Answer: (i) The fifth roots of unity are \(1,\ e^{2\pi i/5},\ e^{4\pi i/5},\ e^{6\pi i/5},\ e^{8\pi i/5}\).
(ii) Let \(w=z^5\). Then \(w^2+w+1=0\), so \(w\) is a non-real cube root of unity:
\(w=e^{2\pi i/3}\) or \(w=e^{4\pi i/3}\).
Hence \(z^5=e^{2\pi i/3}\) or \(z^5=e^{4\pi i/3}\), so the ten roots are
\(z=e^{i(2\pi/15+2k\pi/5)}\) for \(k=0,1,2,3,4\), and \(z=e^{i(4\pi/15+2k\pi/5)}\) for \(k=0,1,2,3,4\).
(i) The fifth roots of unity are the solutions of \(z^5=1\). Writing \(1=e^{2\pi i m}\), we get
\(z=e^{2\pi i m/5}\) for \(m=0,1,2,3,4\).
So the five fifth roots of unity are \(1,\ e^{2\pi i/5},\ e^{4\pi i/5},\ e^{6\pi i/5},\ e^{8\pi i/5}\).
(ii) Let \(w=z^5\). Then the equation becomes
\(w^2+w+1=0\).
Factorising or using the quadratic formula gives
\(w=\frac{-1\pm i\sqrt{3}}{2}\).
These are the two non-real cube roots of unity, so
\(w=e^{2\pi i/3}\) or \(w=e^{4\pi i/3}\).
Therefore
\(z^5=e^{2\pi i/3}\) or \(z^5=e^{4\pi i/3}\).
For \(z^5=e^{2\pi i/3}\), the five roots are
\(z=e^{i\left(\frac{2\pi}{15}+\frac{2k\pi}{5}\right)}\), \(k=0,1,2,3,4\).
For \(z^5=e^{4\pi i/3}\), the five roots are
\(z=e^{i\left(\frac{4\pi}{15}+\frac{2k\pi}{5}\right)}\), \(k=0,1,2,3,4\).
So all ten roots are
\(e^{i(2\pi/15)},\ e^{i(8\pi/15)},\ e^{i(14\pi/15)},\ e^{i(20\pi/15)},\ e^{i(26\pi/15)}\), and \(e^{i(4\pi/15)},\ e^{i(10\pi/15)},\ e^{i(16\pi/15)},\ e^{i(22\pi/15)},\ e^{i(28\pi/15)}\).
