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

Answer: (D)

$\frac{125\pi}{12}$

Answer 2

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: $\boxed{\frac{125\pi}{12}}.$