; q)_\\infty$ is the [[q-Pochhammer symbol|Y1X8ZZQ7VM]].\n\n## Remarks\n\nProof idea: write $\\frac{(b; q)_n}{(c; q)_n} = \\frac{(b; q)_\\infty}{(c; q)_\\infty} \\cdot \\frac{(c q^n; q)_\\infty}{(b q^n; q)_\\infty}$.\nExpand $\\frac{(c q^n; q)_\\infty}{(b q^n; q)_\\infty}$ by the [[q-binomial theorem|VVT27RB46B]] as a series in $b q^n$, exchange the two sums, and sum over $n$ with the $q$-binomial theorem again.\nExample: for $c = b$ the left side is ${}_1\\phi_0(a; -; q, z) = \\frac{(az; q)_\\infty}{(z; q)_\\infty}$, and on the right ${}_2\\phi_1(1, z; az; q, b) = 1$ since $(1; q)_n = 0$ for $n \\ge 1$; both sides agree.\nIterating the transformation gives two more forms, and the three are often called Heine's first, second and third transformations.\nThe third form is ${}_2\\phi_1(a, b; c; q, z) = \\frac{(abz/c; q)_\\infty}{(z; q)_\\infty}\\, {}_2\\phi_1(c/a, c/b; c; q, abz/c)$, valid when both series converge.\n\n## Explanations\n\n### L1 human\n\nHeine's transformation rewrites one basic hypergeometric series as an infinite product times another basic hypergeometric series with permuted parameters.\nThe roles of the variable $z$ and the parameter $b$ are exchanged.\nSuch a change can turn a slowly converging series into a fast one, or a hard sum into an easy one.\n\n### L2 human\n\nIt is the first and most used transformation of the ${}_2\\phi_1$ series, and its third form is the $q$-analogue of Euler's transformation of the [[hypergeometric series|428M453CR2]] ${}_2F_1$.\nSetting $z = c/(ab)$ and summing the right side with the $q$-binomial theorem gives the $q$-Gauss sum ${}_2\\phi_1(a, b; c; q, c/(ab)) = \\frac{(c/a, c/b; q)_\\infty}{(c, c/(ab); q)_\\infty}$ for $\\abs{c/(ab)} \u003c 1$.\nIts terminating case leads to the [[q-Vandermonde identity|MBRSCQRGGH]].\n",
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