<p>If \(|z - i\,\text{Re}(z)| = |z - \text{Im}(z)|\), then prove that \(z\) lies on the bisectors of the quadrants.</p>
Step-by-Step Solution
Key Concept: Convert the modulus condition into coordinates by setting z = x + iy, then expand both sides to eliminate the complex form and identify the geometric locus as straight lines through the origin.
<p><strong>Step 1: Set up coordinates</strong></p><p>Let z = x + iy where x, y ∈ ℝ. Then Re(z) = x and Im(z) = y.</p><p><strong>Step 2: Expand the modulus condition</strong></p><p>Given: |z - i·Re(z)| = |z - Im(z)|</p><p>Substitute: |(x + iy) - ix| = |(x + iy) - y|</p><p>Simplify: |x + i(y - x)| = |(x - y) + iy|</p><p><strong>Step 3: Apply modulus formula</strong></p><p>√[x² + (y - x)²] = √[(x - y)² + y²]</p><p><strong>Step 4: Square both sides</strong></p><p>x² + (y - x)² = (x - y)² + y²</p><p>x² + y² - 2xy + x² = x² - 2xy + y² + y²</p><p>x² + y² - 2xy + x² = x² - 2xy + 2y²</p><p><strong>Step 5: Simplify</strong></p><p>x² = y²</p><p>|x| = |y|</p><p><strong>Step 6: Interpret geometrically</strong></p><p>This gives x = y or x = -y, which are the two bisectors of the quadrants:</p><p>• y = x (bisector of 1st and 3rd quadrants)</p><p>• y = -x (bisector of 2nd and 4th quadrants)</p><p>∴ z lies on the bisectors of the quadrants. ✓</p>
Correct Answer: Proof