A community for students.
Here's the question you clicked on:
 0 viewing
Loser66
 one year ago
A primitive nth root of unity is a complex number a such that 1,a,a^2,....., a^(n1) are distinct roots of unity. Show that if a, b are primitive nth and mth roots of unity, respectively, then ab is a kth root of unity for some integer k. What is the smallest value of k? What cn be said if a and b are nonprimitive roots of unity?
Please, help
Loser66
 one year ago
A primitive nth root of unity is a complex number a such that 1,a,a^2,....., a^(n1) are distinct roots of unity. Show that if a, b are primitive nth and mth roots of unity, respectively, then ab is a kth root of unity for some integer k. What is the smallest value of k? What cn be said if a and b are nonprimitive roots of unity? Please, help

This Question is Closed

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I don't know why I feel it is trivial:) \(a=a e^{i\theta}\\b=be^{i\tau}\\ab= abe^{i(\theta +\tau)}\) it shows that ab is a root of unity, right?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0?i don't get the question :) unity is 1 nd roots of real numbers r real

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0"A primitive nth root of unity is a COMPLEX number a............"

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1If a is \(a = e^{i\phi}\) is a primitive root of unity then \(\frac{\phi}{2\pi}=\frac{p}{q}\) such that p and q are natural numbers and p<q.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1And gcd(p,q) = 1

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0My computer is on and off!! if you don't see I respond, please forgive me.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1BTW, want to point out that here: \[a=a e^{i\theta}\\b=be^{i\tau}\\ab= abe^{i(\theta +\tau)}\] a = b = 1, since they are roots of unity. Anyway, we write \[a=e^{i\theta} = e^{i\frac{p}{q}2\pi} \] \[b=e^{i\tau} = e^{i\frac{r}{s}2\pi} \] \[ab= e^{i2\pi*(\frac{p}{q}+\frac{r}{s})} = e^{i2\pi*\frac{ps+rq}{sq}}\] So ab is a (sq)th root of unity.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1p,q,r,s are natural number s.t p<q, r<s

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1\[a^n= 1 = e^{2 \pi \ i} \\ a= e^{\frac{2\pi}{n} i }\] Did you mean: \[ \Large a= e^{\frac{2\pi k}{n} i }\] and now the constraint that a is a primitive root is equivalent to gcd(k,n) =1

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1Anyway, I think if m and n are distinct a*b will be a primitive root of (m*n)th order. I don't want to go through this because it's a pain to write out in latex. There's also some value in thinking it through for yourself. But really, it's mostly because I'm lazy :) I'd rather go through the main idea again: There is a bijection between primitive roots of unity and proper fractions k/n, where gcd(k,n) = 1. Multiplying two roots of unity is equivalent to adding the fractions they are paired with. So if a and b are primitive roots, associated with k/n and l/m respectively, ab is associated with (k/n + l/m). To find the order of this primitive root write this fraction (km+ln)/mn and write it in its simplest form. The order of ab is the denominator. In particular, when I said that a*b will have an order of (m*n), when m =/= n, I might as well have said that: gcd(k,n) = gcd(l,m) = 1 and m =/= n implies gcd((km+ln)/(mn)) = 1

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1"i.e. k=8 = least common multiple of 4 and 8" You're right. "also, because we have 3 in the numerator this is not a "primitive root"" The definition for primitive root is that it's a root for some degree n but not for any degree m<n. The numerator doesn't need to be 1. In the e^(i*2pi*(k/n)) form it's equivalent to gcd(k,n) = 1.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1There is no contradiction. Take the equation z^4 = 1. The solution set for this is {1,i,1,i}. Of course i is a primitive root. But so is i, which you can check 1 i (i)^2 = 1 (i)^3 = i That's 4 distinct roots of unity, exactly as the definition requires.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1"I guess you are missing the point. The definition says "a" is the primitive nth root. in the list of roots: a^0,a^1, a^2, ... a^(n1)" No, that's not what it says. Read it again carefully. It says that you can check if a number is a number is a primitive nth root by veriftying that its first n powers are all distinct (and roots of unity). If you don't believe read the wikipedia article on roots of unity or the planetmath article I linked earlier. It's a standard definition, not a word made up for specifically this problem.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.