Question

If the locus of $ z \in \mathbb{C} $, such that $ \text{Re} \left( \frac{z - 1}{2z + i} \right) + \text{Re} \left( \frac{ \bar{z} - 1}{2 \bar{z} - i} \right) = 2, $ is a circle of radius $ r $ and center $ (a, b) $, then $ \frac{15ab}{r^2} \text{ is equal to:} $

• A. 24
• B. 12
• C. 18
• D. 16

Hint:

When solving problems involving complex numbers and geometric loci, express complex numbers in terms of their real and imaginary parts. This will help you identify the equation of the circle and solve for the required values.

Answer 1

The Correct Option is C

We are given the equation: \[ \text{Re} \left( \frac{z - 1}{2z + i} \right) + \text{Re} \left( \frac{ \bar{z} - 1}{2 \bar{z} - i} \right) = 2. \] To solve this equation, let \( z = x + iy \), where \( x \) and \( y \) are real numbers representing the real and imaginary parts of \( z \). Substituting \( z = x + iy \) into the equation, we can express the real part of the complex numbers in terms of \( x \) and \( y \). The real part of the first term is: \[ \text{Re} \left( \frac{z - 1}{2z + i} \right) = \text{Re} \left( \frac{(x + iy) - 1}{2(x + iy) + i} \right). \] Simplify the numerator and denominator: \[ = \text{Re} \left( \frac{(x - 1) + iy}{(2x) + (2y + 1)i} \right). \] For the second term, we use the conjugate of \( z \), \( \bar{z} = x - iy \), and substitute in a similar manner. Now simplify both expressions and equate the sum of the real parts to 2.
Step 1: Find the center and radius of the circle.
The resulting equation represents the equation of a circle in the complex plane. From the real and imaginary parts, we can find the center \( (a, b) \) and the radius \( r \).
Step 2: Calculate the required expression.
Once we have the center \( (a, b) \) and radius \( r \), we use the formula: \[ \frac{15ab}{r^2}. \] Substitute the values of \( a \), \( b \), and \( r \) into this formula to find the value of the expression. \[ \frac{15ab}{r^2} = 18. \]