Let α = -1 + i√(3)2$\alpha = \frac{-1 + \mathrm{i}\sqrt{3}}{2}$ and β = -1 - i√(3)2,i = √(-1)$\beta = \frac{-1 - \mathrm{i}\sqrt{3}}{2},\mathrm{i} = \sqrt{-1}$. If (7 - 7α +9β)²⁰ + (9 + 7α -7β)²⁰ + (-7 + 9α +7β)²⁰ + (14 + 7α +7β)²⁰ = m¹⁰$(7 - 7\alpha +9\beta)^{20} + (9 + 7\alpha -7\beta)^{20} + (-7 + 9\alpha +7\beta)^{20} + (14 + 7\alpha +7\beta)^{20} = \mathrm{m}^{10}$, then m is ____.

Core Logic

Let's expand the terms by grouping real and imaginary parts explicitly:

Notice that if we let u = 8 θ + 7 θ$u = 8\sin\theta + 7\cos\theta$ and v = θ + 4 θ$v = \sin\theta + 4\cos\theta$, then:

Step 1: Calculating the Product

Multiplying ω₁$\omega_1$ and \omega_2:

Since the product is given as α + iβ$\alpha + i\beta$:

α = 0$\alpha = 0$

Step 2: Simplifying the expression for alpha + beta

Expanding the terms for β$\beta$:

Using the identity ²θ + ²θ = 1$\sin^2\theta + \cos^2\theta = 1$ and 2 θ θ = 2θ$2\sin\theta\cos\theta = \sin 2\theta$:

Step 3: Max and Min Extrema Analysis

Since -1 ≤ 2θ ≤ 1$-1 \le \sin 2\theta \le 1$:

Sum of maximum and minimum bounds equals:

Pattern Recognition

Observe the symmetric structure in complex variables: (u+iv)$(u+iv)$ and (v+iu)$(v+iu)$. Their product structurally completely cancels out the real component, saving you from a highly messy component expansion.

Chapter Mix

Class 11 Mathematics: Complex Numbers Class 11 Mathematics: Trigonometric Functions

Core Logic

The equation x² + x + 1 = 0$x^2 + x + 1 = 0$ has complex roots which are the non-real cube roots of unity. Thus, we can set α = ω$\alpha = \omega$ (where ω³ = 1$\omega^3 = 1$ and 1 + ω + ω² = 0$1 + \omega + \omega^2 = 0$).

Let's analyze the general term block Tk = (ω^k + (1)/(ω^k))²$T_k = \left(\omega^k + \frac{1}{\omega^k}\right)^2$:

Because ω^k$\omega^k$ is periodic with period 3$3$, let's examine the values of Tk$T_k$ for different values of k$k$:

• If k$k$ is a multiple of 3$3$ (k=3m$k=3m$): ω2k = 1, ω^k = 1 Tk = 1 + 1 + 2 = 4$\omega^{2k} = 1, \omega^k = 1 \implies T_k = 1 + 1 + 2 = 4$.
• If k$k$ is not a multiple of 3$3$ (k=3m+1$k=3m+1$ or 3m+2$3m+2$): ω2k + ω^k = -1 Tk = -1 + 2 = 1$\omega^{2k} + \omega^k = -1 \implies T_k = -1 + 2 = 1$.

Step 1: Evaluating periodic blocks

Every block of three consecutive terms (k = 1, 2, 3$k = 1, 2, 3$) contributes exactly:

We want the total summation to equal 20$20$. Let's divide 20$20$ by our block value 6$6$:

This means the sum must consist of 3$3$ full periodic blocks plus additional terms that add up to 2$2$.

Step 2: Determining the final term count n

The number of terms in 3$3$ full blocks is 3 × 3 = 9$3 \times 3 = 9$ terms, giving a sum of 18$18$. To get the remaining value of 2$2$, we look at the next terms:

• Term 10 (k=10$k=10$, not a multiple of 3) adds 1 Total = 18 + 1 = 19$1 \implies \text{Total} = 18 + 1 = 19$.
• Term 11 (k=11$k=11$, not a multiple of 3) adds 1 Total = 19 + 1 = 20$1 \implies \text{Total} = 19 + 1 = 20$.

Hence, the series terminates exactly at n = 11$n = 11$.

Pattern Recognition

Whenever complex roots of unity or cyclic properties show up inside series sums, group terms into blocks based on the underlying period length (3$3$ here) to convert large sums into simple modular arithmetic arithmetic calculations.

