<p>Given \(z = f(x) + i\,g(x)\) where \(f, g: (0,1) \to (0,1)\) are real valued functions. Then which of the following does not hold good?</p>
<p>(1) \(z = \dfrac{1}{1-ix} + i\left(\dfrac{1}{1+ix}\right)\)</p>
<p>(2) \(z = \dfrac{1}{1+ix} + i\left(\dfrac{1}{1-ix}\right)\)</p>
<p>(3) \(z = \dfrac{1}{1+ix} + i\left(\dfrac{1}{1+ix}\right)\)</p>
<p>(4) \(z = \dfrac{1}{1-ix} + i\left(\dfrac{1}{1-ix}\right)\)</p>
Step-by-Step Solution
Key Concept: For a complex number z = f(x) + ig(x) where both f and g map (0,1) to (0,1), the modulus |z| = √(f²+g²) is bounded by √2, and z must lie strictly inside the square [0,1]×[0,1] in the complex plane. Key insight: both real and imaginary parts are independently constrained to (0,1).
<p><strong>Given:</strong> z = f(x) + ig(x) where f, g: (0,1) → (0,1)</p><p><strong>Step 1:</strong> Since f(x) ∈ (0,1) and g(x) ∈ (0,1), we have 0 < f(x) < 1 and 0 < g(x) < 1 for all x ∈ (0,1).</p><p><strong>Step 2:</strong> Therefore z lies strictly inside the open square with vertices at 0, 1, 1+i, and i in the complex plane.</p><p><strong>Step 3:</strong> Consequences that HOLD GOOD:</p><p>• |z| < √(1² + 1²) = √2 (strict inequality)</p><p>• Re(z) ∈ (0,1) and Im(z) ∈ (0,1)</p><p>• |z| > 0 (z ≠ 0)</p><p>• z cannot equal any point on the boundary of the unit square</p><p><strong>Step 4:</strong> Statement that does NOT hold good:</p><p>• |z| = √2 (impossible, maximum is approached but never reached)</p><p>• |z| ≤ 1 (false; z can have |z| > 1, e.g., when f = g = 0.9)</p><p>• Re(z) = 0 or Im(z) = 0 (impossible since domain is open)</p><p>∴ Answer: The statement(s) claiming |z| can equal √2 OR that |z| ≤ 1 OR that z touches the boundary do NOT hold good.</p>
Correct Answer: A, B