Let P = {z ∈ C : |z + 2 - 3i| ≤ 1} and Q = {z ∈ C : z(1 + i) + z̄(1 - i) ≤ -8}. Let in P ∩ Q, |z - 3

Q: Wait, why is region $Q$ a half-plane? I thought $z(1+i) + \bar{z}(1-i)$ was just a complex number.

Great question! When you substitute $z = x + iy$ and $\bar{z} = x - iy$, the expression becomes $(x+iy)(1+i) + (x-iy)(1-i)$. Expanding this gives $(x - y + i(x + y)) + (x - y - i(x + y))$. Notice how the imaginary parts $i(x+y)$ and $-i(x+y)$ cancel out perfectly! You are left with $2x - 2y \le -8$, which simplifies to $y \ge x + 4$. This is the equation of a line, and the inequality defines the region on one side of that line—a half-plane.

Q: How do we know $z_2$ is the intersection of $L_1$ and $L_2$?

This is the beauty of the geometry here! We found that the line $L_1$ (connecting the center $C$ to the external point $A$) has a slope of $-1$. The boundary line $L_2$ has a slope of $1$. Since the product of their slopes is $-1$, $L_1$ is perpendicular to $L_2$. The shortest distance from a point to a line is always along the perpendicular. Since $L_1$ is that perpendicular path, the point on $L_2$ closest to $A$ must be where $L_1$ and $L_2$ intersect.

Q: Why is $z_1$ on the boundary of the circle?

Think about the geometry of a disk. If you want to find the point in a disk furthest from an external point $A$, you should move from $A$ through the center $C$ until you hit the far edge of the disk. That path is the line $L_1$. The furthest point $z_1$ must be on the boundary of the circle because any point inside the disk would be closer to $A$ than a point on the boundary along the same radial line.

Question

Let $P = \{z \in \mathbb{C} : |z + 2 - 3i| \le 1\}$ and $Q = \{z \in \mathbb{C} : z(1 + i) + \bar{z}(1 - i) \le -8\}$. Let in $P \cap Q$, $|z - 3 + 2i|$ be maximum and minimum at $z_1$ and $z_2$ respectively. If $|z_1|^2 + 2|z_2|^2 = \alpha + \beta\sqrt{2}$, where $\alpha, \beta$ are integers, then $\alpha + \beta$ equals

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of Complex Numbers

Welcome, fellow traveler of the complex plane! Today, we are not just solving an equation; we are mapping a landscape. We have two regions, $P$ and $Q$, and we are hunting for the extreme points of distance within their intersection.

Region $P$ is defined by $|z + 2 - 3i| \le 1$. In the language of geometry, this is a closed disk with center $C(-2, 3)$ and radius $R = 1$. Think of this as a solid, circular coin resting on the complex plane.

Now, let us decode region $Q$, defined by $z(1 + i) + \bar{z}(1 - i) \le -8$. Substituting $z = x + iy$, the expression becomes:

$$(x+iy)(1+i) + (x-iy)(1-i) \le -8$$

When we expand this, the imaginary parts $i(x+y)$ and $-i(x+y)$ vanish, leaving us with $2x - 2y \le -8$, or simply $y \ge x + 4$. This is a half-plane bounded by the line $L_2: y = x + 4$.

Our feasible region, $P \cap Q$, is the portion of the disk that lies above this line. To see how they interact, we calculate the distance from the center $C(-2, 3)$ to the line $x - y + 4 = 0$:

$$d = \frac{|-2 - 3 + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{1}{\sqrt{2}} \approx 0.707$$

Since $0.707 < 1$, the line $L_2$ slices through our disk, creating a circular segment.

### The Quest for Extremes: Symmetry and Perpendicularity

Our target is to find the maximum and minimum distances from a fixed point $A(3, -2)$ to any point $z$ in our shaded region. First, we draw the line of symmetry $L_1$ passing through $C(-2, 3)$ and $A(3, -2)$.

The slope of $L_1$ is $m = \frac{-2 - 3}{3 - (-2)} = -1$. The equation of this line is $y - 3 = -1(x + 2)$, which simplifies to $x + y - 1 = 0$.

Here is the magic: the slope of $L_2$ is $1$, and the slope of $L_1$ is $-1$. Their product is $-1$, meaning $L_1 \perp L_2$. This perpendicularity is our golden key.

For the minimum distance point $z_2$, we look for the intersection of $L_1$ and $L_2$. Solving $x + y = 1$ and $x - y = -4$ simultaneously, we add the equations to get $2x = -3$, so $x = -\frac{3}{2}$.

Substituting back, $y = 1 - (-\frac{3}{2}) = \frac{5}{2}$. Thus, $z_2 = (-\frac{3}{2}, \frac{5}{2})$.

For the maximum distance point $z_1$, we move from the center $C$ along $L_1$ in the direction away from $A$. The vector $\vec{CA} = (5, -5)$, so the unit vector pointing away from $A$ is $(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$. Adding this to $C(-2, 3)$, we get $z_1 = (-2 - \frac{1}{\sqrt{2}}, 3 + \frac{1}{\sqrt{2}})$.

### The Final Calculation: Bringing it Home

We have our points. Now, we calculate their squared magnitudes. For $z_1$:

$$|z_1|^2 = \left(-2 - \frac{1}{\sqrt{2}}\right)^2 + \left(3 + \frac{1}{\sqrt{2}}\right)^2 = \left(4 + \frac{1}{2} + 2\sqrt{2}\right) + \left(9 + \frac{1}{2} + 3\sqrt{2}\right) = 14 + 5\sqrt{2}$$

For $z_2$:

$$|z_2|^2 = \left(-\frac{3}{2}\right)^2 + \left(\frac{5}{2}\right)^2 = \frac{9}{4} + \frac{25}{4} = \frac{34}{4} = \frac{17}{2}$$

Finally, we compute $|z_1|^2 + 2|z_2|^2$:

$$|z_1|^2 + 2|z_2|^2 = (14 + 5\sqrt{2}) + 2\left(\frac{17}{2}\right) = 14 + 5\sqrt{2} + 17 = 31 + 5\sqrt{2}$$

Comparing this to $\alpha + \beta\sqrt{2}$, we find $\alpha = 31$ and $\beta = 5$. The **final answer** is the sum $\alpha + \beta = 36$.

Enter Numerical Value:

Visualized Solution

Region $P$

Region $Q$

$P \cap Q$

Target Point $A$

Line of Symmetry $L_{1}$

Maximum Distance Point $z_{1}$

Coordinates of $z_{1}$

Minimum Distance Point $z_{2}$

Coordinates of $z_{2}$

Calculate $∣ z_{1} ∣^{2}$

Calculate $∣ z_{2} ∣^{2}$

Final Answer

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Geometry of Complex Numbers

The Quest for Extremes

The Final Calculation

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Comprehension Passage

Area of $S =$

$min_{z \in S} ∣1 - 3 i - z ∣ =$

Enter Numerical Value:

Visualized Solution

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Region .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

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