1
$\begingroup$

The sets $A = \{z : z^{18} = 1\} $and $ B = \{w : w^{48} = 1\}$ are both sets of complex roots of unity. The set $C = \{zw : z \in A \ \text{and} \ w \in B\}$ is also a set of complex roots of unity. How many distinct elements are in $C$?

Erm I'm wondering where to start this? Does the stuff in the set equal the ratio of $z$ to $z^{18}$ which is equal to 1 so $z^{18}= z$? and $w = w^{48}$? Help is appreciated, thank you.

$\endgroup$
  • 2
    $\begingroup$ The gcd or lcm of exponents say something to you? $\endgroup$ – Martín-Blas Pérez Pinilla Jan 27 '14 at 13:54
  • 1
    $\begingroup$ You might like to think about this problem geometrically. Remember, multiplying complex numbers adds the angles. $\endgroup$ – Zach L. Jan 27 '14 at 13:54
2
$\begingroup$

A hint that is a bit too long for a comment, but should get you going along one possible path:

The elements of the set $A$ are of the form $e^{2s\pi i/18}$, and the elements of $B$ are of the form $e^{2t\pi i/48}$, so that when you multiply an element from $A$ by an element from $B$, what you get is a number of the form:

$$e^{(s/18+t/48).2\pi i}$$

so now you need to look at the possible values of $s/18+t/48 = (8s+3t)/144$.

$\endgroup$
  • $\begingroup$ I was wondering how you got that the elements of set A and B are of that form? $\endgroup$ – Freedom Jan 27 '14 at 18:11
  • $\begingroup$ @LoyalKnight It follows from De Moivre's theorem, as on this Wikipedia page, and it is one of the things that once you have used it a couple of times, it becomes a really useful fact to recall and use when needed. $\endgroup$ – Old John Jan 27 '14 at 18:52
2
$\begingroup$

HINT:

As gcd$(18,48)=6,$lcm $(18,48)=144$

The minimum positive integer of $n$ such that $(zw)^n=1$ is $144$ for all possible combinations of $z,w$

Now utilize Complex numbers and Roots of unity

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.