Question

If $\omega$ is a complex cube root of unity, find the value of: $(1+\omega )(1+{\omega }^2)(1+{\omega }^4)(1+{\omega }^8)$ ...... to n factors

• A. always 1
• B. 1 if n is even; ${\omega }^2$ if n is odd
• C. 1 if n is odd; $-{\omega }^2$ if n is even
• D. 1 if n is even; $-{\omega }^2$ if n is odd

Answer 1

The correct option is D 1 if n is even; $-{\omega }^2$ if n is odd

$(1+w)(1+w^2)(1+w^4)(1+w^8)...ntimes$

Now

$w^3=1$ and $1+w+w^2=0$

Hence

$(1+w)(1+w^2)(1+w.w^3)(1+w^6.w^2)...ntimes$

$=(1+w)(1+w^2)(1+w)(1+w^2)...ntimes$

If n is even, we get

$\left[\right(1+w\left)\right(1+w^2\left)\right]\left[\right(1+w\left)\right(1+w^2]...ntimes$

$=\left[\right(1+w\left)\right(1+w^2){]}^{\frac{n}{2}}$

$=[1+w+w^2+w^3{]}^{\frac{n}{2}}$

$=1$

If n is odd, we get

$\left[\right(1+w\left)\right(1+w^2\left){]}^{\frac{n-1}{2}}\right(1+w)$

$=1.(1+w)$

$=(-w^2)$

$=-w^2$