Question

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

Answer 1

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?

Answer 2

?i don't get the question :) unity is 1 nd roots of real numbers r real

Answer 3

"A primitive nth root of unity is a COMPLEX number a............"

Answer 4

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<q.

Answer 5

And gcd(p,q) = 1

Answer 6

then?

Answer 7

My computer is on and off!! if you don't see I respond, please forgive me.

Answer 8

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.

Answer 9

p,q,r,s are natural number s.t p<q, r<s

Answer 10

\[a^n= 1 = e^{2 \pi \ i} \\ a= e^{\frac{2\pi}{n} i }\]

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

Answer 11

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

Answer 12

"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.

Answer 13

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.

Answer 14

"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.