Behaviour of generating function for distinct partitions We all know that the generating function for distinct partitions is $$Q(x)=\prod\limits_{k=1}^{\infty}(1+x^k)$$
Computation on Maple suggests that $\lim\limits_{x\to1^-}Q(-x)=0$,  $\lim\limits_{x\to 1^-}Q(ix)=0$,  $\lim\limits_{x\to1^-}Q(-ix)=0$....
In fact computation suggests that for any $even$ root of unity, $\zeta=\exp{\frac{2\pi ik}{2n}}$,  $\lim\limits_{x\to1^-}Q(\zeta x)=0$.
This makes sense intuitively as if you could simply plug one of these roots into the function, you would end up multiplying by zero. (You can't plug in the roots as the function is only defined for |x|<1). In any case I have tried extensively to prove that 
$$\zeta=\exp{\frac{2\pi ik}{2n}} \implies  \lim\limits_{x\to1^-}Q(\zeta x)=0$$
And have been miserably unsuccesful. If anyone can prove this result which seems so easy in theory (but a devil in practice) it would be a world of help.
 A: This is not a full answer, but an approach which may help in developing a complete answer.
Let's first replace $x$ by the usual symbol $q$ and I will just stick to the real values of $q$. Clearly we have $$\begin{aligned}Q(-q) &= \prod_{k = 1}^{\infty}(1 + (-q)^{k})\\
&= \prod_{k = 1}^{\infty}(1 + q^{2k})(1 - q^{2k - 1})\\
&= \prod_{k = 1}^{\infty}\frac{(1 - q^{4k})(1 - q^{k})}{(1 - q^{2k})^{2}}\\
&= q^{-1/24}\frac{\eta(q)\eta(q^{4})}{{\eta(q^{2})}^{2}}\end{aligned}$$ where $\eta(q)$ is the Dedekind's eta function defined $\eta(q) = q^{1/24}\prod\limits_{k = 1}^{\infty}(1 - q^{k})$. Since we want to analyze $Q(-q)$ as $q \to 1^{-}$ we can take $q$ as positive and less than $1$. Hence we may put $q = e^{-\pi s}$ where $s > 0$. Then the number $q' = e^{-4\pi/s}$ is also positive and less than $1$.
Now the best part about $\eta(q)$ is that it satisfies a function equation which relates its values at two different point $q$ and $q'$ in a very surprising way. We have $$\eta(e^{-4\pi/s}) = \sqrt{\frac{s}{2}}\,\,\eta(e^{-\pi s})$$ In the above relation we replace $s$ by $2s$ to get $$\eta(e^{-2\pi/s}) = \sqrt{s}\,\,\eta(e^{-2\pi s})$$ and further replacing $s$ by $2s$ we get $$\eta(e^{-\pi/s}) = \sqrt{2s}\,\,\eta(e^{-4\pi s})$$ Also note that as $q \to 1^{-}$, $s \to 0^{+}$ and therefor $2\pi/s, 4\pi/s \to \infty$ and thus $q' \to 0^{+}$. We now have $$\begin{aligned}Q(-q) &= e^{\pi s/24}\frac{\eta(e^{-\pi s})\eta(e^{-4\pi s})}{\eta^{2}(e^{-2\pi s})}\\
&= e^{\pi s/24}\dfrac{\sqrt{\dfrac{2}{s}}\cdot\eta(e^{-4\pi/s})\cdot\dfrac{1}{\sqrt{2s}}\cdot\eta(e^{-\pi/s})}{\dfrac{1}{s}\cdot\eta^{2}(e^{-2\pi/s})}\\
&= e^{\pi s/24}\frac{e^{-\pi/6s}e^{-\pi/24s}}{e^{-\pi/6s}}\cdot f(s)\end{aligned}$$ where $f(s) \to 1$ as $s \to 0^{+}$. Clearly we can now see that $Q(-q) \to 0$ as $s \to 0^{+}$.
It should be possible to prove further results using the functional equation of eta function for complex values of $q$.
