Let O be the origin, the point A be z₁ = √(3) + 2√(2)i$z_1 = \sqrt{3} + 2\sqrt{2}i$, the point B(z₂)$B(z_2)$ be such that √(3)|z₂| = |z₁|$\sqrt{3}\left|z_2\right| = \left|z_1\right|$ and (z₂) = (z₁) + (π)/(6)$\arg (z_2) = \arg (z_1) + \frac{\pi}{6}$. Then

Related Formula

Using Euler's formula:

Core Logic

Solving the quadratic root configurations for x² - √(6)x + 3 = 0$x^2 - \sqrt{6}x + 3 = 0$:

Given Im(α) > Im(β)$\operatorname{Im}(\alpha) > \operatorname{Im}(\beta)$, we set:

Step 1: Simplify Target Expression

Let us factor out common variables:

Since α + β = √(6)$\alpha + \beta = \sqrt{6}$:

Evaluate ei 99π/4$e^{i 99\pi/4}$:

Substituting this back:

Step 2: Resolving Constants

Comparing with the given expression 3ⁿ(a + ib)$3^n(a + ib)$:

Therefore:

Pattern Recognition

Convert complex expressions into polar form r eiθ$r e^{i\theta}$ early. Power scaling like α⁹⁸$\alpha^{98}$ becomes simple multiplication under Euler structures.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations

Related Formula

Analyze structural equations through modular constraints (e.g., modulo 3 or 4) to limit potential bounds.

Core Logic

Given the equation: x² - 2^y = 2023$x^2 - 2^y = 2023$. Let us inspect the numbers modulo 8:

A perfect square x²$x^2$ can only be congruent to 0, 1, 4 8$0, 1, 4 \pmod 8$.

• If y ≥ 3$y \geq 3$, then 2^y ≡ 0 8 x² ≡ 7 8$2^y \equiv 0 \pmod 8 \implies x^2 \equiv 7 \pmod 8$, which is impossible.

Therefore, y$y$ must be less than 3. Since y in N$y \in \mathbb{N}$, the only choices are y = 1$y = 1$ or y = 2$y = 2$.

Step 1: Testing Small Exponent Valuations

• Case 1: y = 1$y = 1$

This gives a valid natural solution pair: (45, 1)$(45, 1)$.

• Case 2: y = 2$y = 2$

Since 2027 is not a perfect square, this yields no natural solutions.

Thus, the only valid point element inside set C$C$ is (45, 1)$(45, 1)$.

Step 2: Sum Evaluation

Evaluating the required target accumulation:

Pattern Recognition

Exponential Diophantine equations (equations with variables in exponents) are best analyzed using modular arithmetic constraints to quickly find small finite upper bounds.

Chapter Mix

Class 11 Mathematics: Principle of Mathematical Induction / Number Theory

Related Formula

This represents the perpendicular bisector of the line segment joining the points z₁$z_1$ and z₂$z_2$ in the complex plane.

Core Logic

The given set defines a complex number z$z$ that is equidistant from three fixed points: A ≡ (0, 1)$A \equiv (0, 1)$ corresponding to i$i$ B ≡ (0, -1)$B \equiv (0, -1)$ corresponding to -i$-i$ C ≡ (1, 0)$C \equiv (1, 0)$ corresponding to 1$1$

The condition |z-i|=|z+i|=|z-1|$|z-i|=|z+i|=|z-1|$ implies that z$z$ is the point of intersection of the perpendicular bisectors of the sides of the triangle formed by A$A$, B$B$, and C$C$.

Step 1: Finding the Circumcenter

The point of intersection of the perpendicular bisectors of a triangle is its circumcenter. Since A(0,1)$A(0,1)$, B(0,-1)$B(0,-1)$, and C(1,0)$C(1,0)$ form a unique, non-degenerate triangle, they have exactly one unique circumcenter.

Step 2: Final Conclusion

Therefore, there is only one such complex number z$z$ that satisfies the condition. n(S) = 1$n(S) = 1$

Pattern Recognition

Recognize that |z - z₁| = |z - z₂| = |z - z₃|$|z - z_1| = |z - z_2| = |z - z_3|$ is geometrically identical to finding the circumcenter of a triangle with vertices at z₁$z_1$, z₂$z_2$, and z₃$z_3$. A non-collinear set of three points always yields exactly 1 circumcenter.

Chapter Mix

Class 11 Maths: Complex Numbers Class 11 Maths: Straight Lines

Related Formula

ω³ = 1$\omega^3 = 1$

Core Logic

The equation x² + x + 1 = 0$x^2 + x + 1 = 0$ is the standard identity whose roots are the non-real cube roots of unity, ω$\omega$ and ω²$\omega^2$. Let us assign α = ω$\alpha = \omega$. We are given the expression (1+α)⁷$(1+\alpha)^7$. Substituting the root: (1+ω)⁷$(1+\omega)^7$.

Step 1: Simplify using Unity Properties

From the identity 1 + ω + ω² = 0$1 + \omega + \omega^2 = 0$, we extract:

Substitute this into the expression:

Step 2: Cyclical Reduction

Using ω³ = 1$\omega^3 = 1$, reduce the exponent 14 modulo 3:

Thus, the expression reduces to -ω²$-\omega^2$. Rewrite this back to its linear form using 1 + ω + ω² = 0$1 + \omega + \omega^2 = 0$:

Step 3: Finding Co-efficients

We compare 1 + α$1 + \alpha$ with A + Bα + Cα²$A + B\alpha + C\alpha^2$. Notice that 1 + α$1 + \alpha$ can be directly represented without any α²$\alpha^2$ term (and we must keep A, B, C ≥ 0$\ge 0$). So, A = 1$A = 1$, B = 1$B = 1$, C = 0$C = 0$.

Step 4: Final Output Evaluation

Substitute these constants into the required equation 5(3A - 2B - C)$5(3A - 2B - C)$:

Pattern Recognition

The roots of x²+x+1=0$x^2+x+1=0$ are always ω, ω²$\omega, \omega^2$. Expressions of the form (1+ω)^k$(1+\omega)^k$ rapidly collapse down to single variables via the 1+ω+ω²=0$1+\omega+\omega^2=0$ rule, making multi-variable polynomial equations instantly trivial.

Chapter Mix

Class 11 Maths: Complex Numbers

Related Formula

Two complex numbers are equal if and only if their real and imaginary parts are respectively equal.

Core Logic

Given z = (1)/(2) - 2i$z = \frac{1}{2} - 2i$. Calculate |z + 1|$|z + 1|$:

Now substitute z$z$ and |z + 1|$|z + 1|$ into the original equation:

Expand and group real and imaginary components on the RHS:

Step 1: Equate Parts

By equating the real and imaginary parts from both sides, we get a system of linear equations:

Imaginary part:

Real part:

Substitute β = 2α$\beta = 2\alpha$ into the real part equation:

α = 1$\alpha = 1$

Using α = 1$\alpha = 1$, find β$\beta$:

Step 2: Final Calculation

Calculate the final requested value:

Pattern Recognition

Equating complex parts reduces single complex equations into two simultaneous linear equations. Treat |z+1|$|z+1|$ strictly as a scalar magnitude and parse directly into algebraic components.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations