Question

If \( a^2 + b^2 = 1 \), then:

\[ \frac{1 + (a - ib)}{1 + (a + ib)} \] is equal to:

• A. \( a - ib \)
• B. \( a + ib \)
• C. \( -a + ib \)
• D. \( -a - ib \)
• E. \( b + ia \)

Hint:

When simplifying complex rational expressions, multiply both the numerator and denominator by the conjugate of the denominator to eliminate the imaginary part in the denominator.

Answer 1

The Correct Option is A

We are asked to simplify the expression:

\[ \frac{1 + (a - ib)}{1 + (a + ib)}. \] Step 1: Simplify the numerator and denominator:

\[ \text{Numerator: } 1 + (a - ib) = 1 + a - ib. \] \[ \text{Denominator: } 1 + (a + ib) = 1 + a + ib. \] Thus, the expression becomes:

\[ \frac{1 + a - ib}{1 + a + ib}. \] Step 2: Multiply both the numerator and denominator by the conjugate of the denominator:

\[ \frac{(1 + a - ib)}{(1 + a + ib)} \times \frac{(1 + a - ib)}{(1 + a - ib)} = \frac{(1 + a - ib)^2}{(1 + a)^2 + b^2}. \] Step 3: Use the given condition \( a^2 + b^2 = 1 \). Thus, the denominator simplifies to:

\[ (1 + a)^2 + b^2 = 1 + 2a + a^2 + b^2 = 1 + 2a + 1 = 2 + 2a. \] Step 4: Now, expand the numerator:

\[ (1 + a - ib)^2 = (1 + a)^2 - 2i b(1 + a) + (-ib)^2 = (1 + 2a + a^2) - 2i b(1 + a) - b^2. \] Since \( b^2 = 1 - a^2 \), we get:

\[ (1 + a - ib)^2 = 1 + 2a + a^2 - 2ib(1 + a) - (1 - a^2) = 2a + 2a^2 - 2ib(1 + a). \] Step 5: Substitute back into the expression:

\[ \frac{2a + 2a^2 - 2ib(1 + a)}{2 + 2a}. \] Now, factor out the common terms:

\[ \frac{2a(1 + a) - 2ib(1 + a)}{2(1 + a)} = \frac{2a - 2ib}{2} = a - ib. \] Thus, the correct answer is option (A), \( a - ib \).