Here, $z_r$ are $n$-arbitrary complex numbers, $\omega=e^{2i\pi/n}$ and $T_k=\sum_{r=0}^{n-1} \omega^{rk} z_r$, then I prove that
$$S_n=\sum_{k=0}^n |T_k|^2= \sum_{k=0}^{n} \sum_{r=0}^{n-1} \sum_{s=0}^{n-1} e^{2i\pi(r-s)k/n} z_r ~\bar z_s=\sum_{r=0}^{n-1} \sum_{s=0} ^{n-1} z_r ~\bar z_s \sum_{k=0}^n (e^{2i\pi (r-s)/n})^k$$ $$\Rightarrow S_n= \sum_{r=0}^{n-1} \sum_{s=0}^{n-1} z_r ~\bar z_s \frac{e^{2i\pi (r-s)(n+1)/n}-1}{e^{2i\pi (r-s)/n}-1}= \sum_{r=0}^{n-1} \sum_{s=0}^{n-1} z_r ~\bar z_s \frac{e^{2i\pi (r-s)/n}-1}{e^{2i\pi (r-s)/n}-1},~ \mbox{as}~ e^{2i\pi(r-s)}=1.$$ $$\Rightarrow S_n=\sum_{r=0}^{n-1}z_r \sum_{s=0}^{n-1} \bar z_s = \left |\sum_{r=0}^{n-1} z_r \right|^2.$$ Curiously, somehow the numerical experiments do not agree with this result. The question is what is amiss here? Please help.
Thanks @Anurag A, now I get the correct result
Here, $z_r$ are $n$-arbitrary complex numbers, $\omega=e^{2i\pi/n}$ and $T_k=\sum_{r=0}^{n-1} \omega^{rk} z_r$, then I prove that
$$S_n=\sum_{k=0}^n |T_k|^2= \sum_{k=0}^{n} \sum_{r=0}^{n-1} \sum_{s=0}^{n-1} e^{2i\pi(r-s)k/n} z_r ~\bar z_s=\sum_{r=0}^{n-1} \sum_{s=0} ^{n-1} z_r ~\bar z_s \left[\sum_{k=0}^{n-1} (e^{2i\pi (r-s)/n})^k+1\right]$$
$$\Rightarrow S_n= \sum_{r=0}^{n-1} \sum_{s=0}^{n-1} z_r ~\bar z_s \left [ 1+ \frac{e^{2i\pi (r-s)}-1}{e^{2i\pi (r-s)/n}-1} \right]= \sum_{r=0}^{n-1} \sum_{s=0}^{n-1} z_r ~\bar z_s [1+n\delta_{r,s}] ~as~ e^{2i\pi(r-s)}=1. $$
$$\Rightarrow S_n=\left(\sum_{r=0}^{n-1} z_r \sum_{s=0}^{n-1} \ \bar z_s + n\sum_{r=0}^{n-1} z_r \bar z_r \right) = \left| \sum_{r=0}^{n-1} z_r \right|^2+ n \sum_{r=0}^{n-1} |z_r|^2.$$