Show root of unity and order

Question

I have this math problem:

Set $$z=\frac{1}{2}-\frac{\sqrt{3}}{2} i$$ Show that $z$ is a root of unity, find its order, and express $z^{100}$ in the form $a+bi$.

I'm not 100% sure how to do this. This is what I dod so far. I changed $z$ to exponential form and got $z = e^{i\frac{\pi}{3}}$. I know that $z^n=1$ for $z$ to be a root of unity... so doesn't $(e^{i\frac{\pi}{3}})^0=1$? does that prove that it is a root of unity? Also, how would I find the order from this? Thanks

Answer 1

You need a positive $n$ to be a root of unity and the minimum such $n$ is its order. Remember that $1=e^{2k\pi i}$ for any integer $k$. So, what's the smallest positive $n$ that gets you to a multiple of $2\pi i?$

From there, every time you go that many times around you get back to 1, so you can reduce 100 mod the order to simplify your calculation

Answer 2

more to the point $(e^{i\frac{\pi}{3}})^{6n}=1$ for every integer $n$

so $z^{100}= z^{96} z^4=z^4=(e^{i\frac{4\pi}{3}})$

Answer 3

Hint:

$z=-j$, where $j$ is the cubic root of unity: $\;\mathrm e^{\tfrac{2\mathrm i\pi}3}$. Whence …