Complex Numbers
Locus and Reflection
Grade 11
Question:
<p>Let the equation of a ray be |<em>z</em> − 2| − |<em>z</em> − 1 − <em>i</em>| = √2. If it strikes the <em>y</em>-axis, then the equation of reflected ray (including or excluding the point of incidence) is</p>
<p>(1) \(\arg(z - 2i) = \dfrac{\pi}{4}\)</p>
<p>(2) \(|z - 2i| - |z - 1 - 3i| = \sqrt{2}\)</p>
<p>(3) \(\arg(z - 2i) = \dfrac{3\pi}{4}\)</p>
<p>(4) \(|z - 2i| - |z - 1 - 3i| = 2\sqrt{2}\)</p>
Step-by-Step Solution
Key Concept: The equation |z - 2| - |z - 1 - i| = √2 represents a hyperbola branch (difference of distances is constant). Find where this ray intersects the y-axis, then apply the reflection law: angle of incidence equals angle of reflection with respect to the normal (x-axis).
<p><strong>Step 1: Identify the locus.</strong> The equation |z - 2| - |z - 1 - i| = √2 represents one branch of a hyperbola with foci at F₁ = (2, 0) and F₂ = (1, 1).</p><p><strong>Step 2: Find intersection with y-axis.</strong> Set z = iy (where y ∈ ℝ). Then |iy - 2| - |iy - 1 - i| = √2 gives: √(4 + y²) - √((y-1)² + 1) = √2. Solving: 4 + y² - (y-1)² - 1 = 2y√2 ⟹ 2y - 3 = 2y√2 ⟹ y = 3/(2(1 - √2)) = -3(1 + √2)/2. This gives point P on y-axis.</p><p><strong>Step 3: Apply reflection law.</strong> The incident ray comes from the hyperbola branch. When reflecting across the y-axis, the relationship becomes |z - 2| - |z - 1 + i| = -√2 (note: the second focus reflects from (1,1) to (1,-1), and the sign flips because we're on the other side). Equivalently, |z - 1 + i| - |z - 2| = √2.</p><p><strong>Step 4: Verify the reflected ray equation.</strong> The reflected ray satisfies |z - 1 + i| - |z - 2| = √2, which can be verified by the symmetric property of hyperbolic reflection.</p><p>∴ Answer: |z - 1 + i| - |z - 2| = √2</p>
Correct Answer: 2