Key Concepts and Formulas

• Complex numbers: $z = x + iy$, where $x$ is the real part and $y$ is the imaginary part.
• Modulus of a complex number: $|z| = \sqrt{x^2 + y^2}$.
• Area of a sector of a circle: $A = \frac{1}{2}r^2\theta$, where $r$ is the radius and $\theta$ is the angle in radians.

Step-by-Step Solution

Step 1: Understanding the sets

We are given three sets of complex numbers:

• $S_1 = \{z \in \mathbf{C}: |z| \leq 5\}$
• $S_2 = \left\{z \in \mathbf{C}: \operatorname{Im}\left(\frac{z+1-\sqrt{3} i}{1-\sqrt{3} i}\right) \geq 0\right\}$
• $S_3 = \{z \in \mathbf{C}: \operatorname{Re}(z) \geq 0\}$

Our goal is to find the area of the region $S_1 \cap S_2 \cap S_3$. We will analyze each set individually and then find the intersection.

Step 2: Analyzing Set $S_1$

$S_1 = \{z \in \mathbf{C}: |z| \leq 5\}$

• Explanation: This set represents all complex numbers $z$ whose distance from the origin is less than or equal to 5.
• Geometric Interpretation: This is a closed disk centered at the origin with radius $r = 5$.

Step 3: Analyzing Set $S_3$

$S_3 = \{z \in \mathbf{C}: \operatorname{Re}(z) \geq 0\}$

• Explanation: If $z = x + iy$, then $\operatorname{Re}(z) = x$. The inequality $x \geq 0$ means the real part of $z$ is non-negative.
• Geometric Interpretation: This is the right half of the complex plane, including the imaginary axis. This corresponds to angles from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$.

Step 4: Analyzing Set $S_2$

$S_2 = \left\{z \in \mathbf{C}: \operatorname{Im}\left(\frac{z+1-\sqrt{3} i}{1-\sqrt{3} i}\right) \geq 0\right\}$

• Explanation: We need to simplify the complex expression and then apply the imaginary part condition.
• Step-by-step working:
  1. Let $z = x + iy$. Substitute into the expression: $\frac{z+1-\sqrt{3} i}{1-\sqrt{3} i} = \frac{(x+iy)+1-\sqrt{3} i}{1-\sqrt{3} i} = \frac{(x+1)+i(y-\sqrt{3})}{1-\sqrt{3} i}$
  2. Multiply the numerator and denominator by the conjugate of the denominator, which is $1 + \sqrt{3}i$: $\frac{(x+1)+i(y-\sqrt{3})}{1-\sqrt{3} i} \times \frac{1+\sqrt{3} i}{1+\sqrt{3} i}$
  3. Calculate the denominator: $(1-\sqrt{3} i)(1+\sqrt{3} i) = 1^2 - (\sqrt{3} i)^2 = 1 - (-3) = 1+3 = 4$
  4. Calculate the numerator: $((x+1)+i(y-\sqrt{3}))(1+\sqrt{3} i)$ $= (x+1)(1) + (x+1)(\sqrt{3} i) + i(y-\sqrt{3})(1) + i(y-\sqrt{3})(\sqrt{3} i)$ $= (x+1) + i\sqrt{3}(x+1) + iy - i\sqrt{3} + i^2\sqrt{3}(y-\sqrt{3})$ $= (x+1) + i\sqrt{3}x + i\sqrt{3} + iy - i\sqrt{3} - \sqrt{3}y + 3$ $= (x+1-\sqrt{3}y+3) + i(\sqrt{3}x + y)$ $= (x+4-\sqrt{3}y) + i(\sqrt{3}x+y)$
  5. Combine the numerator and denominator: $\frac{z+1-\sqrt{3} i}{1-\sqrt{3} i} = \frac{x+4-\sqrt{3}y}{4} + i\frac{\sqrt{3}x+y}{4}$
  6. Apply the condition $\operatorname{Im}\left(\frac{z+1-\sqrt{3} i}{1-\sqrt{3} i}\right) \geq 0$: $\frac{\sqrt{3}x+y}{4} \geq 0$ $\implies \sqrt{3}x+y \geq 0$ $\implies y \geq -\sqrt{3}x$
• Geometric Interpretation: The inequality $y \geq -\sqrt{3}x$ describes the region above or on the line $y = -\sqrt{3}x$. The slope of this line is $-\sqrt{3}$, corresponding to an angle of $-\frac{\pi}{3}$ with the positive real axis.

Step 5: Finding the Intersection $S_1 \cap S_2 \cap S_3$

We combine all three conditions:

1. $|z| \leq 5$: Inside the circle of radius 5 centered at the origin.
2. $\operatorname{Re}(z) \geq 0 \implies x \geq 0$: The right half-plane (angles from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$).
3. $y \geq -\sqrt{3}x$: The region above or on the line $y = -\sqrt{3}x$ (angles from $-\frac{\pi}{3}$ to $\pi$).

We need to find the angular span of the region $S_2 \cap S_3$.

• $S_3$ restricts us to angles from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$.
• $S_2$ restricts us to angles from $-\frac{\pi}{3}$ upwards.

The intersection of these two angular conditions is the range $[-\frac{\pi}{3}, \frac{\pi}{2}]$.

• The lower bound is the line $y = -\sqrt{3}x$, which makes an angle of $-\frac{\pi}{3}$ with the positive x-axis.
• The upper bound is the positive y-axis, which makes an angle of $\frac{\pi}{2}$ with the positive x-axis.

The region $S_1 \cap S_2 \cap S_3$ is a sector of the circle with radius $r=5$, bounded by the ray $y = -\sqrt{3}x$ for $x \geq 0$ and the positive y-axis.

The total angle $\theta$ of this sector is: $\theta = \frac{\pi}{2} - \left(-\frac{\pi}{3}\right) = \frac{\pi}{2} + \frac{\pi}{3} = \frac{3\pi}{6} + \frac{2\pi}{6} = \frac{5\pi}{6} \text{ radians}$

Step 6: Calculating the Area

The area of a circular sector is given by $\frac{1}{2}r^2\theta$.

• Radius $r=5$.
• Angle $\theta = \frac{5\pi}{6}$ radians.

Area $= \frac{1}{2} (5^2) \left(\frac{5\pi}{6}\right)$ Area $= \frac{1}{2} (25) \left(\frac{5\pi}{6}\right)$ Area $= \frac{125\pi}{12}$

Common Mistakes & Tips

• Be careful when finding the conjugate of a complex number and multiplying complex fractions.
• Remember that the angle in the sector area formula must be in radians.
• Always visualize the regions in the complex plane to help understand the intersection.

Summary

The region $S_1 \cap S_2 \cap S_3$ is a sector of a circle with radius 5. The intersection of the angular conditions derived from $S_2$ and $S_3$ spans from $-\frac{\pi}{3}$ to $\frac{\pi}{2}$, resulting in a total angle of $\frac{5\pi}{6}$ radians. Therefore, the area of the sector is $\frac{1}{2}r^2\theta = \frac{1}{2}(5^2)\left(\frac{5\pi}{6}\right) = \frac{125\pi}{12}$.

The final answer is \boxed{\frac{125 \pi}{12}}, which corresponds to option (C).