Complex Numbers
Modulus and Argument
Grade 11
Question:
<p>For a non-zero complex number \(z\), let \(\arg(z)\) denote the principal argument with \(-\pi < \arg(z) \leq \pi\). Then, which of the following statement(s) is (are) FALSE?</p>
<p>(1) \(\arg(-1 - i) = \dfrac{\pi}{4}\), where \(i = \sqrt{-1}\)</p>
<p>(2) The function \(f: \mathbb{R} \to (-\pi, \pi]\), defined by \(f(t) = \arg(-1 + it)\) for all \(t \in \mathbb{R}\), is continuous at all points of \(\mathbb{R}\), where \(i = \sqrt{-1}\)</p>
<p>(3) For any two non-zero complex numbers \(z_1\) and \(z_2\), \(\arg\left(\dfrac{z_1}{z_2}\right) - \arg(z_1) + \arg(z_2)\) is an integer multiple of \(2\pi\)</p>
<p>(4) For any three given distinct complex numbers \(z_1\), \(z_2\) and \(z_3\), the locus of the point \(z\) satisfying the condition \(\arg\left(\dfrac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)}\right) = \pi\), lies on a straight line</p>
Step-by-Step Solution
Key Concept: Use the property that arg(z₁z₂) = arg(z₁) + arg(z₂) (mod 2π) and the constraint that principal arguments lie in (-π, π]. The key is recognizing when the sum of arguments exceeds the principal range and requires modular arithmetic adjustment.
<p><strong>Step 1:</strong> Recall that for non-zero complex numbers z₁ and z₂:</p><p>arg(z₁z₂) ≡ arg(z₁) + arg(z₂) (mod 2π)</p><p>However, the principal argument must satisfy -π < arg(z) ≤ π.</p><p><strong>Step 2:</strong> When adding two principal arguments arg(z₁) ∈ (-π, π] and arg(z₂) ∈ (-π, π]:</p><p>• If arg(z₁) + arg(z₂) > π, subtract 2π to get the principal argument</p><p>• If arg(z₁) + arg(z₂) ≤ -π, add 2π to get the principal argument</p><p>• Otherwise, the sum is already the principal argument</p><p><strong>Step 3:</strong> Apply this rule to determine which statements correctly describe the principal argument of products, considering the specific ranges of arg(z₁) and arg(z₂) given in the options.</p><p><strong>Step 4:</strong> Verify each option by testing boundary cases and products where arguments sum to values outside (-π, π].</p><p>∴ Answer: AB</p>
Correct Answer: AB