Let \( n \) be a positive odd integer. Determine the greatest number of possible imaginary zeros of \( f(x)=x^{n}-1 \). The greatest number of possible imaginary zeros is

To find the greatest number of possible imaginary zeros of the function \( f(x) = x^n - 1 \) where \( n \) is a positive odd integer, we start by recognizing that the zeros of this polynomial correspond to the \( n \)-th roots of unity, which are given by the formula: \[ x_k = e^{2\pi i k/n} \] for \( k = 0, 1, 2, \ldots, n-1 \). These roots are all located on the unit circle in the complex plane. Among these roots, the imaginary zeros are the ones that lie exclusively on the imaginary axis, i.e., where \( x = iy \). For \( e^{2\pi i k/n} \) to be purely imaginary, the cosine component must be zero, which occurs when: \[ \cos(2\pi k/n) = 0 \] This happens when: \[ 2\pi k/n = \frac{\pi}{2} + m\pi \quad (m \in \mathbb{Z}) \] Simplifying this, we find that: \[ k = \frac{n}{4} + \frac{mn}{2} \] Since \( n \) is an odd integer, the two relevant values of \( k \) that yield purely imaginary roots are \( k = \frac{n}{4} \) (not possible since \( n/4 \) is non-integer) and \( k = \frac{3n}{4} \) (not possible since \( 3n/4 \) is also non-integer). However, we can specifically examine \( n = 2m + 1 \) (where \( m \) is a non-negative integer). Analyzing separately: - For \( n = 1 \), \( x^{1}-1 = 0 \) has 0 imaginary roots. - For \( n = 3 \), \( x^{3}-1 \) has \( \omega, -\omega, 1 \) where \( \omega = e^{2\pi i / 3} \), which include 1 imaginary root \( \omega^1 = e^{2\pi i/3} \). - For \( n = 5 \), \( x^{5}-1 \) includes roots that are \( e^{2\pi i k/5} \), which gives one imaginary root (\( k=2 \rightarrow \omega^2 \)). - Extending this, with \( n = 7, 9, \) etc. still yields only those two distinct rotations \( \frac{n}{4} \) or \( \frac{3n}{4} \) which correlate to 0 or 2 roots. Consequently, we can conclude that: The greatest number of possible imaginary zeros of \( f(x)=x^{n}-1 \) for odd \( n \) is always 2, achieved with the rotations corresponding to \( x = i \) and \( x = -i \), representing the two opposite points on the unit circle. Thus, the answer is: \[ \boxed{2} \]