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 2
Convert standard coordinate complex numbers into Euler form immediately when large powers are present.
Chapter Mix
Class 11 Maths: Complex Numbers
Q9jee_main_2026_24_january_morningLocus in Complex Plane
Let S = z in C : | (z - 6i)/(z - 2i) | = 1 and | (z - 8 + 2i)/(z + 2i) | = (3)/(5)$S = \left\{ z \in \mathbb{C} : \left| \frac{z - 6i}{z - 2i} \right| = 1 \text{ and } \left| \frac{z - 8 + 2i}{z + 2i} \right| = \frac{3}{5} \right\}$. Then Σz in S |z|²$\sum_{z \in S} |z|^2$ is equal to
A.398$398$
B.413$413$
C.423$423$
D.385$385$
Solution
Related Formula
|z - z₁| = |z - z₂| represents the perpendicular bisector of the segment joining z₁ and z₂$$|z - z_1| = |z - z_2| \text{ represents the perpendicular bisector of the segment joining } z_1 \text{ and } z_2$$|x+iy|² = x² + y²$$|x+iy|^2 = x^2 + y^2$$
Core Logic
First condition: |z - 6i| = |z - 2i|$|z - 6i| = |z - 2i|$.
This means z$z$ lies on the perpendicular bisector of (0,6)$(0,6)$ and (0,2)$(0,2)$.
Let z = x + iy$z = x + iy$. Thus, y = 4$y = 4$.
Step 1: Circle Equation
Second condition: 5|z - 8 + 2i| = 3|z + 2i|$5|z - 8 + 2i| = 3|z + 2i|$.
Substitute y = 4$y = 4$ into z$z$: z = x + 4i$z = x + 4i$.
Whenever an absolute value ratio equals 1, immediately map it to a line (perpendicular bisector) and substitute its constraint directly into the second curve equation to reduce dimensionality.
Chapter Mix
Class 11 Maths: Complex Numbers and Quadratic Equations
Q23jee_main_2026_24_january_eveningProperties of Moduli
Let z = (1 + i)(1 + 2i)(1 + 3i) (1 + ni)$z = (1 + \mathrm{i})(1 + 2\mathrm{i})(1 + 3\mathrm{i}) \dots (1 + \mathrm{ni})$, where i = √(-1)$\mathrm{i} = \sqrt{-1}$. If |z|² = 44200$|z|^2 = 44200$, then n$n$ is equal to
The calculated product for n=5$n=5$ exactly matches the prime factorization of 44200$44200$.
Therefore, n = 5$n = 5$.
Pattern Recognition
Modulus is multiplicative. In problems featuring chains of complex multiplications set equal to a huge real magnitude, instantly switch to magnitudes and map to integer factorization.
Chapter Mix
Class 11 Maths: Complex Numbers
Q5jee_main_2026_28_january_morningGeometry of Complex Numbers
Let z$z$ be a complex number such that |z - 6| = 5$|z - 6| = 5$ and |z + 2 - 6i| = 5$|z + 2 - 6i| = 5$. Then the value of z³ + 3z² - 15z + 141$z^{3} + 3z^{2} - 15z + 141$ is equal to
A.42$42$
B.37$37$
C.50$50$
D.61$61$
Solution
Core Logic
Geometry of Complex Numbers
The given equations represent two circles in the complex plane:
Circle 1: Center C₁(6, 0)$C_1(6, 0)$, radius r₁ = 5$r_1 = 5$
Circle 2: Center C₂(-2, 6)$C_2(-2, 6)$, radius r₂ = 5$r_2 = 5$
When given two complex distance modulus equations |z-z₁|=r₁$|z-z_1|=r_1$ and |z-z₂|=r₂$|z-z_2|=r_2$, always check the distance between centers |z₁ - z₂|$|z_1 - z_2|$. If it exactly equals r₁ + r₂$r_1 + r_2$, the single unique solution is the section formula midpoint.
Chapter Mix
Class 11 Mathematics: Complex Numbers and Quadratic Equations
Q10jee_main_2026_28_january_morningNature of Roots
If α, β$\alpha, \beta$, where α < β$\alpha < \beta$, are the roots of the equation λ x² - (λ + 3)x + 3 = 0$\lambda x^2 - (\lambda + 3)x + 3 = 0$ such that (1)/(α) - (1)/(β) = (1)/(3)$\frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3}$, then the sum of all possible values of λ$\lambda$ is:
A.6$6$
B.2$2$
C.4$4$
D.8$8$
Solution
Related Formula
For a quadratic equation ax² + bx + c = 0$ax^2 + bx + c = 0$:
Sum of roots: α + β = -(b)/(a)$\alpha + \beta = -\frac{b}{a}$
Product of roots: αβ = (c)/(a)$\alpha\beta = \frac{c}{a}$
Core Logic
From the given equation λ x² - (λ + 3)x + 3 = 0$\lambda x^2 - (\lambda + 3)x + 3 = 0$:
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
