Complex Numbers
Complex Equations
Grade None
Question:
<p>Let <em>z</em> be a complex number satisfying equation \(z^p = \bar{z}^q\), where \(p, q \in \mathbb{N}\), then</p>
<p>(1) if <em>p</em> = <em>q</em>, then number of solutions of equation will be infinite</p>
<p>(2) if <em>p</em> = <em>q</em>, then number of solutions of equation will be finite</p>
<p>(3) if <em>p</em> ≠ <em>q</em>, then number of solutions of equation will be <em>p</em> + <em>q</em> + 1</p>
<p>(4) if <em>p</em> ≠ <em>q</em>, then number of solutions of equation will be <em>p</em> + <em>q</em></p>
Step-by-Step Solution
Key Concept: Express z in polar form as z = re^(iθ) and use the property that z^p = z̄^q implies both magnitude and argument constraints. The magnitude constraint r^p = r^q forces r = 0 or r = 1, while the argument constraint determines the allowed values of θ.
<p><strong>Step 1:</strong> Let z = re^(iθ) where r ≥ 0 and θ ∈ ℝ. Then z̄ = re^(-iθ).</p><p><strong>Step 2:</strong> The equation z^p = z̄^q becomes: r^p·e^(ipθ) = r^q·e^(-iqθ)</p><p><strong>Step 3:</strong> Comparing magnitudes: r^p = r^q, which gives r = 0 or r = 1 (since p, q ∈ ℕ and r ≥ 0).</p><p><strong>Step 4:</strong> Comparing arguments: pθ = -qθ + 2πk for some integer k, giving (p + q)θ = 2πk, so θ = 2πk/(p+q).</p><p><strong>Step 5:</strong> Therefore, z lies on the unit circle (|z| = 1) at specific angles determined by p and q. For the general case where no specific p, q values are given, the solution set includes the unit circle with restricted arguments, and notably z = 1 (when k = 0) always satisfies the equation.</p><p>∴ Answer: <strong>The solutions form a discrete set on the unit circle; |z| = 1 is the magnitude constraint</strong> (specific solutions depend on p and q values)</p>
Correct Answer: 1