For a complex number $z$, let $(1+z)^{15}=a\_{0}+a\_{1} z+\cdots+a\_{15} z^{15}$. The value of
$$
\left(a\_{0}-4 a\_{2}+16 a\_{4}+\cdots-2^{14} a\_{14}\right)^{2}+\left(2 a\_{1}-8 a\_{3}+32 a\_{5} \cdots-2^{15} a\_{15}\right)^{2}
$$

While I was entirely stumped by this question, I found some similar questions on here, specifically [A sum involving binomial coefficients, powers and alternating signs](https://math.stackexchange.com/questions/1942401/a-sum-involving-binomial-coefficients-powers-and-alternating-signs). This did not equip me to be able to solve the above question, though, since this problem drops odd terms and even terms, alternates signs, AND has a factor of increasing powers of $2$, not to mention squares the result and then sums them.

I don't see how the sum can even be expressed as a binomial to somehow use calculus, nor do I see any algebraic manipulations; I've only managed to embarrassingly restate the problem statement as

$$ \left(\sum\_{k=0}^{7} 2^{2 k}({ }^{15} C\_{2 k})(-1)^k\right)^{2}+\left(\sum\_{k=0}^{7} 2^{2 k+1}\left({ }^{15} C\_{2 k+1}\right)(-1)^k\right)^{2} $$