Loser66
  • Loser66
A primitive nth root of unity is a complex number a such that 1,a,a^2,....., a^(n-1) 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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Loser66
  • Loser66
I don't know why I feel it is trivial:) \(a=|a| e^{i\theta}\\b=|b|e^{i\tau}\\ab= |ab|e^{i(\theta +\tau)}\) it shows that ab is a root of unity, right?
imqwerty
  • imqwerty
?i don't get the question :) unity is 1 nd roots of real numbers r real
Loser66
  • Loser66
"A primitive nth root of unity is a COMPLEX number a............"

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

beginnersmind
  • beginnersmind
If 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
beginnersmind
  • beginnersmind
And gcd(p,q) = 1
Loser66
  • Loser66
then?
Loser66
  • Loser66
My computer is on and off!! if you don't see I respond, please forgive me.
beginnersmind
  • beginnersmind
BTW, want to point out that here: \[a=|a| e^{i\theta}\\b=|b|e^{i\tau}\\ab= |ab|e^{i(\theta +\tau)}\] |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
  • beginnersmind
p,q,r,s are natural number s.t p
beginnersmind
  • beginnersmind
\[a^n= 1 = e^{2 \pi \ i} \\ a= e^{\frac{2\pi}{n} i }\] 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
  • beginnersmind
Anyway, 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
  • beginnersmind
"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
beginnersmind
  • beginnersmind
There 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
  • beginnersmind
"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^(n-1)" 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 n-th 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.