The $n$ roots are given by
\begin{equation}
z = a^{1/n} e^{i(\theta + 2\pi k)/n},
\end{equation}

for $k = 1, 2, \ldots, n - 1$. We now wish to show that
\begin{equation}
\sum_{k=0}^{n-1} \sin \left( \frac{2 \pi k}{n} \right) = 0,
\end{equation}

and similarly for the cosine terms. Let us take the case of $n$ being even first. Then the above sum is
\[
\sin(0) + \sin(\pi) + \sum_{k=1}^{n/2} \sin \left( \frac{2 \pi k}{n} \right) +
\sum_{k=-(n/2)}^{-1} \sin \left( \frac{2 \pi k}{n} \right).
\]

We can use the oddness property of sines to then write
\[
\sin(0) + \sin(\pi) + \sum_{k=1}^{n/2} \sin \left( \frac{2 \pi k}{n} \right) +
\sum_{k=1}^{n/2} \sin \left(-\frac{2 \pi k}{n} \right) = 0.
\]

The proof of the odd version is done similarly and the proof for the cosine terms is done by profiting from the oddness of cosines about the $y$ axis.