nth roots of unity explanation

Question

For some reason I just don't understand nth root of unity...at all. Even when reading resources online and listening in class I don't understand the purpose. And in the question provided as an image, I don't understand how D is the answer

edit - I don't understand the jump from $$e^{2\pi i(3/10 -1)} = w_{100}^{30}$$

where did 10-1 come from?

Answer 1

Let $\omega_n$ be the $n^{th}$ root of unity.

There are $n$ distinct $n^{th}$ roots of unity, so I suspect that there is some previous context which defines $\omega_n = e^{2 \pi i / n}$, otherwise it would be wrong/ambiguous to refer to the $n^{th}$ root of unity.

$$\omega_{100}^{30}=e^{2 \pi i(30/100)}$$

$\omega_{100}=e^{2 \pi i / 100}$ by the previous definition, then $\,\omega_{100}^{30} =\left(e^{2 \pi i / 100}\right)^{30} =e^{30 \,\cdot\, 2 \pi i / 100}=e^{2 \pi i \,\cdot\, 30 / 100}$

$$=e^{2 \pi i(3/10)}$$

$30/100 = 3/10$

$$= e^{2 \pi i(3/10-1)}$$

$e^{2\pi i}=1\,$, so $e^{2 \pi i(3/10)} = e^{2 \pi i(3/10)} / e^{2\pi i} = e^{2 \pi i(3/10) - 2 \pi i} = e^{2 \pi i(3/10-1)}$

$$=\omega_{10}^{-7}$$

$e^{2 \pi i(3/10-1)} = e^{2 \pi i(-7/10)}=\left(e^{2 \pi i /10}\right)^{-7} = \omega_{10}^{-7}$

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[ EDIT ]   The direct proof along the same line as above would be: $\,\omega_{100}^{30}=\omega_{10}^{3}=\omega_{10}^{3-10}=\omega_{10}^{-7}\,$.

Answer 2

It's not $10-1$

its $$e^{\frac{2\pi i\cdot3}{10}}=e^{2\pi i\left(\frac3{10}-1\right)}\cdot e^{2\pi i}$$

Now $e^{2\pi i}=?$