Question

The relation between the real numbers a and b, which satisfy the equation $\frac{1 - ix}{1 + ix}$= a –ib, for some real value of x, is

• A. (a – b) (a + b) = 1
• B. $\left( \frac{a - b}{a + b} \right)$=1
• C. a² + b² = 1
• D. None of these

Answer 1

Answer: (C)

a² + b² = 1

Answer 2

Sol. $\frac{1 - ix}{1 + ix}$= a – ib Ž 1 –ix = (a – ib) (1 + ix)

Ž 1 – ix = a + aix – ib + bx

Ž (1 – a + ib) = x [ai + b + i]

Ž x = $\frac{(1 - a) + ib}{b + i(a + 1)}$× $\frac{b - i(a + 1)}{b - i(a + 1)}$

$\frac{b(1 - a) + b(a + 1) + i(b^{2} - (1 - a^{2}))}{b^{2} + (a + 1)^{2}}$

x is real if b² + a² – 1 = 0

i.e. a² + b² = 1