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Does the set of all #n#th roots of unity form a group under multiplication?

Answer 1

Yes

Identity: #1# is the identity.

Inverse: If #a# is an #n#th root of unity, then so is #1/a#, since:

#(1/a)^n = 1/(a^n) = 1/1 = 1#

Closure under product: If #a# is an #m#th root of unity and #b# an #n#th root of unity, then #ab# is an #mn#th root of unity:

#(ab)^(mn) = (a^m)^n(b^n)^m = 1^n*1^m = 1*1 = 1#

Associativity: Inherited from the complex numbers:

#a(bc) = (ab)c " "# for any #a, b, c#

#color(white)()# Footnote

The elements of this group are all the numbers of the form:

#cos theta + i sin theta#

where #theta# is a rational multiple of #pi#.

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Answer 2

Eva Abramson

Yes, the set of all nth roots of unity forms a group under multiplication.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.