Roots of Unity - Complex Numbers

Question

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.

You might like to think about this problem geometrically. Remember, multiplying complex numbers adds the angles. — Zach L., Jan 27 '14 at 13:54

score 2 · Answer 1 · edited Apr 13 '17 at 12:20

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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

edited Apr 13 '17 at 12:20

Community

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answered Jan 27 '14 at 13:56

lab bhattacharjee

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score 2 · Accepted Answer · answered Jan 27 '14 at 13:58

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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 $s/18+t/48 = (8s+3t)/144$.

answered Jan 27 '14 at 13:58

Old John

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I was wondering how you got that the elements of set A and B are of that form? – Freedom Jan 27 '14 at 18:11
@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. – Old John Jan 27 '14 at 18:52

Roots of Unity - Complex Numbers

2 Answers2