General \(n\)th roots of unity

3. \(n^{\mathrm{th}}\) Roots of Unity

For a positive integer \(n\), \[z^n=1\quad\Longrightarrow\quad z_k=\cos\frac{2k\pi}{n}+i\sin\frac{2k\pi}{n}.\] Taking \(k=0,1,\ldots,n-1\) lists all distinct roots. The following enumeration illustrates the pattern (include only indices less than \(n\)).

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

\[\vdots\]

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

Take note that

1. For \(1\le i\le n-1\), \(z_i\) and \(z_{n-i}\) are conjugates; \(z_0=1\) is its own conjugate. So, \(z_i\) is a reflection of \(z_{n-i}\) in the real axis.
2. if \(\omega\) is a \(n^{\mathrm{th}}\) root of unity, then
   1. \(\omega^m\) is also a \(n^{\mathrm{th}}\) root of unity, for any positive integral value of \(m\)
   2. If \(\omega\ne1\), then \((\omega-1)(1+\omega+\cdots+\omega^{n-1})=\omega^n-1=0\), so \(1+\omega+\cdots+\omega^{n-1}=0\). For \(\omega=1\), the sum is \(n\).
3. 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.
4. For \(n\ge3\), 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 \(\frac{2\pi}{n}\). For \(n=2\), the roots are the opposite points \(1,-1\); for \(n=1\), the only root is \(1\).