Complex Numbers
Conjugate of complex numbers
Grade 11
Question:
<p>The complex numbers \(\sin x + i\sin 2x\) and \(\cos x - i\cos 2x\) are conjugate to each other, for</p>
<p>\(x = n\pi,\; n \in \mathbb{Z}\)</p>
<p>\(x = 0\)</p>
<p>\(x = (n + 1/2)\pi,\; n \in \mathbb{Z}\)</p>
<p>no value of \(x\)</p>
Step-by-Step Solution
Key Concept: Two complex numbers are conjugates if their real parts are equal and imaginary parts are opposite. Here, set the real parts equal and imaginary parts as negatives of each other to find the condition on x.
<p><strong>Step 1:</strong> If z₁ = sin x + i sin 2x and z₂ = cos x - i cos 2x are conjugates, then z₁ = conjugate of z₂.</p><p><strong>Step 2:</strong> The conjugate of z₂ = cos x - i cos 2x is cos x + i cos 2x.</p><p><strong>Step 3:</strong> Setting z₁ equal to this conjugate: sin x + i sin 2x = cos x + i cos 2x</p><p><strong>Step 4:</strong> Equating real and imaginary parts:</p><ul><li>Real: sin x = cos x</li><li>Imaginary: sin 2x = cos 2x</li></ul><p><strong>Step 5:</strong> From sin x = cos x: tan x = 1 ⟹ x = π/4 + nπ</p><p><strong>Step 6:</strong> Verify with sin 2x = cos 2x: When x = π/4, sin(π/2) = 1 and cos(π/2) = 0 ✗</p><p><strong>Step 7:</strong> Solve sin 2x = cos 2x: tan 2x = 1 ⟹ 2x = π/4 + nπ ⟹ x = π/8 + nπ/2</p><p><strong>Step 8:</strong> Find intersection of conditions x = π/4 + nπ and x = π/8 + mπ/2. Both are satisfied when x = π/8 + (2n)π/2 for integer n.</p><p>∴ Answer: <strong>x = nπ ± π/8</strong> or equivalent form depending on options provided (typically x = π/8 + nπ/2 where specific values are selected)</p>
Correct Answer: D