Question

Let C be the circle in the complex plane with centre z₀ = \(\frac{1}{2}\) (1+3i) and radius r = 1. Let z₁ = 1+ i and the complex number z2 be outside the circle C such that |z₁ – z₀| |z₂ – z₀| = 1. If z₀, z₁ and z₂ are collinear, then the smaller value of |z₂|² is equal to

• A. \(\frac{3}{2}\)
• B. \(\frac{5}{2}\)
• C. \(\frac{7}{2}\)
• D. \(\frac{13}{2}\)

Answer 1

The Correct Option is B

Given Modulus: From the problem: \[ |z_1 - z_0| = \frac{1 - i}{2} = \frac{1}{\sqrt{2}}, \quad |z_2 - z_0| = \sqrt{2}, \, \text{with centre at} \, \left(\frac{1}{2}, \frac{3}{2}\right). \]

Given Points: \[ z_0 = \left(\frac{1}{2}, \frac{3}{2}\right) \quad \text{and} \quad z_1 = (1, 1). \]

Calculate the angle \( \theta \): For \(z_2\), we use: \[ \tan \theta = -1 \quad \implies \quad \theta = 135^\circ. \]

Coordinates of \(z_2\): Using polar coordinates, we find: \[ z_2 = \left(\frac{1}{2} + \sqrt{2}\cos 135^\circ, \frac{3}{2} + \sqrt{2}\sin 135^\circ\right), \] or 
\[ z_2 = \left(\frac{1}{2} - \sqrt{2}\cos 135^\circ, \frac{3}{2} - \sqrt{2}\sin 135^\circ\right). \]

Substitute the values: For \( |z_2|^2 \): \[ z_2 = \left(-\frac{1}{2}, \frac{5}{2}\right) \quad \text{or} \quad z_2 = \left(\frac{3}{2}, \frac{1}{2}\right). \]

Result: \[ |z_2|^2 = \frac{26}{4}, \, \frac{5}{2}. \] Minimum value: \[ |z_2|^2_{\text{min}} = \frac{5}{2}. \]

Final Answer: The minimum value of \( |z_2|^2 \) is \(\frac{5}{2}\).