Let P=zin C:|z+2-3i|≤1$P=\{z\in\mathbb{C}:|z+2-3i|\le1\}$ and Q=zin C:z(1+i)+ z(1-i)≤-8$Q=\{z\in\mathbb{C}:z(1+i)+\overline{z}(1-i)\le-8\}$. Let in P Q, |z-3+2i|$P \cap Q, |z-3+2i|$ be maximum and minimum at z₁$z_{1}$ and z₂$z_{2}$ respectively. If |z₁|²+2|z₂|²=α+β√(2)$|z_{1}|^{2}+2|z_{2}|^{2}=\alpha+\beta\sqrt{2}$, where α, β$\alpha, \beta$ are integers, then α+β$\alpha+\beta$ equals
Numerical Answer Type:
Enter a numerical valueAnswer: 36 to 36+4 marks
Solution & Explanation
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$(x+iy)(1+i) + (x-iy)(1-i) \le -8 \implies (x - y + ix + iy) + (x - y - ix - iy) \le -8$
2(x - y) ≤ -8 x - y + 4 ≤ 0$$2(x - y) \le -8 \implies x - y + 4 \le 0$$
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$y - 3 = -1(x + 2) \implies x + y - 1 = 0$.
The graphic details the intersection region bounded by the circular locus and the line inequality, showing points z1 and z2 relative to external reference point P.
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$L_2$:
x - y + 4 = 0 and x + y - 1 = 0 2x + 3 = 0 x = -(3)/(2), y = (5)/(2)$$x - y + 4 = 0 \quad \text{and} \quad x + y - 1 = 0 \implies 2x + 3 = 0 \implies x = -\frac{3}{2}, \, y = \frac{5}{2}$$
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))$\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$:
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
Keywords:#complex locus circle#half plane inequality intersection#maximum minimum distance geometry
More Complex Numbers Previous-Year Questions — Page 9
Q26jee_main_2024_31_jan_morningProperties of Modulus and Argument
If α$\alpha$ denotes the number of solutions of |1 - i|^x = 2^x$|1 - i|^x = 2^x$ and β = ((|z|)/( (z)))$\beta = \left(\frac{|z|}{\arg(z)}\right)$, where z = (π)/(4) (1 + i)⁴ ( 1 - √(π) i√(π) + i + √(π) - i1 + √(π) i), i = √(-1)$z = \frac{\pi}{4} (1 + i)^4 \left(\frac{1 - \sqrt{\pi} i}{\sqrt{\pi} + i} + \frac{\sqrt{\pi} - i}{1 + \sqrt{\pi} i}\right), i = \sqrt{-1}$, then the distance of the point (α, β)$(\alpha, \beta)$ from the line 4x - 3y = 7$4x - 3y = 7$ is
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