PROVE (w/ a rigorous, detailed, proof),
or
DISPROVE (w/ a concrete counterexample)
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Note:
o(z) is the order of z,
ℚ is the set of all rational numbers,
ℝ is the set of all real numbers,
ℂ\{0} is the set of all complex numbers besides 0,
and, "iff" means "if and only if"
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Answers & Comments
The proof isn't hard and you should be able to put it together.
Use the fact that if r is not 1, then |z^n| goes to 0 or infinity.
And eventually nΘ has to be 2kπ for some integer k.
The problem characterizes all finite subgroups.