Evaluate the product (1-omega+omega2)(1-omega2+omega4)(1-omega4+omega8)... to 2n factors.

Question

ScanSolve AI Tutor · Accepted Answer

To evaluate the given product, we will use the properties of the cube roots of unity:

The product is given as $(1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)...$ to $2n$ factors.

Simplify the first factor $(1-\omega+\omega^2)$ .
Using the property $1+\omega+\omega^2=0$ , we can write $1+\omega^2 = -\omega$ .
$(1-\omega+\omega^2) = (1+\omega^2) - \omega = -\omega - \omega = -2\omega$
Simplify the second factor $(1-\omega^2+\omega^4)$ .
Using the property $\omega^3=1$ , we have $\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega$ .
So the second factor becomes $(1-\omega^2+\omega)$ .
Using the property $1+\omega+\omega^2=0$ , we can write $1+\omega = -\omega^2$ .
$(1-\omega^2+\omega) = (1+\omega) - \omega^2 = -\omega^2 - \omega^2 = -2\omega^2$
Simplify the third factor $(1-\omega^4+\omega^8)$ .
Using the property $\omega^3=1$ , we have $\omega^4 = \omega$ and $\omega^8 = \omega^6 \cdot \omega^2 = (\omega^3)^2 \cdot \omega^2 = 1^2 \cdot \omega^2 = \omega^2$ .
So the third factor becomes $(1-\omega+\omega^2)$ . This is the same as the first factor.
Identify the pattern of the factors.
The factors repeat in a cycle of two: $(-2\omega)$ and $(-2\omega^2)$ .
Since there are $2n$ factors in total, there will be $n$ pairs of these two factors.
Calculate the product of $2n$ factors.
The product $P$ is given by:
$P = [(-2\omega) \cdot (-2\omega^2)]^n$ $P = [4 \omega^3]^n$
Using the property $\omega^3=1$ :
$P = [4 \cdot 1]^n$ $P = 4^n$
Express the result in terms of powers of 2.
$4^n = (2^2)^n = 2^{2n}$
Comparing this with the given options, it matches option (b).

The final answer is $\text{(b) } 2^{2n}$ .

Answer