Question:

If \( Z \) is a complex number such that \( |Z| \leq 3 \) and \( -\frac{\pi}{2} \leq \text{arg } Z \leq \frac{\pi}{2} \), then the area of the region formed by the locus of \( Z \) is:

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For complex number loci, use modulus for radius constraints and argument limits to determine angular coverage.
Updated On: Mar 24, 2025
  • \(9\pi\)
  • \( \frac{9\pi}{2} \)
  • \(3\pi\)
  • \( \frac{9\pi}{4} \)
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Given Conditions - The modulus condition \( |Z| \leq 3 \) represents a disk of radius 3 centered at the origin. - The argument constraint \( -\frac{\pi}{2} \leq \text{arg } Z \leq \frac{\pi}{2} \) restricts the region to the right half of the disk.
Step 2: Compute the Area of the Semi-Circle - The total area of a full circle with radius \( 3 \) is: \[ A_{\text{circle}} = \pi r^2 = \pi (3)^2 = 9\pi. \] - Since the region represents a semi-circle, the required area is: \[ A_{\text{semi-circle}} = \frac{1}{2} \times 9\pi = \frac{9\pi}{2}. \] Thus, the correct answer is: \[ \boxed{\frac{9\pi}{2}} \]
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