If $x, y$ and $z$ are the cube roots of unity, then what is the value of $xy + yz + zx$?

Question

Prepp Team · Accepted Answer

Understanding Cube Roots of Unity

The question asks for the value of the expression $xy + yz + zx$, where $x, y,$ and $z$ represent the cube roots of unity.

The cube roots of unity are the solutions to the equation $t^3 = 1$. These roots are:

$1$
$\omega = e^{i \frac{2\pi}{3}} = \cos(\frac{2\pi}{3}) + i \sin(\frac{2\pi}{3}) = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$
$\omega^2 = e^{i \frac{4\pi}{3}} = \cos(\frac{4\pi}{3}) + i \sin(\frac{4\pi}{3}) = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$

Key Properties of Cube Roots of Unity

These roots have important properties that are often used in algebra:

Sum of Roots: The sum of the cube roots of unity is always zero. $1 + \omega + \omega^2 = 0$
Product of Roots: The product of the cube roots is one. $1 \times \omega \times \omega^2 = \omega^3 = 1$
Powers of $\omega$: Any power of $\omega$ can be simplified using $\omega^3 = 1$. For example, $\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega$, and $\omega^5 = \omega^3 \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2$.

Calculating $xy + yz + zx$

Let's assign the cube roots of unity to $x, y,$ and $z$. A common assignment is:

$x = 1$
$y = \omega$
$z = \omega^2$

Now, we can calculate each term in the expression $xy + yz + zx$:

$xy = 1 \times \omega = \omega$
$yz = \omega \times \omega^2 = \omega^3$
$zx = \omega^2 \times 1 = \omega^2$

Using the property $\omega^3 = 1$, the term $yz$ becomes $1$. So, the expression is:

$xy + yz + zx = \omega + \omega^3 + \omega^2$ $xy + yz + zx = \omega + 1 + \omega^2$

From the property of the sum of cube roots of unity ($1 + \omega + \omega^2 = 0$), we get:

$xy + yz + zx = 0$

Alternative Method Using Vieta's Formulas

The equation $t^3 = 1$ can be written as $t^3 - 1 = 0$. This is a cubic equation of the form $at^3 + bt^2 + ct + d = 0$, where $a=1$, $b=0$, $c=0$, and $d=-1$.

If $x, y,$ and $z$ are the roots of this equation, Vieta's formulas relate the coefficients to the sums and products of the roots:

Sum of roots: $x + y + z = -b/a$
Sum of the product of roots taken two at a time: $xy + yz + zx = c/a$
Product of roots: $xyz = -d/a$

Using the coefficients from $t^3 - 1 = 0$ ($a=1, b=0, c=0$):

$xy + yz + zx = \frac{c}{a} = \frac{0}{1} = 0$

Both methods confirm that the value of the expression is $0$. The assignment of $x, y, z$ to $1, \omega, \omega^2$ does not change the result due to the symmetric nature of the expression and the properties of the roots.