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:

• A. \(9\pi\)
• B. \( \frac{9\pi}{2} \)
• C. \(3\pi\)
• D. \( \frac{9\pi}{4} \)

Hint:

For complex number loci, use modulus for radius constraints and argument limits to determine angular coverage.

Answer 1

The Correct Option is B

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: \[ \boxed{\frac{9\pi}{2}} \]

Answer 2

The problem involves finding the area of the region formed by the locus of a complex number \( Z \) where \( |Z| \leq 3 \) and \(-\frac{\pi}{2} \leq \text{arg } Z \leq \frac{\pi}{2} \).

1. A complex number \( Z \) can be represented as \( Z = x + yi \), where \( x \) and \( y \) are real numbers, and \( i \) is the imaginary unit.

2. The magnitude \( |Z| \) or modulus of \( Z \) is given by \( |Z| = \sqrt{x^2 + y^2} \). Here, \( |Z| \leq 3 \) implies the point \( (x, y) \) lies within or on a circle of radius 3 centered at the origin in the complex plane.

3. The argument \(\text{arg } Z\) represents the angle \(\theta\) such that \(\tan(\theta) = \frac{y}{x}\). Hence, the condition \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\) restricts \( Z \) to lie in the right half of the complex plane.

4. Therefore, the locus is a semicircle (right half) with radius 3 centered at the origin. The area \( A \) of a full circle is \( \pi r^2 \), thus area of the semicircle:

\(A = \frac{1}{2}\pi(3)^2 = \frac{1}{2}(9\pi) = \frac{9\pi}{2}\).

The area of the region is therefore \(\frac{9\pi}{2}\).