Question

\(a + ib>c + id\) can be explained only when:

• A. \(b = 0,\; c = 0\)
• B. \(b = 0,\; d = 0\)
• C. \(a = 0,\; c = 0\)
• D. \(a = 0,\; d = 0\)

Hint:

Inequalities ((>, <)) are defined only for real numbers. To compare complex numbers, their imaginary parts must be zero.

Answer 1

The Correct Option is B

Step 1: Recall the order property of numbers. The symbols \(>\) and \(<\) are defined only for real numbers. Complex numbers \((a + ib)\) cannot, in general, be compared using inequalities.

Step 2: Condition for a complex number to be real. A complex number \(a + ib\) is real only if its imaginary part is zero: \[ b = 0 \] Similarly, \(c + id\) is real only if: \[ d = 0 \] 

Step 3: Apply the condition to the given inequality. The inequality \(a + ib>c + id\) is meaningful only when: \[ b = 0 \quad \text{and} \quad d = 0 \] Then it reduces to the real-number inequality: \[ a>c \] 

Step 4: Identify the correct option. Option (B) satisfies the required condition.