Question

Let $S_1 = \{z \in \mathbb{C} : |z| \leq 5\}$,
$S_2 = \left\{z \in \mathbb{C} : \text{Im}\left(\frac{z + 1 - \sqrt{3}i}{1 - \sqrt{3}i}\right) \geq 0\right\}$ and
$S_3 = \{z \in \mathbb{C} : \text{Re}(z) \geq 0\}$. Then

• A. $\frac{125\pi}{6}$
• B. $\frac{125\pi}{24}$
• C. $\frac{125\pi}{4}$
• D. $\frac{125\pi}{12}$

Answer 1

The Correct Option is D

The goal is to find the area of the region formed by the intersection of \( S_1, S_2, \) and \( S_3 \). We evaluate these step by step.

Step 1: Region defined by \( S_1 \)
The condition \( |z| \leq 5 \) implies: \[ x^2 + y^2 \leq 25. \] This represents the interior of a circle with radius 5 centered at the origin.

Step 2: Region defined by \( S_2 \)
The condition \( S_2 \) is given by: \[ \text{Im}\left(\frac{z + (1 - \sqrt{3}i)}{1 - \sqrt{3}i}\right) \geq 0. \] Let \( z = x + iy \). Rewrite the expression: \[ \frac{z + (1 - \sqrt{3}i)}{1 - \sqrt{3}i} = \frac{(x + iy) + (1 - \sqrt{3}i)}{1 - \sqrt{3}i}. \] Multiply numerator and denominator by the conjugate of \( 1 - \sqrt{3}i \), i.e., \( 1 + \sqrt{3}i \): \[ \frac{((x + 1) + i(y - \sqrt{3}))(1 + \sqrt{3}i)}{(1 - \sqrt{3}i)(1 + \sqrt{3}i)}. \] Simplify the denominator: \[ (1 - \sqrt{3}i)(1 + \sqrt{3}i) = 1^2 + 3 = 4. \] Now expand the numerator and focus on the imaginary part: \[ \text{Im}\left(\frac{z + (1 - \sqrt{3}i)}{1 - \sqrt{3}i}\right) = \frac{\sqrt{3}(x + 1) + y - \sqrt{3}}{4}. \] For \( S_2 \), the imaginary part must satisfy: \[ \sqrt{3}(x + 1) + y - \sqrt{3} \geq 0 \implies \sqrt{3}x + y + \sqrt{3} - \sqrt{3} \geq 0 \implies \sqrt{3}x + y \geq \sqrt{3}. \tag{1} \]

Step 3: Region defined by \( S_3 \)
The condition \( S_3 \) is given by: \[ \text{Re}(z) \geq 0 \implies x \geq 0. \tag{2} \]

Step 4: Intersection of \( S_1, S_2, \) and \( S_3 \)
The intersection of these conditions forms a sector of the circle \( x^2 + y^2 \leq 25 \), bounded by the lines \( \sqrt{3}x + y = 0 \) and \( x = 0 \), in the first quadrant.
Angle of the sector: The line \( \sqrt{3}x + y = 0 \) passes through the origin and makes an angle of \( 30^\circ \) (or \( \pi/6 \)) with the negative \( y \)-axis.
Therefore, the angle of the sector in the first quadrant is: \[ \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}. \]

Step 5: Area of the region
The area of the region is the area of the half-circle minus the area of the sector defined by the arc \( AB \).
1. Area of the half-circle:
\[ \text{Area of half-circle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (5)^2 = \frac{25\pi}{2}. \] 2. Area of the sector \( AB \):
\[ \text{Area of sector} = \frac{\theta}{2\pi} \cdot \pi r^2 = \frac{\pi/6}{2\pi} \cdot \pi (5)^2 = \frac{25\pi}{12}. \] 3. Shaded region:
\[ \text{Shaded area} = \text{Area of half-circle} - \text{Area of sector} = \frac{25\pi}{2} - \frac{25\pi}{12}. \] Simplify: \[ \text{Shaded area} = \frac{150\pi}{12} - \frac{25\pi}{12} = \frac{125\pi}{12}. \] Thus, the total area of the region is: \[ \boxed{\frac{125\pi}{12}}. \]

Answer 2

Step 1: Interpret the sets geometrically
S1 = { z : |z| ≤ 5 } is the closed disk of radius 5 centered at the origin.
S3 = { z : Re(z) ≥ 0 } is the right half-plane, i.e., points whose argument lies in [−π/2, π/2].

Step 2: Simplify S2
Let a = 1/(1 − √3 i). Then \[ \text{Im}\!\left(\frac{z + 1 - \sqrt{3}i}{1 - \sqrt{3}i}\right) = \text{Im}(a z + 1) = \text{Im}(a z), \] since 1 is real. Now write \(1 - \sqrt{3}i = 2\,e^{-i\pi/3}\), hence \(a = \frac{1}{2}e^{i\pi/3}\). Therefore \[ \text{Im}(a z) \ge 0 \quad\Longleftrightarrow\quad \text{Im}\!\left(e^{i\pi/3} z\right) \ge 0. \] This means: after rotating z by +π/3, the imaginary part is nonnegative. Equivalently, the allowed arguments of z lie in \[ \arg z \in \left[-\frac{\pi}{3},\,\frac{2\pi}{3}\right]. \] Thus S2 is a half-plane bounded by the line through the origin at angle −π/3.

Step 3: Angular sector for S2 ∩ S3
S3 allows \(\arg z \in [-\pi/2,\,\pi/2]\). Intersecting with S2 gives \[ \arg z \in \left[\max\!\left(-\frac{\pi}{3},-\frac{\pi}{2}\right),\ \min\!\left(\frac{2\pi}{3},\frac{\pi}{2}\right)\right] = \left[-\frac{\pi}{3},\,\frac{\pi}{2}\right], \] which has angular width \[ \theta = \frac{\pi}{2} - \left(-\frac{\pi}{3}\right) = \frac{5\pi}{6}. \] Hence \(S_1 \cap S_2 \cap S_3\) is a circular sector of radius 5 and angle \(5\pi/6\).

Step 4: Compute the area
Area of a sector with angle θ (radians) and radius r is \(A = \tfrac{1}{2}\,r^2\,\theta\). With \(r=5\) and \(\theta=\tfrac{5\pi}{6}\), \[ A = \frac{1}{2}\cdot 25 \cdot \frac{5\pi}{6} = \frac{125\pi}{12}. \]

Final answer

\(\frac{125\pi}{12}\)