Let mathrmA = left\theta in [0,2pi ]:1 + 10operatorname Releft(frac2costheta + mathrmisinthetacostheta - 3mathrmisinthetaright) = 0right\$\mathrm{A} = \left\{\theta \in [0,2\pi ]:1 + 10\operatorname {Re}\left(\frac{2\cos\theta + \mathrm{i}\sin\theta}{\cos\theta - 3\mathrm{i}\sin\theta}\right) = 0\right\}$. Then sum_theta in mathrmAtheta^2$\sum_{\theta \in \mathrm{A}}\theta^2$ is equal to

### Related Formula Using Euler's formula: e^itheta = costheta + isintheta$$e^{i\theta} = \cos\theta + i\sin\theta$$ ### Core Logic Solving the quadratic root configurations for x^2 - sqrt6x + 3 = 0$x^2 - \sqrt{6}x + 3 = 0$: x = fracsqrt6 pm sqrt6 - 122 = fracsqrt6 pm isqrt62 = fracsqrt62(1 pm i)$$x = \frac{\sqrt{6} \pm \sqrt{6 - 12}}{2} = \frac{\sqrt{6} \pm i\sqrt{6}}{2} = \frac{\sqrt{6}}{2}(1 \pm i)$$ Given operatornameIm(alpha) > operatornameIm(beta)$\operatorname{Im}(\alpha) > \operatorname{Im}(\beta)$, we set: alpha = sqrt3 left(frac1+isqrt2right) = sqrt3 e^ipi/4$$\alpha = \sqrt{3} \left(\frac{1+i}{\sqrt{2}}\right) = \sqrt{3} e^{i\pi/4}$$ beta = sqrt3 left(frac1-isqrt2right) = sqrt3 e^-ipi/4$$\beta = \sqrt{3} \left(\frac{1-i}{\sqrt{2}}\right) = \sqrt{3} e^{-i\pi/4}$$ ### Step 1: Simplify Target Expression Let us factor out common variables: fracalpha^99beta + alpha^98 = alpha^98 left( fracalphabeta + 1 right) = fracalpha^98(alpha + beta)beta$$\frac{\alpha^{99}}{\beta} + \alpha^{98} = \alpha^{98} \left( \frac{\alpha}{\beta} + 1 \right) = \frac{\alpha^{98}(\alpha + \beta)}{\beta}$$ Since alpha + beta = sqrt6$\alpha + \beta = \sqrt{6}$: textValue = frac(sqrt3e^ipi/4)^98 cdot sqrt6sqrt3e^-ipi/4 = 3^49 e^i 98pi/4 cdot sqrt2 e^ipi/4 = 3^49 cdot sqrt2 e^i 99pi/4$$\text{Value} = \frac{(\sqrt{3}e^{i\pi/4})^{98} \cdot \sqrt{6}}{\sqrt{3}e^{-i\pi/4}} = 3^{49} e^{i 98\pi/4} \cdot \sqrt{2} e^{i\pi/4} = 3^{49} \cdot \sqrt{2} e^{i 99\pi/4}$$ Evaluate e^i 99pi/4$e^{i 99\pi/4}$: 99fracpi4 = 24pi + frac3pi4 implies e^i 99pi/4 = e^i 3pi/4 = frac-1+isqrt2$$99\frac{\pi}{4} = 24\pi + \frac{3\pi}{4} \implies e^{i 99\pi/4} = e^{i 3\pi/4} = \frac{-1+i}{\sqrt{2}}$$ Substituting this back: textValue = 3^49 cdot sqrt2 left( frac-1+isqrt2 right) = 3^49(-1 + i)$$\text{Value} = 3^{49} \cdot \sqrt{2} \left( \frac{-1+i}{\sqrt{2}} \right) = 3^{49}(-1 + i)$$ ### Step 2: Resolving Constants Comparing with the given expression 3^n(a + ib)$3^n(a + ib)$: n = 49, quad a = -1, quad b = 1$$n = 49, \quad a = -1, \quad b = 1$$ Therefore: n + a + b = 49 - 1 + 1 = 49$$n + a + b = 49 - 1 + 1 = 49$$ ### Pattern Recognition Convert complex expressions into polar form r e^itheta$r e^{i\theta}$ early. Power scaling like alpha^98$\alpha^{98}$ becomes simple multiplication under Euler structures. ### Evaluation Rubric / Model Answer null ### 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: x^2 - 2^y = 2023$x^2 - 2^y = 2023$. Let us inspect the numbers modulo 8: 2023 equiv 7 pmod 8$$2023 \equiv 7 \pmod 8$$ A perfect square x^2$x^2$ can only be congruent to 0, 1, 4 pmod 8$0, 1, 4 \pmod 8$. * If y geq 3$y \geq 3$, then 2^y equiv 0 pmod 8 implies x^2 equiv 7 pmod 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 mathbbN$y \in \mathbb{N}$, the only choices are y = 1$y = 1$ or y = 2$y = 2$. ### Step 1: Testing Small Exponent Valuations * Case 1: y = 1$y = 1$ x^2 - 2^1 = 2023 implies x^2 = 2025 implies x = 45 quad (textsince 45^2 = 2025)$$x^2 - 2^1 = 2023 \implies x^2 = 2025 \implies x = 45 \quad (\text{since } 45^2 = 2025)$$ This gives a valid natural solution pair: (45, 1)$(45, 1)$. * Case 2: y = 2$y = 2$ x^2 - 2^2 = 2023 implies x^2 = 2027$$x^2 - 2^2 = 2023 \implies x^2 = 2027$$ 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: sum (x + y) = 45 + 1 = 46$$\sum (x + y) = 45 + 1 = 46$$ ### Pattern Recognition Exponential Diophantine equations (equations with variables in exponents) are best analyzed using modular arithmetic constraints to quickly find small finite upper bounds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Principle of Mathematical Induction / Number Theory

### Related Formula |z - z_1| = |z - z_2|$$|z - z_1| = |z - z_2|$$ This represents the perpendicular bisector of the line segment joining the points z_1$z_1$ and z_2$z_2$ in the complex plane. ### Core Logic The given set defines a complex number z$z$ that is equidistant from three fixed points: A equiv (0, 1)$A \equiv (0, 1)$ corresponding to i$i$ B equiv (0, -1)$B \equiv (0, -1)$ corresponding to -i$-i$ C equiv (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 - z_1| = |z - z_2| = |z - z_3|$|z - z_1| = |z - z_2| = |z - z_3|$ is geometrically identical to finding the circumcenter of a triangle with vertices at z_1$z_1$, z_2$z_2$, and z_3$z_3$. A non-collinear set of three points always yields exactly 1 circumcenter. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers Class 11 Maths: Straight Lines

### Related Formula 1 + omega + omega^2 = 0$$1 + \omega + \omega^2 = 0$$ omega^3 = 1$\omega^3 = 1$ ### Core Logic The equation x^2 + x + 1 = 0$x^2 + x + 1 = 0$ is the standard identity whose roots are the non-real cube roots of unity, omega$\omega$ and omega^2$\omega^2$. Let us assign alpha = omega$\alpha = \omega$. We are given the expression (1+alpha)^7$(1+\alpha)^7$. Substituting the root: (1+omega)^7$(1+\omega)^7$. ### Step 1: Simplify using Unity Properties From the identity 1 + omega + omega^2 = 0$1 + \omega + \omega^2 = 0$, we extract: 1 + omega = -omega^2$$1 + \omega = -\omega^2$$ Substitute this into the expression: (1+omega)^7 = (-omega^2)^7 = -omega^14$$(1+\omega)^7 = (-\omega^2)^7 = -\omega^{14}$$ ### Step 2: Cyclical Reduction Using omega^3 = 1$\omega^3 = 1$, reduce the exponent 14 modulo 3: 14 = 3(4) + 2 Rightarrow omega^14 = (omega^3)^4 cdot omega^2 = 1 cdot omega^2 = omega^2$$14 = 3(4) + 2 \Rightarrow \omega^{14} = (\omega^3)^4 \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2$$ Thus, the expression