Let A = z in C : |z - 2| ≤ 4$A = \{z \in \mathbb{C} : |z - 2| \le 4\}$ and B = z in C : |z - 2| + |z + 2| = 5$B = \{z \in \mathbb{C} : |z - 2| + |z + 2| = 5\}$. Then the max |z₁ - z₂| : z₁ in A and z₂ in B$\{|z_{1} - z_{2}| : z_{1} \in A \text{ and } z_{2} \in B\}$ is

Related Formula

For a quadratic equation Ax² + Bx + C = 0$Ax^2 + Bx + C = 0$, roots can be obtained via the quadratic formula:

Core Logic

Given quadratic equation:

Using the quadratic formula where A=1, B=-(3-2i), C=-(2i-2)$A=1, B=-(3-2i), C=-(2i-2)$:

Step 1: Simplify the Discriminant

We need to find √(-3-4i)$\sqrt{-3-4i}$. Let it be written as a perfect square:

Thus, √(D) = ±(1-2i)$\sqrt{D} = \pm(1-2i)$.

Step 2: Find the Roots

Boxedx = ((3-2i) ± (1-2i))/(2)$Boxed{x = \frac{(3-2i) \pm (1-2i)}{2}}$

Case 1 (+ sign):

Case 2 (- sign):

Let the roots be α + iβ = 2 - 2i α=2, β=-2$\alpha + i\beta = 2 - 2i \implies \alpha=2, \beta=-2$ and γ + iδ = 1 + 0i γ=1, δ=0$\gamma + i\delta = 1 + 0i \implies \gamma=1, \delta=0$

Step 3: Evaluate Target Expression

Pattern Recognition

Always try to express the complex number under the square root in the form (a + bi)²$(a + bi)^2$ by matching the imaginary part 2ab = -4i ab = -2$2ab = -4i \implies ab = -2$, and a² - b² = -3$a^2 - b^2 = -3$. This avoids long calculations.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations

Related Formula

Core Logic

The expressions define two circular disc fields in the complex plane: Circle 1: Center C₁(8, 2)$C_1(8, 2)$, radius r₁ = 1$r_1 = 1$ Circle 2: Center C₂(2, -6)$C_2(2, -6)$, radius r₂ = 2$r_2 = 2$

Step 1: Calculate Center Distance

Using coordinate distance formulation:

Step 2: Find Minimum Separation

Pattern Recognition

Always interpret modulus circle properties geometrically rather than algebraically. Disconnecting complex plane variables into simple 2D analytical geometry centers avoids calculation mistakes entirely.

Chapter Mix

Class 11 Mathematics: Complex Numbers Class 11 Mathematics: Coordinate Geometry

Related Formula

For a complex coordinate transformation, substituting z = x + iy$z = x + iy$ and its conjugate z = x - iy$\overline{z} = x - iy$ maps a complex condition directly into rectangular Cartesian coordinates.

Core Logic

Let's translate the complex set properties into Cartesian geometry:

• Set P$P$: |z - (-2 + 3i)| ≤ 1$|z - (-2 + 3i)| \le 1 \implies$ Interior and boundary of a circle with center C(-2, 3)$C(-2, 3)$ and radius r = 1$r = 1$.
• Set Q$Q$: (x+iy)(1+i) + (x-iy)(1-i) ≤ -8 (x - y + ix + iy) + (x - y - ix - iy) ≤ -8$(x+iy)(1+i) + (x-iy)(1-i) \le -8 \implies (x - y + ix + iy) + (x - y - ix - iy) \le -8$

This defines a half-plane below or to the left of the boundary line L₂: x - y + 4 = 0$L_2: x - y + 4 = 0$.

Step 1: Identify Extreme Points for Distance from Point A

We want to find points in the region P Q$P \cap Q$ that minimize and maximize the distance to the external point A(3, -2)$A(3, -2)$, corresponding to |z - (3 - 2i)|$|z - (3 - 2i)|$.

The line L₁$L_1$ connecting center C(-2, 3)$C(-2, 3)$ and point A(3, -2)$A(3, -2)$ has slope:

Equation of line L₁$L_1$: y - 3 = -1(x + 2) x + y - 1 = 0$y - 3 = -1(x + 2) \implies x + y - 1 = 0$.

Step 2: Calculate Coordinates for z₁$z_1$ and z₂$z_2$

• Minimum Distance Point (z₂$z_2$): By geometric observation, the minimum distance from A$A$ to the bounded region is the intersection point of lines L₁$L_1$ and L₂$L_2$:

So, z₂ = (-(3)/(2), (5)/(2))$z_2 = \left(-\frac{3}{2}, \frac{5}{2}\right)$.

• Maximum Distance Point (z₁$z_1$): The maximum distance is at the far boundary edge of the circle along line L₁$L_1$. The vector path from C(-2, 3)$C(-2, 3)$ opposite to A$A$ has unit direction (- 1√(2), 1√(2))$\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$:

Step 3: Evaluate Magnitudes and Sum Parameters

Calculate the squares of the moduli:

Now compute the total requested term:

Matching with α + β√(2)$\alpha + \beta\sqrt{2}$ gives α = 31$\alpha = 31$ and β = 5$\beta = 5$. Thus:

Pattern Recognition

Sees: Locus intersection involving geometric complex inequalities. Shortcut: Translating complex equations into standard 2D graphs reveals the geometry instantly, mapping extreme distances to line intersections or boundary nodes cleanly.

Chapter Mix

Class 11 Complex Numbers: Geometry Class 11 Coordinate Geometry: Straight Lines

Related Formula

For a complex number z = x + iy$z = x + iy$:

• |z| = √(x² + y²)$|z| = \sqrt{x^2 + y^2}$
• z + z = 2x$z + \overline{z} = 2x$
• z - z = 2iy$z - \overline{z} = 2iy$

Core Logic

Let z = x + iy$z = x + iy$. The first condition |z - 1| = 1$|z - 1| = 1$ describes a circle:

The second condition gives a straight line:

The set S$S$ contains the intersection points of this circle and line.

Step 1: Solve for Intersection Points

From (2), y = √(2) - (√(2)-1)x$y = \sqrt{2} - (\sqrt{2}-1)x$. Substitute this into (1):

Solving this quadratic equation gives two values of x$x$:

• Case A: x = 1 y = √(2) - (√(2)-1)(1) = 1$x = 1 \implies y = \sqrt{2} - (\sqrt{2}-1)(1) = 1$.

So, one point is zA = 1 + i$z_A = 1 + i$. Its magnitude is |zA| = √(1² + 1²) = √(2)$|z_A| = \sqrt{1^2 + 1^2} = \sqrt{2}$.

• Case B: x = 12-√(2) = 2+√(2)2 = 1 + 1√(2)$x = \frac{1}{2-\sqrt{2}} = \frac{2+\sqrt{2}}{2} = 1 + \frac{1}{\sqrt{2}}$.

Then y = √(2) - (√(2)-1)(1+ 1√(2)) = √(2) - (√(2) + 1 - 1 - 1√(2)) = 1√(2)$y = \sqrt{2} - (\sqrt{2}-1)\left(1+\frac{1}{\sqrt{2}}\right) = \sqrt{2} - \left(\sqrt{2} + 1 - 1 - \frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}$. So, the other point is zB = (1 + 1√(2)) + i√(2)$z_B = \left(1 + \frac{1}{\sqrt{2}}\right) + \frac{i}{\sqrt{2}}$. Its magnitude is |zB| = (1+ 1√(2))² + ( 1√(2))² = 1 + √(2) + (1)/(2) + (1)/(2) = 2 + √(2)$|z_B| = \sqrt{\left(1+\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2} = \sqrt{1 + \sqrt{2} + \frac{1}{2} + \frac{1}{2}} = \sqrt{2 + \sqrt{2}}$.

Step 2: Evaluate Min and Max Magnitude Expressions

Comparing magnitudes, |zB| > |zA|$|z_B| > |z_A|$, so:

Now, calculate |√(2)z₁-z₂|²$|\sqrt{2}z_{1}-z_{2}|^{2}$:

Pattern Recognition

Sees: Geometric constraint mapping to a circle and line intersection in the complex plane. Shortcut: Rationalizing terms like 12-√(2)$\frac{1}{2-\sqrt{2}}$ immediately into standard form 1+ 1√(2)$1+\frac{1}{\sqrt{2}}$ saves you from handling layered fraction algebra down the line.

Chapter Mix

Class 11 Complex Numbers: Geometry Class 10 Coordinate Geometry: Lines and Circles

Related Formula

For z = x + iy$z = x + iy$, modulus r = √(x² + y²)$r = \sqrt{x^2 + y^2}$ and argument θ$\theta$ depends on the quadrant location.

Core Logic

Given z = 2 - i(2 (5π)/(8))$z = 2 - i\left(2 \tan \frac{5\pi}{8}\right)$. Note that (5π)/(8)$\frac{5\pi}{8}$ lies in the second quadrant, so (5π)/(8) < 0$\tan \frac{5\pi}{8} < 0$. Let's write r$r$:

Since (5π)/(8)$\sec \frac{5\pi}{8}$ is negative:

Step 1: Finding the Amplitude

Since x = 2 > 0$x = 2 > 0$ and y = -2 (5π)/(8) > 0$y = -2 \tan \frac{5\pi}{8} > 0$, the complex number lies in the first quadrant.

Pattern Recognition

Always absolute-value trigonometric terms coming out of square roots (e.g., √( ² φ) = | φ|$\sqrt{\sec^2 \phi} = |\sec \phi|$). Knowing the precise quadrant prevents incorrect signs.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations