# nth roots of unity explanation

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?

• I'm not sure what there is to be said about "understanding the purpose"... at any rate: tell us more about this answer. What exactly aren't you understanding? Do you understand the equations presented going from left to right? If not, then what's the confusing step? – Omnomnomnom Feb 15 '17 at 4:29
• Someday you may be sitting in a lecture when the professor says something like "because they are all the $10$th roots of unity blah blah blah," and everyone who didn't study $n$th roots of unity is instantly totally lost while you (hopefully) remember enough of this topic to continue following the lecture. Is that a "purpose"? – David K Feb 15 '17 at 5:05
• OK, snark aside, $n$th roots of unity do neat things with rotational symmetries, for example. Besides, isn't it mind-blowing that you let *one non real number* into your number system, namely, you allow $-1$ to have a square root, and suddenly you have a huge number of cube roots, fourth roots, fifth roots, sixth roots, seventeenth roots, and other roots of $1$ (and $-1$) that you didn't have before? – David K Feb 15 '17 at 5:09

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

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

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

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

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