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I have stumbled upon the following fact, easily confirmed numerically: The $q$-Pochhammer symbol $(a;z)_L$ with $z$ given by the $L$th root of unity, $$ (a;\mathrm{e}^{2\pi i/L})_L = \prod_{n=0}^{L-1} \left(1 - a \, \mathrm{e}^{2\pi i n/L}\right)\,\text{,} $$ obeys $$ (a;\mathrm{e}^{2\pi i/L})_L = 1 - a^L\,\text{,} $$ apparently for all complex $a$ and positive integer $L$.

Does this identity have a name? Can anyone provide a proof? (I can prove it when $L$ is an exact power of $2$, but the result is more general. I also have a "proof" using contour integration, but it involves a couple of dubious steps.)

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Let $r_1,\cdots,r_{L-1}$ be the complex $L$-th roots of unity. We have that $$ r_1^k+r_2^k+\cdots +r_{L-1}^k = -1$$ for all $k=1,\cdots, L-1.$ From Newton's identities we can then prove that $$ e_k(r_1,\cdots, r_{L-1})=(-1)^k$$ where $e_k$ is the $k$-th elementary symmetric polynomial. Now, observe that (setting $r_0=1$)\begin{align*} \prod\limits_{k=0}^{L-1}(1-ar_k) &= a^{L-1}(1-a)\prod\limits_{k=1}^{L-1}(\frac{1}{a}-r_k) \\ &= a^{L-1}(1-a)\left(\sum_{k=0}^{L-1}(-1)^{L-1+k}e_{L-1-k}\frac{1}{a^k}\right) \\ &= a^{L-1}(1-a)\left(\sum_{k=0}^{L-1}(-1)^{2(L-1)}\frac{1}{a^k}\right) \\ &= 1-a^L. \end{align*}

Edit:

Here is a quick proof.

Define the polynomial $$P(X) = \prod\limits_{k=0}^{L-1}(1-r_k X)$$ this is of degree $L$ and has roots exactly the $L$-th roots of unity. Hence $P(X) = 1-X^L$.

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  • $\begingroup$ Would you mind splitting the quick proof into a separate answer, so I can accept it? (I think it's also necessary to observe that $P(0)=1$ to fix the overall factor and complete the proof.) $\endgroup$ – Stephen Powell Jan 4 '17 at 11:12
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For the record, here's a sketch of my derivation using contour integration, which I believe is valid at least for $0 < a < 1$. Define the function $$ f(z) = \frac{L}{z}\frac{1}{z^L - 1}\,\text{,} $$ which has poles at $z = e^{2\pi i n / L}$ with residue $+1$ and at $z = 0$ with residue $-L$. Then \begin{align} \log [ (a;e^{2\pi i n/L})_L] &= \sum_{n=0}^{L-1}\log (1 - a e^{2\pi i n / L})\\ &= \oint \frac{dz}{2\pi i}\, f(z) \log(1 - a z)\,\text{,} \end{align} where the integration contour surrounds the unit circle.

For $0<a<1$ and $L \ge 1$, the integration contour can be deformed so that it surrounds the branch cut of $\log(1-a z)$ on the real axis for $z > 1/a$. The discontinuity across the branch cut is $2\pi i$, so the integral becomes \begin{align} \log [ (a;e^{2\pi i n/L})_L] &= \int_{1/a}^{\infty} dx\, f(x)\\ &= \log (1 - a^L)\,\text{.} \end{align}

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