<p>Find the number of roots of the equation <span>\(z^{15} = 1\)</span> satisfying <span>\(|\arg z| < \pi/2\)</span>.</p>
Step-by-Step Solution
Key Concept: The 15th roots of unity are equally spaced on the unit circle at angles 2πk/15 (k=0,1,...,14). We must count how many of these satisfy |arg z| < π/3, meaning the argument lies strictly between -π/3 and π/3.
<p><strong>Step 1:</strong> The solutions to z^15 = 1 are the 15th roots of unity: z = e^(2πik/15) where k = 0, 1, 2, ..., 14.</p><p><strong>Step 2:</strong> The argument of z_k = e^(2πik/15) is arg(z_k) = 2πk/15 (taking principal values in [0, 2π)).</p><p><strong>Step 3:</strong> We need |arg z| < π/3. Using the standard convention where arg ∈ (-π, π], we rewrite this as: -π/3 < arg z < π/3.</p><p><strong>Step 4:</strong> For k = 0, 1, 2, ..., 14, the arguments are: 0, 2π/15, 4π/15, 6π/15, 8π/15, 10π/15, 12π/15, 14π/15, ... (and values > π wrap to negative).</p><p><strong>Step 5:</strong> Converting to (-π, π]: Arguments ≥ π become arg(z) = 2πk/15 - 2π. We need -π/3 < arg(z) < π/3.</p><p><strong>Step 6:</strong> π/3 = 5π/15, so we need arguments in (-5π/15, 5π/15). This gives us k values where 2πk/15 ∈ (0, 5π/15) or equivalently 2πk/15 ∈ (-5π/15, 0) [from wrapping].</p><p><strong>Step 7:</strong> Directly: k = 0 gives arg = 0 ✓; k = 1 gives 2π/15 ✓; k = 2 gives 4π/15 ✓; k = 3 gives 6π/15 = 2π/5 > π/3 ✗. For k = 14, 13, 12: these give 28π/15, 26π/15, 24π/15, which mod 2π give negative arguments. k = 14 gives -2π/15 ✓; k = 13 gives -4π/15 ✓; k = 12 gives -6π/15 < -π/3 ✗.</p><p>∴ <strong>Answer: 5 roots</strong> (corresponding to k = 0, 1, 2, 13, 14)</p>
Correct Answer: 5