Cube roots of unity - Practice Questions & MCQ

Let $z$ be the cube root of unity (1)

So, $z^3=1$

$
\begin{aligned}
& \Rightarrow z^3-1=0 \\
& \Rightarrow(z-1)\left(z^2+z+1\right)=0
\end{aligned}
$
$
\Rightarrow z-1=0 \text { or } z^2+z+1=0
$

$
\therefore \mathrm{z}=1 \text { or } \mathrm{z}=\frac{-1 \pm \sqrt{(1-4)}}{2}=\frac{-1 \pm \mathrm{i} \sqrt{3}}{2}
$
Therefore, $z=1, z=\frac{-1+i \sqrt{3}}{2}$ and $z=\frac{-1-i \sqrt{3}}{2}$
If the second root is represented by $\omega$, then the third root will be represented by $\omega^2$ (we can check that by squaring the second root, we get the third root)

$
\omega=\frac{-1+\mathrm{i} \sqrt{3}}{2}, \omega^2=\frac{-1-\mathrm{i} \sqrt{3}}{2}
$

So, $1, \omega, \omega^2$ are cube roots of unity and $\omega, \omega^2$ are the non-real complex root of unity.

Properties of Cube roots of unity

i) $1+\omega+\omega^2=0$ and $\omega^3=1$ (Using sum and product of roots relations for the equation $z^3-1=0$ )
ii) To find $\omega^n$, first we write $\omega$ in multiple of 3 with the remainder being 0 or 1 or 2 .

Now $\omega^{\mathrm{n}}=\omega^{3 \mathrm{q}}+\mathrm{r}=\left(\omega^3\right)^{\mathrm{q}} \cdot \omega^{\mathrm{r}}=\omega^{\mathrm{r}}($ Where r is from $0,1,2)$
$\mathrm{Eg}, \omega^{121}=\omega^{3.40+1}=\left(\omega^3\right)^{40} \cdot \omega^1=\omega$
iii) $|\omega|=\left|\omega^2\right|=1, \arg (\omega)=2 \pi / 3, \arg \left(\omega^2\right)=4 \pi / 3$ or $-2 \pi / 3$
iv) We can see that $\omega$ and $\omega^2$ differ by the minus sign of the imaginary part hence $\bar{\omega}=\omega^2$
v) Cube roots of -1 are $-1,-\omega,-\omega^2$
vi) The cube roots of unity when represented on the complex plane have their point on vertices of a triangle circumscribed by a unit circle whose one vertices lies on the +ve X-axis.