Question:
For \( |z| \ge 2 \), if \( \left|z + \frac{1}{2}\right| \ge k \), the minimum possible value of \(k\) is
For \( |z| \ge 2 \), if \( \left|z + \frac{1}{2}\right| \ge k \), the minimum possible value of \(k\) is
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Minimum of expression occurs at boundary value of \(|z|\).
Updated On: Apr 30, 2026
- \(1/2\)
- \(3/2\)
- \(2\)
- \(5/2\)
- \(3\)
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The Correct Option is B
Solution and Explanation
Concept:
Use triangle inequality:
\[
|z+a| \ge ||z| - |a||
\]
Step 1: Apply inequality. \[ \left|z + \frac{1}{2}\right| \ge \left||z| - \frac{1}{2}\right| \]
Step 2: Use condition \( |z|\ge 2 \). Minimum occurs at smallest \(|z|\): \[ = \left|2 - \frac{1}{2}\right| = \frac{3}{2} \]
Step 1: Apply inequality. \[ \left|z + \frac{1}{2}\right| \ge \left||z| - \frac{1}{2}\right| \]
Step 2: Use condition \( |z|\ge 2 \). Minimum occurs at smallest \(|z|\): \[ = \left|2 - \frac{1}{2}\right| = \frac{3}{2} \]
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