<ol class="i-list" start="3"><li><div class="list-item-paragraph">\(n^{\mathrm{th}}\) Roots of Unity</div></li></ol>
<p>\[z^n = 1 \implies z^n = \cos 2k\pi + i\sin 2k\pi \implies z = \cos \frac{2k\pi}{n} + i\sin \frac{2k\pi}{n} \text{ for}\hspace{0.5em} k \in \mathbb{Z} .\]</p>
<p>When&nbsp;\(k = 0\), \(z_0 = \cos 0 + i\sin 0 = 1\);</p>
<p>When&nbsp;\(k = 1\), \(z_1 = \cos \frac{2\pi}{n} + i\sin \frac{2\pi}{n}\)</p>
<p>When&nbsp;\(k = 2\), \(z_2 = \cos \frac{4\pi}{n} + i\sin \frac{4\pi}{n}\);</p>
<p>\[\vdots\]</p>
<p>When&nbsp;\(k = n - 1\), \(z_{n-1} = \cos \frac{2(n - 1)\pi}{n} + i\sin \frac{2(n - 1)\pi}{n} .\)</p>
<p>Take note that</p>
<ol class="abc-list"><li><div class="list-item-paragraph">\(z_i\) and \(z_{n-i}\) are conjugates. So, \(z_i\) is a reflection of \(z_{n-i}\) in the real axis.</div></li><li><div class="list-item-paragraph">if&nbsp;\(\omega\) is a \(n^{\mathrm{th}}\) root of unity, then</div><ol class="i-list"><li><div class="list-item-paragraph">\(\omega^m\) is also a \(n^{\mathrm{th}}\) root of unity, for any positive integral value of \(m\)</div></li><li><div class="list-item-paragraph">Spread across a box: \(1 + \omega + \omega^2 + \omega^3 + \ldots + \omega^{n-1} = 0\)</div></li></ol></li><li><div class="list-item-paragraph">If&nbsp;\(n\) is even, then&nbsp;\(z^n = 1\) has two real roots (\(z = 1\) and \(z = -1\)) and \(\frac{n-2}{2}\) conjugate pairs.If&nbsp;\(n\) is odd, then&nbsp;\(z^n = 1\) has one real root (\(z = 1\)) and \(\frac{n-1}{2}\) conjugate pairs.</div></li><li><div class="list-item-paragraph">The \(n^{\mathrm{th}}\) roots of unity are represented by the vertices of a regular \(n\)-sided polygon inscribed in a unit circle with the centre at the origin, with \(z_0\) on the real axis, and \(z_0, z_1, z_2, \ldots, z_{n-1}\) separated by equal angles of \(\frac{2\pi}{n}\).</div></li></ol>

\[z^n = 1 \implies z^n = \cos 2k\pi + i\sin 2k\pi \implies z = \cos \frac{2k\pi}{n} + i\sin \frac{2k\pi}{n} \text{ for } k \in \mathbb{Z} .\]

When \(k = 0\), \(z_0 = \cos 0 + i\sin 0 = 1\);

When \(k = 1\), \(z_1 = \cos \frac{2\pi}{n} + i\sin \frac{2\pi}{n}\)

When \(k = 2\), \(z_2 = \cos \frac{4\pi}{n} + i\sin \frac{4\pi}{n}\);

When \(k = n - 1\), \(z_{n-1} = \cos \frac{2(n - 1)\pi}{n} + i\sin \frac{2(n - 1)\pi}{n} .\)

\(z_i\) and \(z_{n-i}\) are conjugates. So, \(z_i\) is a reflection of \(z_{n-i}\) in the real axis.

if \(\omega\) is a \(n^{\mathrm{th}}\) root of unity, then

If \(n\) is even, then \(z^n = 1\) has two real roots (\(z = 1\) and \(z = -1\)) and \(\frac{n-2}{2}\) conjugate pairs.If \(n\) is odd, then \(z^n = 1\) has one real root (\(z = 1\)) and \(\frac{n-1}{2}\) conjugate pairs.

The \(n^{\mathrm{th}}\) roots of unity are represented by the vertices of a regular \(n\)-sided polygon inscribed in a unit circle with the centre at the origin, with \(z_0\) on the real axis, and \(z_0, z_1, z_2, \ldots, z_{n-1}\) separated by equal angles of \(\frac{2\pi}{n}\).

General \(n\)th roots of unity

General \(n\)th roots of unity