Discussion
Let m be an even positive integer. What is the product of all mth roots of unity in C ?
Answer: B
For even m the product is -1.
The mth roots of unity are the roots of \(z^{m} - 1 = 0\). For a monic polynomial of degree m, the product of the roots is \((-1)^{m}\) times the constant term, here \((-1)^{m}(-1) = (-1)^{m + 1}\). When m is even this is \(-1\).
A direct check: pair each root \(\zeta\) with its conjugate \(\bar{\zeta} = \zeta^{-1}\); each pair multiplies to 1. The only roots equal to their own inverses are 1 and \(-1\), and \(-1\) is a root exactly when m is even. So the product is \(1 \cdot (-1) = -1\). For m = 2, the roots 1 and \(-1\) give \(-1\); for m = 4, \(1 \cdot i \cdot (-1) \cdot (-i) = -1\).
1 is the answer for odd m, and \(\pm i\) cannot occur because the non-real roots come in conjugate pairs.
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