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

Related Formula

Properties of cube roots of unity: ω³ = 1$\omega^3 = 1$ and 1 + ω + ω² = 0$1 + \omega + \omega^2 = 0$.

Core Logic

Let α = ω$\alpha = \omega$. Then (1)/(x) = (1)/(ω) = ω²$\frac{1}{x} = \frac{1}{\omega} = \omega^2$. The series is Σk=1²⁵ (ω^k + ω2k)⁴$\sum_{k=1}^{25} (\omega^k + \omega^{2k})^4$. Evaluate the term Tk = (ω^k + ω2k)⁴$T_k = (\omega^k + \omega^{2k})^4$ based on the modulo of k$k$ with 3.

Step 1: Cyclic Evaluation

Case 1: k = 3m$k = 3m$ (multiples of 3)

There are 8$8$ such multiples up to 25 (3, 6, 9, , 24$3, 6, 9, \dots, 24$).

Case 2: k ≠ 3m$k \neq 3m$ (non-multiples of 3) For k = 1, 2, 4, 5,$k = 1, 2, 4, 5, \dots$ ω^k + ω2k$\omega^k + \omega^{2k}$ will always be ω + ω²$\omega + \omega^2$ or ω² + ω$\omega^2 + \omega$. Since 1 + ω + ω² = 0$1 + \omega + \omega^2 = 0$, we have ω + ω² = -1$\omega + \omega^2 = -1$.

There are 25 - 8 = 17$25 - 8 = 17$ such non-multiples up to 25.

Step 2: Total Sum

Pattern Recognition

Powers of x+1/x$x+1/x$ when x$x$ solves x² ± x + 1 = 0$x^2 \pm x + 1 = 0$ perfectly orbit around periods of 3 or 6. Isolate the 3m$3m$ resonant beats (which hit pure scalars like 1+1=2$1+1=2$) versus the out-of-phase beats (which collapse to -1$-1$ or 1$1$ via basic ω$\omega$ identities).

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations

Related Formula

Core Logic

Step 1: Locate the Point z

From geometry, hypotenuse c = 5$c = 5$, opposite (radius) r = 3$r = 3$. The adjacent side (length of tangent) is √(5² - 3²) = 4$\sqrt{5^2 - 3^2} = 4$. The angle of tangency θ$\theta$ satisfies θ = (3)/(5)$\sin\theta = \frac{3}{5}$ and θ = (4)/(5)$\cos\theta = \frac{4}{5}$. The point P(z)$P(z)$ lies on the circle and the tangent ray:

Step 2: Evaluate the Target Expression

We need 34|(5z - 12)/(5iz + 16)|²$34\left|\frac{5z - 12}{5iz + 16}\right|^2$. Substitute 5z = 16 + 12i$5z = 16 + 12i$: Numerator: 5z - 12 = 16 + 12i - 12 = 4 + 12i$5z - 12 = 16 + 12i - 12 = 4 + 12i$ Denominator: 5iz + 16 = i(16 + 12i) + 16 = 16i - 12 + 16 = 4 + 16i$5iz + 16 = i(16 + 12i) + 16 = 16i - 12 + 16 = 4 + 16i$ Expression:

Pattern Recognition

For maximum (z)$\arg(z)$ on |z-c| = r$|z-c| = r$ where c$c$ is real, z$z$ coordinates are given by geometric projection: z = √(c²-r²)( θ + i θ)$z = \sqrt{c^2-r^2}(\cos\theta + i\sin\theta)$ where θ = r/c$\sin\theta = r/c$.

Chapter Mix

Class 11 Maths: Complex Numbers

Related Formula

Core Logic

Substitute ω$\omega$ and ω²$\omega^2$ into the terms:

Let T₁ = 7 - 7ω + 9ω²$T_1 = 7 - 7\omega + 9\omega^2$ Let T₂ = 9 + 7ω - 7ω²$T_2 = 9 + 7\omega - 7\omega^2$ Let T₃ = -7 + 9ω + 7ω²$T_3 = -7 + 9\omega + 7\omega^2$ Let T₄ = 14 + 7ω + 7ω²$T_4 = 14 + 7\omega + 7\omega^2$

Step 1: Simplify Each Term Using Properties of Omega

Notice relationships between the terms by factoring powers of ω$\omega$:

Observe that:

So, T₁²⁰ + T₂²⁰ + T₃²⁰ = (ω² T₂)²⁰ + T₂²⁰ + (ω T₂)²⁰$T_1^{20} + T_2^{20} + T_3^{20} = (\omega^2 T_2)^{20} + T_2^{20} + (\omega T_2)^{20}$

Since ω³ = 1$\omega^3 = 1$, we have ω⁴⁰ = ω$\omega^{40} = \omega$ and ω²⁰ = ω²$\omega^{20} = \omega^2$.

