Let C be the set of all complex numbers. Let S₁ = {z ∈ C : |z - 2| ≤ 1} and S₂ = {z ∈ C : z(1 + i) +

Q: Wait, why does the line $y = x - 2$ pass through the center of the circle?

Great observation! The center of the circle $S_1$ is $(2, 0)$. If you plug $x = 2$ and $y = 0$ into the equation $y = x - 2$, you get $0 = 2 - 2$, which is $0 = 0$. This confirms the line passes exactly through the center, effectively cutting the disk into two equal semi-disks.

Q: How do I know which side of the line $y = x - 2$ to shade?

Test a point! Take the center $(2, 0)$ and plug it into the inequality $x - y \ge 2$. You get $2 - 0 \ge 2$, which is $2 \ge 2$. This is true, so the center is included. However, since the line passes through the center, you should test another point like $(3, 0)$. $3 - 0 \ge 2$ is $3 \ge 2$, which is also true. This means the region $x - y \ge 2$ is the half-plane containing the point $(3, 0)$.

Q: Why did we use parameterization instead of just the distance formula?

You could use the distance formula, but parameterization turns a geometric problem into a simple trigonometric optimization. By expressing $z$ in terms of $\theta$, we reduce the problem to finding the extremum of a single-variable function, which is much less prone to algebraic errors than working with Cartesian coordinates directly.

Question

Let $\mathbb{C}$ be the set of all complex numbers. Let $S_1 = \{z \in \mathbb{C} : |z - 2| \le 1\}$ and $S_2 = \{z \in \mathbb{C} : z(1 + i) + \bar{z} (1 - i) \ge 4\}$. Then, the maximum value of $|z - \frac{5}{2}|^2$ for $z \in S_1 \cap S_2$ is equal to :

Options:

(A) $\frac{3+2\sqrt{2}}{4}$

(B) $\frac{5+2\sqrt{2}}{2}$

(C) $\frac{3+2\sqrt{2}}{2}$

(D) $\frac{5+2\sqrt{2}}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The region $S_1$ is defined by $|z - 2| \le 1$. This represents a solid disk in the complex plane centered at $(2, 0)$ with a radius of $1$.

The region $S_2$ is defined by the inequality $z(1 + i) + \bar{z} (1 - i) \ge 4$. By substituting $z = x + iy$ and $\bar{z} = x - iy$, we expand the expression:

$$(x + iy)(1 + i) + (x - iy)(1 - i) = (x + ix + iy - y) + (x - ix - iy - y) = 2x - 2y$$

Thus, the inequality simplifies to $2x - 2y \ge 4$, or:

$$x - y \ge 2$$

This defines a half-plane bounded by the line $y = x - 2$.

### The Intersection

We seek the intersection $S_1 \cap S_2$. Note that the line $y = x - 2$ passes through the center of the disk, $(2, 0)$, since $0 = 2 - 2$.

Because the line passes through the center, it bisects the disk perfectly. The inequality $x - y \ge 2$ selects the half-disk region. Any point $z$ in our set must lie within this semi-disk.

### The Objective Function

We aim to maximize $|z - \frac{5}{2}|^2$. Geometrically, this represents the square of the distance from any point $z$ in the semi-disk to the fixed point $P(\frac{5}{2}, 0)$.

Since $P$ lies on the real axis just to the right of the disk's center, the maximum distance must occur on the curved boundary of the semi-disk.

### The Trigonometric Dance

We parameterize the boundary of the circle as $z = 2 + \cos	heta + i\sin	heta$, where $x = 2 + \cos	heta$ and $y = \sin	heta$. The constraint $x - y \ge 2$ becomes:

$$(2 + \cos	heta) - \sin	heta \ge 2 \implies \cos	heta \ge \sin	heta$$

On the unit circle, $\cos	heta = \sin	heta$ at $	heta = \frac{\pi}{4}$ and $	heta = -\frac{3\pi}{4}$. The condition $\cos	heta \ge \sin	heta$ holds for $	heta \in [-\frac{3\pi}{4}, \frac{\pi}{4}]$.

Now, we evaluate the objective function:

$$|z - \frac{5}{2}|^2 = |(2 + \cos	heta + i\sin	heta) - \frac{5}{2}|^2 = |(\cos	heta - \frac{1}{2}) + i\sin	heta|^2$$

Expanding this expression, we obtain:

$$(\cos	heta - \frac{1}{2})^2 + \sin^2	heta = \cos^2	heta - \cos	heta + \frac{1}{4} + \sin^2	heta$$

Using the identity $\cos^2	heta + \sin^2	heta = 1$, the expression simplifies to:

$$\frac{5}{4} - \cos	heta$$

### The Final Climax

To maximize $\frac{5}{4} - \cos	heta$, we must minimize $\cos	heta$ within the interval $	heta \in [-\frac{3\pi}{4}, \frac{\pi}{4}]$. The minimum value of $\cos	heta$ in this range occurs at the boundary $	heta = -\frac{3\pi}{4}$.

At this point, $\cos(-\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}}$. Substituting this into our simplified objective function:

$$\frac{5}{4} - (-\frac{1}{\sqrt{2}}) = \frac{5}{4} + \frac{1}{\sqrt{2}} = \frac{5 + 2\sqrt{2}}{4}$$

The maximum value is **$\frac{5 + 2\sqrt{2}}{4}$**.

Select Answer:

Visualized Solution

Visualizing the Constraints

Analyzing Set $S_{1}$

Simplifying Set $S_{2}$

The Cartesian Form of $S_{2}$

Identifying the Intersection Region

Defining the Objective Function

Parameterizing the Boundary

Applying the Constraint to $θ$

Simplifying the Distance Formula

Optimizing the Function

The Final Calculation

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Intersection

The Objective Function

The Trigonometric Dance

The Final Climax

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