<p>Let <math>n</math> be a positive integer and a complex number with unit modulus is a solution of the equation <math>z^n + z + 1 = 0</math>. Then the value of <math>n</math> can be:</p>
Step-by-Step Solution
Key Concept: Use the unit modulus condition and the equation to determine that z must be a specific root, then find which values of n satisfy the resulting congruence.
<p>If <math>|z| = 1</math> and <math>z^n + z + 1 = 0</math>, then <math>z^n = -(z + 1)</math>.</p><p>Taking modulus: <math>|z^n| = |z + 1|</math>, so <math>1 = |z + 1|</math>.</p><p>With <math>z = e^{i\theta}</math>, we have <math>|e^{i\theta} + 1| = 1</math>.</p><p><math>|e^{i\theta} + 1|^2 = (\cos\theta + 1)^2 + \sin^2\theta = 2 + 2\cos\theta = 1</math></p><p>Thus <math>\cos\theta = -\frac{1}{2}</math>, giving <math>\theta = \frac{2\pi}{3}</math> or <math>\theta = \frac{4\pi}{3}</math>.</p><p>From <math>z^n = -(z+1)</math> and <math>z = e^{i2\pi/3}</math>, we get <math>e^{i2\pi n/3} = e^{i\pi}\cdot e^{i2\pi/3} = e^{i5\pi/3}</math>.</p><p>Thus <math>\frac{2\pi n}{3} \equiv \frac{5\pi}{3} \pmod{2\pi}</math>, giving <math>2n \equiv 5 \pmod{6}</math>.</p><p>This is satisfied by <math>n = 62, 155, 221</math> but not 196.</p>
Correct Answer: A, B, C