<p>The \(n^{\text{th}}\) roots of unity are the complex numbers that satisfy the equation \(z^n = 1\).</p>
<p>From the Fundamental Theorem of Algebra, a polynomial of degree \(n\) will have exactly \(n\) roots (whether real or complex). <br><br>For example, \(ax^3 + bx^2 + cx + d = 0\) will have&nbsp;\(3\) solutions for&nbsp;\(x\).</p>
<p>Hence, for&nbsp;\(x^3 = 1\), \(x = 1\) is not the only answer when we consider complex numbers. There will be \(2\) other complex numbers to work out.</p>
<p>Re-writing \(x\) as&nbsp;\(z\) to account for complex numbers, \(z^3 = 1\) will have&nbsp;\(3\) roots to be found in the method below.</p>
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<ol class="i-list"><li><div class="list-item-paragraph"><strong>Square Roots of Unity</strong><br>\[z^2 = 1 \implies z = \pm 1\]</div></li><li><div class="list-item-paragraph"><strong>Cube Roots of Unity</strong><br>\[z^3 = 1 \implies z^3 = \cos 2k\pi + i\sin 2k\pi \implies z = \cos \frac{2k\pi}{3} + i\sin \frac{2k\pi}{3} \text{ for}\hspace{0.5em} k \in \mathbb{Z} \]</div></li></ol>
<p>When&nbsp;\(k = 0\), \(z_0 = \cos 0 + i\sin 0 = 1\);<br><br>When&nbsp;\(k = 1\), \(z_1 = \cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i\);<br><br>When&nbsp;\(k = 2\), \(z_2 = \cos \frac{4\pi}{3} + i\sin \frac{4\pi}{3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i\).</p>
<p>Note that we would get the same answers even if we had chosen \(k = -1, 0, 1\) or \(k = 1, 2, 3\).</p>
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The \(n^{\text{th}}\) roots of unity are the complex numbers that satisfy the equation \(z^n = 1\).

From the Fundamental Theorem of Algebra, a polynomial of degree \(n\) will have exactly \(n\) roots (whether real or complex). 

For example, \(ax^3 + bx^2 + cx + d = 0\) will have \(3\) solutions for \(x\).

Hence, for \(x^3 = 1\), \(x = 1\) is not the only answer when we consider complex numbers. There will be \(2\) other complex numbers to work out.

Re-writing \(x\) as \(z\) to account for complex numbers, \(z^3 = 1\) will have \(3\) roots to be found in the method below.

**Square Roots of Unity**
\[z^2 = 1 \implies z = \pm 1\]

**Cube Roots of Unity**
\[z^3 = 1 \implies z^3 = \cos 2k\pi + i\sin 2k\pi \implies z = \cos \frac{2k\pi}{3} + i\sin \frac{2k\pi}{3} \text{ for } k \in \mathbb{Z} \]

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

When \(k = 1\), \(z_1 = \cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i\);

When \(k = 2\), \(z_2 = \cos \frac{4\pi}{3} + i\sin \frac{4\pi}{3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i\).

Note that we would get the same answers even if we had chosen \(k = -1, 0, 1\) or \(k = 1, 2, 3\).

Roots of unity

Roots of unity