### 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 |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

### Related Formula Since alpha$\alpha$ is a root of x^2 - x + 2 = 0$x^2 - x + 2 = 0$, it must satisfy the equation exactly: alpha^2 - alpha + 2 = 0 Rightarrow alpha^2 = alpha - 2$\alpha^2 - \alpha + 2 = 0 \Rightarrow \alpha^2 = \alpha - 2$ This is an essential root reduction property allowing polynomials of high degrees to be collapsed linearly. ### Core Logic We need to evaluate the expression E = alpha^6 + alpha^4 + beta^4 - 5alpha^2$E = \alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2$. Use the substitution alpha^2 = alpha - 2$\alpha^2 = \alpha - 2$ to iteratively depress the powers of alpha$\alpha$. alpha^4 = (alpha^2)^2 = (alpha - 2)^2 = alpha^2 - 4alpha + 4$\alpha^4 = (\alpha^2)^2 = (\alpha - 2)^2 = \alpha^2 - 4\alpha + 4$ Substitute alpha^2 = alpha - 2$\alpha^2 = \alpha - 2$ again into the result: alpha^4 = (alpha - 2) - 4alpha + 4 = -3alpha + 2$\alpha^4 = (\alpha - 2) - 4\alpha + 4 = -3\alpha + 2$ Now, generate alpha^6$\alpha^6$ using alpha^4$\alpha^4$: alpha^6 = alpha^4 cdot alpha^2 = (-3alpha + 2)(alpha - 2) = -3alpha^2 + 6alpha + 2alpha - 4 = -3alpha^2 + 8alpha - 4$\alpha^6 = \alpha^4 \cdot \alpha^2 = (-3\alpha + 2)(\alpha - 2) = -3\alpha^2 + 6\alpha + 2\alpha - 4 = -3\alpha^2 + 8\alpha - 4$ Substitute alpha^2 = alpha - 2$\alpha^2 = \alpha - 2$ into the result again: alpha^6 = -3(alpha - 2) + 8alpha - 4 = -3alpha + 6 + 8alpha - 4 = 5alpha + 2$\alpha^6 = -3(\alpha - 2) + 8\alpha - 4 = -3\alpha + 6 + 8\alpha - 4 = 5\alpha + 2$ ### Step 1: Simplify the Full Expression The symmetry of the roots dictates that beta^4$\beta^4$ behaves identically to alpha^4$\alpha^4$. Thus: beta^4 = -3beta + 2$\beta^4 = -3\beta + 2$ Substitute all depressed linear forms back into E = alpha^6 + alpha^4 + beta^4 - 5alpha^2$E = \alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2$: E = (5alpha + 2) + (-3alpha + 2) + (-3beta + 2) - 5(alpha - 2)$E = (5\alpha + 2) + (-3\alpha + 2) + (-3\beta + 2) - 5(\alpha - 2)$ E = 5alpha - 3alpha - 5alpha - 3beta + 2 + 2 + 2 + 10$E = 5\alpha - 3\alpha - 5\alpha - 3\beta + 2 + 2 + 2 + 10$ E = -3alpha - 3beta + 16$E = -3\alpha - 3\beta + 16$ E = -3(alpha + beta) + 16$E = -3(\alpha + \beta) + 16$ ### Step 2: Apply Sum of Roots From the original quadratic equation x^2 - x + 2 = 0$x^2 - x + 2 = 0$, the sum of roots is: alpha + beta = -frac-11 = 1$\alpha + \beta = -\frac{-1}{1} = 1$ Substitute this back: E = -3(1) + 16 = 13$E = -3(1) + 16 = 13$ (Note: The condition Im(alpha) gt Im(beta)$Im(\alpha) \gt Im(\beta)$ was a distractor since the expression simplified perfectly symmetrically into alpha + beta$\alpha + \beta$ without needing the individual complex values of the roots). ### Pattern Recognition Never compute De Moivre polar forms for high root powers unless the quadratic has roots like omega$\omega$ or i$i$. Always use the characteristic quadratic relation alpha^2 = palpha + q$\alpha^2 = p\alpha + q$ to rapidly step down degrees until everything is strictly linear. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers and Quadratic Equations