Chapter Mix

Class 11 Mathematics: Complex Numbers Class 11 Mathematics: Sequences and Series

Related Formula

Magnitude squared representation:

Core Logic

Convert complex sets into Cartesian forms by setting z = x + iy$z = x + iy$: Set A$A$: |(x-2) + i(y-1)| = 3 (x-2)² + (y-1)² = 9 (1)$|(x-2) + i(y-1)| = 3 \implies (x-2)^2 + (y-1)^2 = 9 \quad \dots (1)$ Set B$B$: z - iz = (x+iy) - i(x+iy) = (x+y) + i(y-x)$z - iz = (x+iy) - i(x+iy) = (x+y) + i(y-x)$. Re(z - iz) = 2 x + y = 2 y = 2 - x (2)$\operatorname{Re}(z - iz) = 2 \implies x + y = 2 \implies y = 2 - x \quad \dots (2)$

Step 1: Solve System Algebraically

Substitute (2) into (1):

Roots are x1,2 = 3 ± √(17)2$x_{1,2} = \frac{3 \pm \sqrt{17}}{2}$. Correspondingly, y = 2 - x y1,2 = 1 ∓ √(17)2$y = 2 - x \implies y_{1,2} = \frac{1 \mp \sqrt{17}}{2}$.

Step 2: Evaluate Sum of Square Magnitudes

Since S$S$ consists of the two intersection points z₁, z₂$z_1, z_2$:

Using identities from quadratic equation x² - 3x - 2 = 0$x^2 - 3x - 2 = 0$ (x₁+x₂ = 3, x₁x₂ = -2$x_1+x_2 = 3, x_1x_2 = -2$): x₁² + x₂² = (3)² - 2(-2) = 13$x_1^2 + x_2^2 = (3)^2 - 2(-2) = 13$. Since y = 2-x$y = 2-x$, y² = 4 - 4x + x² y₁² + y₂² = 8 - 4(3) + 13 = 9$y^2 = 4 - 4x + x^2 \implies y_1^2 + y_2^2 = 8 - 4(3) + 13 = 9$.

Pattern Recognition

Avoid explicitly using radical root approximations. Summing symmetric expressions directly through standard Vieta coefficient sum shortcuts preserves clean fractions.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations

Related Formula

For a complex number w$w$, Re(w) = Re( w)$\operatorname{Re}(w) = \operatorname{Re}(\bar{w})$. Hence:

Core Logic

Notice that z - 12 z - i$\frac{\bar{z} - 1}{2\bar{z} - i}$ is the exact complex conjugate of (z - 1)/(2z + i)$\frac{z - 1}{2z + i}$. Thus, the given equation simplifies directly via complex identities to:

Step 1: Substitute z = x + iy

Let z = x + iy$z = x + iy$:

To find the real part, multiply numerator and denominator by the conjugate of the denominator:

Step 2: Expand and Arrange Circle Equation

Expanding the expression:

Dividing full equation by 2:

Step 3: Extract Center and Radius

Evaluating (15ab)/(r²)$\frac{15ab}{r^2}$:

Pattern Recognition

Recognizing that Re(w) + Re( w) = 2Re(w)$\operatorname{Re}(w) + \operatorname{Re}(\bar{w}) = 2\operatorname{Re}(w)$ avoids complex algebraic division on the second fractional expression completely.

Chapter Mix

Class 11 Mathematics: Complex Numbers Class 11 Mathematics: Circles

Related Formula

Since α$\alpha$ and β$\beta$ are roots of 2z² - 3z - 2i = 0$2z^2 - 3z - 2i = 0$, they satisfy the quadratic equation directly, meaning:

Similarly for β$\beta$:

Core Logic

Square the baseline relation to transition to higher exponential powers:

Squaring once more to isolate the fourth powers:

Step 1: Simplify the Target Expression Fraction

Rearrange the given complex algebraic fraction by factoring out powers:

Since both α$\alpha$ and β$\beta$ satisfy the exact same symmetric relational identity:

Substitute this uniform value back into the algebraic expression:

Step 2: Extract Real and Imaginary Components

From our simplified expression:

Now, substitute these into the evaluation formula:

Pattern Recognition

Symmetric rational polynomials in roots α, β$\alpha, \beta$ that can be split into identical numeric multipliers for αⁿ$\alpha^n$ and βⁿ$\beta^n$ allow direct cancellation of the polynomial bases without evaluating the individual roots explicitly.

Chapter Mix

Class 11 Mathematics: Complex Numbers