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

The given problem involves a circle in the complex plane and uses properties of collinear points along with complex numbers to find the square of the modulus of a complex number. Here's the step-by-step solution:

We are given:

• The circle \( C \) with center \( z_0 = \frac{1}{2}(1+3i) = \frac{1}{2} + \frac{3}{2}i \) and radius \( r = 1 \).
• A point \( z_1 = 1+i \) inside the circle.
• A point \( z_2 \) outside the circle such that the points \( z_0 \), \( z_1 \), and \( z_2 \) are collinear.
• The condition \( |z_1 - z_0| \cdot |z_2 - z_0| = 1 \). 

First, calculate \( |z_1 - z_0| \):

\[ z_1 - z_0 = (1+i) - \left( \frac{1}{2} + \frac{3}{2}i \right) = \frac{1}{2} - \frac{1}{2}i \]

\[ |z_1 - z_0| = \left| \frac{1}{2} - \frac{1}{2}i \right| = \sqrt{\left( \frac{1}{2} \right)^2 + \left( -\frac{1}{2} \right)^2} = \frac{\sqrt{2}}{2} \]

From the given condition, we know:

\[ \left(\frac{\sqrt{2}}{2}\right) \cdot |z_2 - z_0| = 1 \]

\[ |z_2 - z_0| = \sqrt{2} \]

Since \( z_0 \), \( z_1 \), and \( z_2 \) are collinear, \( z_2 \) must be on the line extending from \( z_0 \) through \( z_1 \). We can express \( z_2 \) as:

\[ z_2 = z_1 + k(z_1 - z_0) \]

For \( z_2 \) to be outside the circle, \( k > 1 \). Let's assume \( z_2 = z_1 + (z_1 - z_0) \):

\[ z_2 = (1+i) + \left(\frac{1}{2} - \frac{1}{2}i\right) = \frac{3}{2} + \frac{1}{2}i \]

Now, calculate \( |z_2 - z_0| \):

\[ z_2 - z_0 = \left(\frac{3}{2} + \frac{1}{2}i\right) - \left(\frac{1}{2} + \frac{3}{2}i\right) = 1 - i \]

\[ |z_2 - z_0| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \]

Verify if \( |z_2|^2 \) equals to one of the given options:

\[ |z_2| = \sqrt{\left( \frac{3}{2} \right)^2 + \left( \frac{1}{2} \right)^2} = \sqrt{\frac{9}{4} + \frac{1}{4}} = \sqrt{2.5} \]

\[ |z_2|^2 = 2.5 = \frac{5}{2} \]

Therefore, the smaller value of \( |z_2|^2 \) is \( \frac{5}{2} \) which matches the correct answer.