The Cube Roots of Unity

An important type of complex number arises when we consider the cube roots of unity.

\[\begin{array}{l} \text{Let} & \sqrt[3]{1} &= z \\ \text{or,} & 1 &= z^3 \\ \text{or,} & z^3-1 &= 0 \\ \text{or,} & (z-1)(z^2+z+1) &= 0 \end{array}\] \[\text{Either,}\;\; z-1=0\;\;\text{or}\;\;z^2+z+1=0\]

\[\begin{array}{l} \therefore & z=1 & \text{or} & z^2+z+1= 0 \\ \text{i.e.} & z=1 &\text{or} & z= \frac{-1\pm\sqrt{1-4}}{2} \\ \text{i.e.} & z=1 & \text{or} & z=\frac{-1\pm\sqrt{-3}}{2} \\ \text{i.e.} & z=1 & \text{or} & z=\frac{-1\pm\sqrt{3}\,i}{2} \end{array}\]

Thus, $1$, $-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $-\frac{1}{2}-\frac{\sqrt{3}}{2}i$ are the three cube roots of unity.

The first root is a real number and the other two are imaginary or complex numbers. These complex roots are often known as the imaginary cube roots of unity, any one of which is denoted by the Greek letter $\omega$ (omega).

Properties of the Cube Roots of Unity

P1: Each complex cube root of unity is the square of the other.

Here, $-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $-\frac{1}{2}-\frac{\sqrt{3}}{2}i$ are the complex cube roots of unity.

Now, \[\begin{array}{l} -\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)^2 &= \frac{1}{4}(1-2\sqrt{3}i+3i^2) \\ &= \frac{1}{4}(-2-2\sqrt{3}\,i) \\ &= -\frac{1}{2}-\frac{\sqrt{3}}{2}\,i \end{array}\]

Also, \[\begin{array}{l} -\left(\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)^2 &= \frac{1}{4}(1+2\sqrt{3}i+3i^2) \\ &= \frac{1}{4}(-2+2\sqrt{3}\,i) \\ &= -\frac{1}{2}+\frac{\sqrt{3}}{2}\,i \end{array}\]

Thus, if we denote any one of the imaginary cube roots by $\omega$, then the other will be $\omega^2$. Hence, the three cube roots of unity are $1$, $\omega$ and $\omega^2$.

P2: The sum of the three cube roots of unity is zero.

Here, \[\begin{array}{l} 1+\omega+\omega^2 &= 1-\frac{1}{2}+\frac{\sqrt{3}}{2}i-\frac{1}{2}-\frac{\sqrt{3}}{2}i \\ &= 1-1 \\ &= 0 \\ \therefore 1+\omega+\omega^2 &= 0 \end{array}\]

P3: The product of the two complex roots of unity is 1.

Here, $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $\omega^2=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$ are the complex cube roots of unity.

Now, \[\begin{array}{l} \omega\cdot\omega^2 &= \left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right) \left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right) \\ &= \frac{1}{4}-\frac{3}{4}i \\ &= \frac{1}{4}+\frac{3}{4} \\ &= 1 \end{array}\] \[\therefore\omega\cdot\omega^2=\omega^3=1\]

Also, \[\omega=\frac{1}{\omega^2}\;\;\text{and}\;\;\omega^2=\frac{1}{\omega}\] Thus, one imaginary cube root of unity is the reciprocal of the other.

Powers of $\omega$

Since $\omega^3=1$, $\omega^{3n}=1$ for any integral value of $n$.

\[\begin{array}{l}\omega^4 &= \omega^3\cdot\omega &= \omega \\ \omega^5 &= \omega^3\cdot \omega^2 &= \omega^2 \\ \omega^6 &= \omega^3\cdot\omega^3 &= 1 \\ \omega^{20} &= \omega^{18} \cdot\omega^2 &= (\omega^3)^6\omega^2 &= \omega^2 \\ \omega^{-1} &= \frac{1}{\omega} &= \frac{\omega^3}{\omega} &= \omega^2 \\ \omega^{-10} &= \frac{1}{\omega^{10}} &= \frac{1}{\omega} &= \omega^2, \;\;\text{etc.} \end{array}\]

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