Consider the complex number $\omega = \frac{1+i}{\sqrt{2}} $. Find the value of $(\sum_{k=0}^n z^{k^2}) (\sum_{j=0}^n z^{-j^2})$
I started by collecting the terms with similar but opposite sign indices and writing the sum in polar form, but I am unable to proceed with the other terms
Denote $g_n=\displaystyle\sum_{k=0}^n \omega^{k^2}$ where $\omega=e^{\pi i/4}=\frac{1}{\sqrt{2}}(1+i)$ is a primitive $8$th root of unity.
Note that $\omega^{-1}=\overline{\omega}$ so you're interested in the value of $g_n\overline{g_n}=|g_n|^2$.
Observe that if $n=8q+r$ with $r<8$, then $g_n=qg_8+g_r$ because the squares $k^2$ in the exponent of $\omega$ witihin the $g_n$ sum are periodic modulo $8$.
Therefore, it suffices to compute $g_n$ for $n=0,1,\cdots,7$ and $n=8$.
