Question

Linear Algebra:
Consider the sixth roots of unity and choose ALL correct responses

A. The argument of any sixth root of unity is a multiple of 3
B. If z is a sixth root of unity, then z2 is also a sixth root of unity
C. If z is a sixth root of unity, then z is also a 12th root of unity
D. If z is a sixth root of unity, then z is a cube root of unity
E. If z is a sixth root of unity, then −z is also a sixth root of unity

Answer 1

Roots of unity always have magnitude 1, so it's the angles that distinguish them from each other. Remember that when you multiply two complex numbers the angles add, so a solution to \(x^6=1\) (i.e., a sixth root of unity) will be of magnitude 1 with an angle \(\theta_x\) that when multiplied by 6 lands back at zero degrees (which is equal to any integer \(n\) multiple of \(2\pi\). So:\[6\theta_x=2\pi n \longrightarrow \theta_x = \frac{\pi n}{3} \text{ for any natural number $n$}\]See if that helps you.

Answer 2

well I know these things but what drives me crazy is the options below which one is wrong I chose that C and D are wrong but it is not true

Answer 3

12th roots unity have angles \(\theta_x=\dfrac{\pi n}{6}\). Any 6th root of unity can have its angle expressed in that form for a natural number \(n\). So, C is thus true.

D is undoubtedly false, though. Take the complex number with an angle \(\dfrac{\pi}{3}\) and magnitude. It's cube will be \(-2\), so it is not a cube root of unity.

Answer 4

For C, you can see it easier by squaring both sides.\[x^6 = 1 \longrightarrow x^{12}=1\]However for D, if you square root both sides, you're not guaranteed to still have a root of unity.\[x^6 = 1 \longrightarrow x^3 = \pm 1\]

Answer 5

so all are true but D

Answer 6

Thanks a lot :)

Answer 7

I'd look at A again if I were you.

Answer 8

A is true

Answer 9

Look at my first post...the arguments of the sixth roots of unity are multiples of \(\dfrac \pi 3\) not 3.

Answer 10

no no it is pi/3 but u know I copied it

Answer 11

?

Answer 12

it is just a typo

Answer 13

oh ok