Question

If $S_k=\sin \left(\frac{2\pi k}{n+1}\right)-i\cos \left(\frac{2\pi k}{n+1}\right)$ and $P_k=\cos \left(\frac{\pi }{2^k}\right)+i\sin \left(\frac{\pi }{2^k}\right),$
Let $m=|\sum _{k=1}^n{S}_k+\prod _{k=1}^{\infty }{P}_k|$, then the value of $2m^2$ is

Answer 1

$S_k=\sin \left(\frac{2\pi k}{n+1}\right)-i\cos \left(\frac{2\pi k}{n+1}\right)$
$=-i[\cos \left(\frac{2\pi k}{n+1}\right)+i\sin \left(\frac{2\pi k}{n+1}\right)]$
$=-i{e}^{\frac{i2\pi k}{n+1}}$
$\sum _{k=1}^n{S}_k=-i\sum _{k=1}^n{z}^k,[z=e^{\frac{i2\pi }{n+1}}]$
$=-i(z+z^2+z^3+\cdots +z^n)$
$=-i\frac{z(z^n-1)}{z-1}$
$=-i\left(\frac{z^{n+1}-z}{z-1}\right)$
$\text{Now,}{z}^{n+1}=e^{\left(\frac{i2\pi }{n+1}\right)(n+1)}=e^{i2\pi }=1$
$\therefore \sum _{k=1}^n{S}_k=-i\frac{(1-z)}{z-1}=i$

$P_k=\cos \left(\frac{\pi }{2^k}\right)+i\sin \left(\frac{\pi }{2^k}\right)=e^{\frac{i\pi }{2^k}}$
$\prod _{k=1}^{\infty }{P}_k=\prod _{k=1}^{\infty }{e}^{\frac{i\pi }{2^k}}$
$=e^{\frac{i\pi }{2}}\cdot e^{\frac{i\pi }{2^2}}\cdot e^{\frac{i\pi }{2^3}}\cdots$
$=\text{exp}\left[i\pi (\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots )\right]$
$=e^{i\pi }=-1$
$|\sum _{k=1}^n{S}_k+\prod _{k=1}^{\infty }{P}_k|=|i-1|=\sqrt{2}$