The area (in sq. units) of the region S = {z ∈ C : |z - 1| ≤ 2; (z + z̄) + i(z - z̄) ≤ 2, Im(z) ≥ 0}

Q: Wait, why does $|z - 1| \le 2$ represent a disk?

In the complex plane, the equation $|z - z_0| = R$ defines a circle centered at $z_0$ with radius $R$. Since we have an inequality $|z - 1| \le 2$, it includes all points whose distance from $(1, 0)$ is less than or equal to $2$, which forms a solid disk.

Q: How do I simplify $(z + \bar{z}) + i(z - \bar{z}) \le 2$?

Use the fundamental definitions $z = x + iy$ and $\bar{z} = x - iy$. Then $z + \bar{z} = 2x$ and $z - \bar{z} = 2iy$. Substituting these gives $2x + i(2iy) = 2x - 2y$. Thus, $2x - 2y \le 2$ simplifies to $y \ge x - 1$.

Q: Why is the area a sector?

The line $y = x - 1$ passes through the center $(1, 0)$ of the circle. Any line passing through the center of a circle divides it into two semicircles. Since we are restricted to the upper half-plane ($y \ge 0$), we are looking at a specific portion of the disk bounded by the line and the x-axis, which is a circular sector.

Question

The area (in sq. units) of the region $S = \{z \in \mathbb{C} : |z - 1| \le 2; (z + \bar{z}) + i(z - \bar{z}) \le 2, 	ext{Im}(z) \ge 0\}$ is

Options:

(A) $\frac{7\pi}{3}$

(B) $\frac{7\pi}{4}$

(C) $\frac{17\pi}{8}$

(D) $\frac{3\pi}{2}$

Sigma Percentile · Accepted Answer

### The Geometry of Complex Numbers

Welcome, fellow explorer of the mathematical universe! Today, we are going to peel back the layers of a seemingly intimidating complex numbers problem.

Often, when we see $z \in \mathbb{C}$, our minds jump to algebraic manipulation. However, the true secret to mastering JEE Advanced problems is to see the geometry hidden beneath the algebra. Let us embark on this journey together.

### Phase 1: The Disk

Our first constraint is $|z - 1| \le 2$. In the complex plane, the modulus $|z - z_0|$ represents the distance between $z$ and $z_0$.

So, $|z - 1| \le 2$ is simply the set of all points $z$ that are at most $2$ units away from the point $(1, 0)$. If you visualize this, you are looking at a solid disk centered at $(1, 0)$ with a radius $R = 2$.

### Phase 2: The Algebraic Inequality

Next, we encounter the expression $(z + \bar{z}) + i(z - \bar{z}) \le 2$. It looks like a jumble of symbols, but let us apply our toolkit.

We know that for any complex number $z = x + iy$, its conjugate is $\bar{z} = x - iy$. Therefore, $z + \bar{z} = 2x$ and $z - \bar{z} = 2iy$.

Substituting these into our inequality, we get $2x + i(2iy) \le 2$. Since $i^2 = -1$, this simplifies elegantly to:

$$2x - 2y \le 2$$

Dividing by $2$, we find $x - y \le 1$, or $y \ge x - 1$. This is the equation of a line with a slope of $1$ passing through $(1, 0)$.

### Phase 3: The Upper Half-Plane

Finally, we have the constraint $	ext{Im}(z) \ge 0$, which simply means $y \ge 0$. This restricts our entire world to the upper half of the Cartesian plane.

We are now looking for the intersection of a disk, the region above the line $y = x - 1$, and the region above the x-axis.

### Phase 4: The Final Calculation

When you plot these, you will see that the line $y = x - 1$ passes exactly through the center of our disk $(1, 0)$. The line makes an angle of $\frac{\pi}{4}$ with the positive x-axis.

The region we are interested in is the sector of the circle bounded by this line and the negative x-axis. The angle of this sector is $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

The area of a circular sector is given by $A = \frac{1}{2} r^2 	heta$. Substituting $r = 2$ and $	heta = \frac{3\pi}{4}$, we get:

$$A = \frac{1}{2} (2)^2 \left( \frac{3\pi}{4} ight) = \frac{3\pi}{2}$$

And there it is! The complexity melts away, leaving us with a clean, elegant result of **$\frac{3\pi}{2}$** square units. Keep practicing, and remember: geometry is the key to unlocking the most difficult problems.

Animated Solution for Mathematics - Complex Numbers: The area (in sq. units) of the region $S = {z \in C : ∣ z - 1∣ \leq 2; (z + \overset{z}{ˉ}) + i (z - \overset{z}{ˉ}) \leq 2, Im (z) \geq 0}$ is

Select Answer:

Visualized Solution

Introduction to the Complex Plane

Constraint 1: The Modulus

Cartesian Equation of the Disk

Constraint 2: The Algebraic Inequality

Substituting Real and Imaginary Parts

Equation of the Half-Plane

Constraint 3: The Upper Half-Plane

Visualizing the Intersection

Analyzing the Boundary Line

Finding the Initial Angle

Calculating the Sector Angle

Setting Up the Area Formula

Final Area Calculation

The Sigma Insight: Geometrical Applications of Complex Numbers

The Geometry of Complex Numbers

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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Comprehension Passage

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area (in sq. units) of the region S={z∈C:∣z−1∣≤2;(z+zˉ)+i(z−zˉ)≤2,Im(z)≥0} is