Complex Numbers
Modulus of Complex Numbers
Grade None
Question:
<p>If \(z\) is a complex number such that \(\left|z - \dfrac{25}{|z|}\right| \leq 24\), then which of the following is/are correct?</p>
<p>\(|z|^2 - 24|z| - 25 \leq 0\)</p>
<p>\(|z| \leq 25\)</p>
<p>\(1 \leq |z| \leq 25\)</p>
<p>\(|z| \geq 1\)</p>
Step-by-Step Solution
Key Concept: Convert the modulus inequality into a real constraint by letting |z| = r, then use the triangle inequality |a - b| ≤ |a| + |b| strategically to bound r, recognizing that z and 25/|z| can be positioned to either reinforce or cancel their magnitudes.
<p><strong>Step 1:</strong> Let |z| = r where r > 0. We need to analyze |z - 25/r| ≤ 24.</p><p><strong>Step 2:</strong> By the triangle inequality: |z - 25/r| ≥ ||z| - |25/r|| = |r - 25/r|.</p><p>Therefore: |r - 25/r| ≤ 24</p><p><strong>Step 3:</strong> This gives us: -24 ≤ r - 25/r ≤ 24</p><p><strong>Step 4:</strong> From r - 25/r ≤ 24: r² - 24r - 25 ≤ 0 → (r - 25)(r + 1) ≤ 0 → r ≤ 25 (since r > 0)</p><p><strong>Step 5:</strong> From r - 25/r ≥ -24: r² + 24r - 25 ≥ 0 → (r + 25)(r - 1) ≥ 0 → r ≥ 1 (since r > 0)</p><p><strong>Step 6:</strong> Therefore: 1 ≤ |z| ≤ 25. The equality |z - 25/|z|| = 24 occurs when z and 25/|z| are oppositely directed, achievable for all r in [1, 25].</p><p>∴ Answer: B,C</p>
Correct Answer: B,C