Question

If \( z_1, z_2 \) are complex numbers such that \( \frac{z_1}{3z_2} \) is a purely imaginary number, then the value of \( \left| \frac{z_1 - z_2}{z_1 + z_2} \right| \) is:

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

For complex numbers, use the condition of purely imaginary ratios to derive real part constraints, then compute moduli using conjugates and magnitudes.

Answer 1

The Correct Option is A

Step 1: Define the given condition.
Let \( z_1 = x_1 + i y_1 \) and \( z_2 = x_2 + i y_2 \), where \( x_1, y_1, x_2, y_2 \) are real numbers. The problem states \( \frac{z_1}{3z_2} \) is purely imaginary. This means its real part is zero. Therefore: \[\n\frac{z_1}{3z_2} = \frac{x_1 + i y_1}{3 (x_2 + i y_2)}.\n\] The real part of this expression must equal zero.
Step 2: Compute the real part.
Multiply the numerator and denominator by the conjugate of \( 3z_2 \): \[\n\frac{z_1}{3z_2} = \frac{(x_1 + i y_1) (x_2 - i y_2)}{3 (x_2^2 + y_2^2)} = \frac{(x_1 x_2 + y_1 y_2) + i (y_1 x_2 - x_1 y_2)}{3 (x_2^2 + y_2^2)}.\n\] The real part is: \[\n\frac{x_1 x_2 + y_1 y_2}{3 (x_2^2 + y_2^2)} = 0,\n\] assuming \( z_2 \neq 0 \) (i.e., \( x_2^2 + y_2^2 \neq 0 \)). Thus: \[\nx_1 x_2 + y_1 y_2 = 0. \quad (1)\n\]
Step 3: Evaluate the expression \( \left| \frac{z_1 - z_2}{z_1 + z_2} \right| \).
Calculate the numerator and denominator: \( z_1 - z_2 = (x_1 - x_2) + i (y_1 - y_2) \), \( z_1 + z_2 = (x_1 + x_2) + i (y_1 + y_2) \). The modulus is: \[\n\left| \frac{z_1 - z_2}{z_1 + z_2} \right| = \frac{|z_1 - z_2|}{|z_1 + z_2|},\n\] where \( |z| = \sqrt{x^2 + y^2} \) for \( z = x + i y \). \( |z_1 - z_2|^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 \), \( |z_1 + z_2|^2 = (x_1 + x_2)^2 + (y_1 + y_2)^2 \).
Step 4: Use the condition to simplify.
From equation (1), \( x_1 x_2 + y_1 y_2 = 0 \), which implies \( \text{Re}(z_1 \overline{z_2}) = 0 \), or \( z_1 \overline{z_2} \) is purely imaginary. This is equivalent to: \[\nz_1 \overline{z_2} = - \overline{z_1 z_2},\n\] but we directly use \( \text{Re}(z_1 \overline{z_2}) = 0 \). Consider the ratio: \[\n\frac{z_1 - z_2}{z_1 + z_2} = \frac{(x_1 - x_2) + i (y_1 - y_2)}{(x_1 + x_2) + i (y_1 + y_2)}.\n\] The modulus squared is: \[\n\left| \frac{z_1 - z_2}{z_1 + z_2} \right|^2 = \frac{(x_1 - x_2)^2 + (y_1 - y_2)^2}{(x_1 + x_2)^2 + (y_1 + y_2)^2}.\n\] Using \( x_1 x_2 + y_1 y_2 = 0 \): Expand: \( (x_1 + x_2)^2 + (y_1 + y_2)^2 = x_1^2 + 2x_1 x_2 + x_2^2 + y_1^2 + 2y_1 y_2 + y_2^2 \), Since \( 2(x_1 x_2 + y_1 y_2) = 0 \), this simplifies to \( x_1^2 + x_2^2 + y_1^2 + y_2^2 + 0 = |z_1|^2 + |z_2|^2 \). Similarly, \( (x_1 - x_2)^2 + (y_1 - y_2)^2 = x_1^2 - 2x_1 x_2 + x_2^2 + y_1^2 - 2y_1 y_2 + y_2^2 = |z_1|^2 + |z_2|^2 - 2(x_1 x_2 + y_1 y_2) = |z_1|^2 + |z_2|^2 \). Thus: \[\n\left| \frac{z_1 - z_2}{z_1 + z_2} \right|^2 = \frac{|z_1|^2 + |z_2|^2}{|z_1|^2 + |z_2|^2} = 1,\n\] so \( \left| \frac{z_1 - z_2}{z_1 + z_2} \right| = 1 \), assuming \( z_1 + z_2 \neq 0 \).
Step 5: Verify and select the answer.
The condition \( \frac{z_1}{3z_2} \) purely imaginary implies \( x_1 x_2 + y_1 y_2 = 0 \), which confirms the modulus result. The value is 1, matching option (A).