Relating $\sum_{k=1}^N a_k^2 e^{\frac{2\pi i}{N}k}$ to $(\;\sum_{k=1}^N a_k e^{\frac{2\pi i}{N}k}\;)^2$

Question

Consider the following expression $$ \sum_{k=1}^N a_k^2 e^{\frac{2\pi i}{N}k}\tag{1} $$ where $i$ is the imaginary number. How may I relate it to the following expression $$ \left(\sum_{k=1}^N a_k e^{\frac{2\pi i}{N}k}\right)^2 $$ My attempt: I know that $$ \left( \sum_{k=1}^N x_k \right)^2 = \sum_{k=1}^N x_k^2 + 2 \sum_{k=1}^{N}\sum_{j=1}^{k-1} x_k x_j $$ With $x_k\equiv a_k e^{\frac{2\pi i}{N}k}$ I get $$ \left( \sum_{k=1}^N a_k e^{\frac{2\pi i}{N}k} \right)^2 = \sum_{k=1}^N a_k^2 e^{\frac{4\pi i}{N}k} + 2 \sum_{k=1}^{N}\sum_{j=1}^{k-1} a_ka_j e^{\frac{2\pi i}{N}(k+j)} $$ where the first term of the right-hand side is close to my initial term $(1)$. Any ideas on how to proceed from here? I wonder if there are some special properties about $e^{\frac{2\pi i}{N}k}$ and Fourier transforms that I could use.