Question

dSVDSfbf

Answer 1

@KingGeorge

Answer 2

Have you learned about different ways of writing complex numbers? For example, \(x+iy\), \(r(\cos(\theta)+i\sin(\theta))\), or \(re^{i\theta}\)?

Answer 3

I suppose this might be possible using Descarte's rule of signs. Have you learned about that?

Answer 4

yes i'm supposed to use descarte's rule

Answer 5

Let's use that then. The first thing we need to check, is if 0 is a root. since \(0^n-1=-1\neq0\), we can continue. Can you tell me how many positive roots \(x^n-1\) should have?

Answer 6

it should have 1?

Answer 7

Right. Since there is only a single sign-change, there should be exactly one positive root. Now we have to look at the negative roots. This is where we use the fact that \(n\) is even.

Answer 8

ooohhh ok. i get how it changes the number of the roots but why n-2 nonreal complex roots?

Answer 9

Well, when \(n\) is even, you get that there is also exactly one negative root. Thus, we have exactly one real positive root, and one real negative root. This leaves \(n-2\) roots left. They can't be real, so they must be non-real complex roots.

Answer 10

wow i feel kinda dumb right now. but when n is odd there are no negative roots so it would be n-1?

Answer 11

Correct.

Answer 12

And don't worry about feeling dumb. It happens to all of us at some point or another. And in situations like this, often the best way to learn how to do these problems is to see it done by someone else.