Is e^i recognized as a root of unity? This is a complex number, and the square of the real part and the square of the imaginary part equal 1 when added together.
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No. Roots of unity are complex number z such that z^n = 1 for some positive integer n. Here's a proof that e^i is not a root of unity:
Suppose e^(ix) = 1, where x is real. Then:
e^(ix) = cos(x) + i sin(x) = 1
===> cos(x) = 1 AND sin(x) = 0
===> x = 2pi * k ... for some integer k
Now, suppose e^i is a root of unity. Then, for some integer n > 0:
(e^i)^n = e^(in) = 1
===> n = 2pi * k
Since n > 0, we cannot have k = 0. Thus:
pi = n / (2k)
which would imply pi is rational. But, pi is not rational, so e^i is not a root of unity.
Suppose e^(ix) = 1, where x is real. Then:
e^(ix) = cos(x) + i sin(x) = 1
===> cos(x) = 1 AND sin(x) = 0
===> x = 2pi * k ... for some integer k
Now, suppose e^i is a root of unity. Then, for some integer n > 0:
(e^i)^n = e^(in) = 1
===> n = 2pi * k
Since n > 0, we cannot have k = 0. Thus:
pi = n / (2k)
which would imply pi is rational. But, pi is not rational, so e^i is not a root of unity.