Thus, T₂²⁰ (1 + ω + ω²) = T₂²⁰ (0) = 0$T_2^{20} (1 + \omega + \omega^2) = T_2^{20} (0) = 0$.

Step 2: Evaluating the Non-Zero Term

The only remaining term is T₄²⁰$T_4^{20}$:

Since ω + ω² = -1$\omega + \omega^2 = -1$,

So the entire expression sum is 0 + 7²⁰ = 7²⁰$0 + 7^{20} = 7^{20}$.

Step 3: Finding m

Given that the sum equals m¹⁰$m^{10}$:

Hence, m = 49$m = 49$.

Pattern Recognition

Complex expressions with ω$\omega$ raised to large powers almost always exploit rotational symmetry. If coefficients are permuted circularly (A, B, C) → (C, A, B) → (B, C, A)$(A, B, C) \to (C, A, B) \to (B, C, A)$, multiplying by ω$\omega$ maps them to one another, meaning their sum of 3n$3n$ or symmetric powers identically nullifies.

Chapter Mix

Class 11 Maths: Complex Numbers

Related Formula

For z = x + iy$z = x + iy$, z = x - iy$\overline{z} = x - iy$ and |z|² = x² + y²$|z|^2 = x^2 + y^2$.

Core Logic

Substitute z = x + iy$z = x + iy$ into 4z² + z = 0$4z^2 + \overline{z} = 0$:

Separating real and imaginary parts:

Case 1: y = 0 4x² + x = 0 x = 0$y = 0 \implies 4x^2 + x = 0 \implies x = 0$ or x = -1/4$x = -1/4$. Solutions: z₁ = 0 |z₁|² = 0$z_1 = 0 \implies |z_1|^2 = 0$, z₂ = -1/4 |z₂|² = 1/16$z_2 = -1/4 \implies |z_2|^2 = 1/16$.

Case 2: x = 1/8 4(1/64) - 4y² + 1/8 = 0 4y² = 3/16 y = ± √(3)8$x = 1/8 \implies 4(1/64) - 4y^2 + 1/8 = 0 \implies 4y^2 = 3/16 \implies y = \pm \frac{\sqrt{3}}{8}$. Solutions: z₃, z₄ = (1)/(8) ± i √(3)8 |z₃|² = |z₄|² = (1)/(64) + (3)/(64) = (1)/(16)$z_3, z_4 = \frac{1}{8} \pm i \frac{\sqrt{3}}{8} \implies |z_3|^2 = |z_4|^2 = \frac{1}{64} + \frac{3}{64} = \frac{1}{16}$.

Step 1: Sum of Modulus Squared

Pattern Recognition

Separate complex equation into real and imaginary components to systematically find all roots.

Chapter Mix

Class 11 Maths: Complex Numbers

Core Logic

Simplify the condition for set S$S$:

This represents an annular region bounded by two concentric circles centered at C = (3)/(2) + (3)/(2)i$C = \frac{3}{2} + \frac{3}{2}i$ with radii r₁ = (3)/(2)$r_1 = \frac{3}{2}$ and r₂ = (7)/(2)$r_2 = \frac{7}{2}$.

Step 1: Identify the Target Point

We need to minimize | z - ( -(5)/(2) - (3)/(2)i ) |$\left| z - \left( -\frac{5}{2} - \frac{3}{2}i \right) \right|$. Let P$P$ be the point -(5)/(2) - (3)/(2)i$-\frac{5}{2} - \frac{3}{2}i$. The expression represents the distance from point P$P$ to a point z$z$ in the set S$S$.

Step 2: Distance from Center to P

Calculate the distance PC$PC$ between the center of the circles C((3)/(2), (3)/(2))$C\left(\frac{3}{2}, \frac{3}{2}\right)$ and the point P(-(5)/(2), -(3)/(2))$P\left(-\frac{5}{2}, -\frac{3}{2}\right)$:

Since 5 > (7)/(2)$5 > \frac{7}{2}$, the point P$P$ lies outside the outer circle.

Step 3: Minimum Distance Calculation

The shortest distance from an external point to an annular region is the distance to the outer boundary along the line connecting the point to the center.

Pattern Recognition

Transforming |az - b|$|az - b|$ by factoring out a$a$ immediately reveals the true geometric center. Shortest distance to any ring/circle from an external point is always collinear with the center: d - Router$d - R_{outer}$.

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations