Complex Numbers
Geometry in Argand plane
Grade 11
Question:
<p>The points \(z_1 = 3 + \sqrt{3}\,i\) and \(z_2 = 2\sqrt{3} + 6i\) are given on a complex plane. The complex number lying on the bisector of the angle formed by the vectors \(z_1\) and \(z_2\) is</p>
<p>\(z = \dfrac{(3+2\sqrt{3})}{2} + \dfrac{\sqrt{3}+2}{2}\,i\)</p>
<p>\(z = 5 + 5i\)</p>
<p>\(z = -1 - i\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: The angle bisector of two complex numbers z₁ and z₂ from origin passes through points of the form t(z₁/|z₁| + z₂/|z₂|), where we normalize each vector by its modulus before adding them.
<p><strong>Step 1:</strong> Calculate |z₁| and |z₂|</p><p>|z₁| = √[(3)² + (√3)²] = √[9 + 3] = √12 = 2√3</p><p>|z₂| = √[(2√3)² + 6²] = √[12 + 36] = √48 = 4√3</p><p><strong>Step 2:</strong> Normalize each complex number (unit vectors)</p><p>z₁/|z₁| = (3 + √3 i)/(2√3) = (3/(2√3)) + (√3/(2√3))i = (√3/2) + (1/2)i</p><p>z₂/|z₂| = (2√3 + 6i)/(4√3) = (2√3/(4√3)) + (6/(4√3))i = (1/2) + (√3/2)i</p><p><strong>Step 3:</strong> Add the unit vectors to get the angle bisector direction</p><p>z₁/|z₁| + z₂/|z₂| = (√3/2 + 1/2) + (1/2 + √3/2)i = ((√3 + 1)/2) + ((1 + √3)/2)i</p><p><strong>Step 4:</strong> The angle bisector passes through points t·[(√3 + 1)/2 + (1 + √3)/2 i] for t > 0, or equivalently through (√3 + 1) + (√3 + 1)i up to scaling</p><p>∴ Answer: A</p>
Correct Answer: A