More fun in 2015, Part 18

Since we are not allowed to use calculus, we will present a somewhat lengthy algebraic solution.

We will at first include the root of unity $w_{2015}=1$ . At the end we need to remember to add $1/4$ to our answer.

Let's make a change of variables, $z=\frac{1}{1+w}$ . Since the $w_k$ are the solutions of $w^{2015}=1$ , the corresponding $z_k$ will be the solutions of $\left(\frac{1-z}{z}\right)^{2015}=1$ or $z^{2015}+(z-1)^{2015}=0$ or $2z^{2015}-2015z^{2014}+\frac{2015\times{2014}}{2}z^{2013}-...=0$ . We seek $\sum_{k=1}^{2015}z_k^2=\left(\sum_{k=1}^{2015}z_k\right)^2-2\sum_{i<j}z_iz_j$ $=\left(\frac{2015}{2}\right)^2-2\left(\frac{2015\times{2014}}{4}\right)=-1014048.75$ , by Viete. Changing the sign, as required, and adding the $1/4$ stemming form $w_{2015}=1$ , we find the final answer, $\boxed{1014049}$ .

Let us first consider the following manipulation: $\dfrac{1}{(1+e^{2 i z})^2}= \dfrac{e^{-2 i z}}{(e^{-i z})^2(1+e^{2 i z})^2}$ $\hspace{17mm} =\dfrac{e^{-2 i z}}{(e^{-i z}+e^{i z})^2}$ $\hspace{17mm} =\dfrac{e^{-2 i z}}{2cos^2(z)}=\frac{1}{4} [\frac{cos(2z)}{cos^2(z)}-i \frac{sin(2z)}{cos^2(z)}]$ $\hspace{17mm}=\frac{1}{4} (2-\sec^2(z)-2 i \tan(z))$ $\hspace{17mm}=\frac{1}{4} (1-\tan^2(z)-2 i \tan(z))$ Because the roots of unity can be written as $w_k=e^{\frac{2 i \pi k}{2015}}$ , our question can be rephrased as the following sum, with $z=\frac{\pi k}{2015}$ : $\frac{1}{4}(\sum_{k=1}^{2014} 1-\sum_{k=1}^{2014}\tan^2(\frac{\pi k}{2015})-2i\sum_{k=1}^{2014}\tan(\frac{\pi k}{2015}))$ Now, lets first look at the imaginary part of our problem, namely $\sum_{k=1}^{2014}\tan(\frac{\pi k}{2015})$ . . Using the identity $\tan(\frac{\pi k}{2015})=-tan(\pi-\frac{\pi k}{2015})=-\tan(\frac{\pi(2015-k)}{2015}) \hspace{5mm} (1)$ , we see that that $\sum_{k=1}^{2014} \tan(\frac{\pi k}{2015})=0$ , as the terms can be paired off to cancel out

The real part is $\frac{1}{4}\sum_{k=1}^{2014} (1-\tan^2(\frac{\pi k}{2015}))$ . Using $(1)$ we have: $\frac{1}{4}(\sum_{k=1}^{2014} 1-2\sum_{k=1}^{1007}\tan^2(\frac{\pi k}{2015}))$ . Using the following identity, with m=1007: $\sum_{k=1}^{m}\tan^2(\frac{\pi k}{2m+1})=2m^2+m$ , we finally have: $\frac{1}{4}(2014-2(2\times 1007^2+1007)=-1007^2=-1014049$