More Fun in 2015, Part 40

$(1-\omega)^{2015}=(\omega(\bar{\omega}-1))^{2015}=\omega^{2015}(\bar{\omega}-1)^{2015}=-(1-\bar{\omega})^{2015}$ . Thus the summands of $S$ cancel out in conjugate pairs, $\frac{1}{(1-\omega)^{2015}}+\frac{1}{(1-\bar{\omega})^{2015}}=0$ , and $S=\boxed{0}$ .

After multiplying with conjugate complex and application of some addition theorem we write for kth root: $\frac{1}{1-w}=\frac{1}{2}(1+icot(\frac{k\pi}{2015}))$ Now we determine that pairs of roots (1st/2014th, 2nd/2013th etc.) lead to conjugate complex sum members:

$(|z|e^{i\phi})^{2015}+(|z|e^{-i\phi})^{2015}=2|z|^{2015}cos(2015\phi)$

So we get: $\sum_{k=1}^{1007}\sqrt{1+cot^{2}(\frac{k\pi}{2015})}^{2015}cos[2015arctan(cot(\frac{k\pi}{2015}))]$

For the last term we may write: $cos[sgn(tan(\frac{k\pi}{2015}))\frac{\pi}{2}2015-k\pi]$

Since the argument of cos is an odd multiple of $\frac{\pi}{2}$ for all k the result of the sum is 0.