reduces to -omega^2$-\omega^2$. Rewrite this back to its linear form using 1 + omega + omega^2 = 0$1 + \omega + \omega^2 = 0$: -omega^2 = 1 + omega = 1 + alpha$$-\omega^2 = 1 + \omega = 1 + \alpha$$ ### Step 3: Finding Co-efficients We compare 1 + alpha$1 + \alpha$ with A + Balpha + Calpha^2$A + B\alpha + C\alpha^2$. Notice that 1 + alpha$1 + \alpha$ can be directly represented without any alpha^2$\alpha^2$ term (and we must keep A, B, C ge 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)$5(3A - 2B - C)$: 5(3(1) - 2(1) - 0)$$5(3(1) - 2(1) - 0)$$ 5(3 - 2) = 5(1) = 5$$5(3 - 2) = 5(1) = 5$$ ### Pattern Recognition The roots of x^2+x+1=0$x^2+x+1=0$ are always omega, omega^2$\omega, \omega^2$. Expressions of the form (1+omega)^k$(1+\omega)^k$ rapidly collapse down to single variables via the 1+omega+omega^2=0$1+\omega+\omega^2=0$ rule, making multi-variable polynomial equations instantly trivial. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers

### Related Formula |x + iy| = sqrtx^2 + y^2$$|x + iy| = \sqrt{x^2 + y^2}$$ Two complex numbers are equal if and only if their real and imaginary parts are respectively equal. ### Core Logic Given z = frac12 - 2i$z = \frac{1}{2} - 2i$. Calculate |z + 1|$|z + 1|$: |z + 1| = left| left(frac12 - 2iright) + 1 right|$$|z + 1| = \left| \left(\frac{1}{2} - 2i\right) + 1 \right|$$ = left| frac32 - 2i right|$$= \left| \frac{3}{2} - 2i \right|$$ = sqrtleft(frac32right)^2 + (-2)^2 = sqrtfrac94 + 4 = sqrtfrac254 = frac52$$= \sqrt{\left(\frac{3}{2}\right)^2 + (-2)^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$$ Now substitute z$z$ and |z + 1|$|z + 1|$ into the original equation: frac52 = alphaleft(frac12 - 2iright) + beta(1 + i)$$\frac{5}{2} = \alpha\left(\frac{1}{2} - 2i\right) + \beta(1 + i)$$ Expand and group real and imaginary components on the RHS: frac52 = left(fracalpha2 - 2alpha iright) + (beta + beta i)$$\frac{5}{2} = \left(\frac{\alpha}{2} - 2\alpha i\right) + (\beta + \beta i)$$ frac52 = left(fracalpha2 + betaright) + i(beta - 2alpha)$$\frac{5}{2} = \left(\frac{\alpha}{2} + \beta\right) + i(\beta - 2\alpha)$$ ### Step 1: Equate Parts By equating the real and imaginary parts from both sides, we get a system of linear equations: Imaginary part: 0 = beta - 2alpha Rightarrow beta = 2alpha$$0 = \beta - 2\alpha \Rightarrow \beta = 2\alpha$$ Real part: frac52 = fracalpha2 + beta$$\frac{5}{2} = \frac{\alpha}{2} + \beta$$ Substitute beta = 2alpha$\beta = 2\alpha$ into the real part equation: frac52 = fracalpha2 + 2alpha$$\frac{5}{2} = \frac{\alpha}{2} + 2\alpha$$ frac52 = frac5alpha2$$\frac{5}{2} = \frac{5\alpha}{2}$$ alpha = 1$\alpha = 1$ Using alpha = 1$\alpha = 1$, find beta$\beta$: beta = 2(1) = 2$$\beta = 2(1) = 2$$ ### Step 2: Final Calculation Calculate the final requested value: alpha + beta = 1 + 2 = 3$$\alpha + \beta = 1 + 2 = 3$$ ### 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers and Quadratic Equations