Let alpha = frac-1 + mathrmisqrt32$\alpha = \frac{-1 + \mathrm{i}\sqrt{3}}{2}$ and beta = frac-1 - mathrmisqrt32,mathrmi = sqrt-1$\beta = \frac{-1 - \mathrm{i}\sqrt{3}}{2},\mathrm{i} = \sqrt{-1}$. If (7 - 7alpha +9beta)^20 + (9 + 7alpha -7beta)^20 + (-7 + 9alpha +7beta)^20 + (14 + 7alpha +7beta)^20 = mathrmm^10$(7 - 7\alpha +9\beta)^{20} + (9 + 7\alpha -7\beta)^{20} + (-7 + 9\alpha +7\beta)^{20} + (14 + 7\alpha +7\beta)^{20} = \mathrm{m}^{10}$, then m is ____.

### Related Formula x^2 + x + 1 = 0 Rightarrow x = omega, omega^2$$x^2 + x + 1 = 0 \Rightarrow x = \omega, \omega^2$$ Properties of cube roots of unity: omega^3 = 1$\omega^3 = 1$ and 1 + omega + omega^2 = 0$1 + \omega + \omega^2 = 0$. ### Core Logic Let alpha = omega$\alpha = \omega$. Then frac1x = frac1omega = omega^2$\frac{1}{x} = \frac{1}{\omega} = \omega^2$. The series is sum_k=1^25 (omega^k + omega^2k)^4$\sum_{k=1}^{25} (\omega^k + \omega^{2k})^4$. Evaluate the term T_k = (omega^k + omega^2k)^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) T_3m = (omega^3m + omega^6m)^4 = (1 + 1)^4 = 2^4 = 16$$T_{3m} = (\omega^{3m} + \omega^{6m})^4 = (1 + 1)^4 = 2^4 = 16$$ There are 8$8$ such multiples up to 25 (3, 6, 9, dots, 24$3, 6, 9, \dots, 24$). Case 2: k neq 3m$k \neq 3m$ (non-multiples of 3) For k = 1, 2, 4, 5, dots$k = 1, 2, 4, 5, \dots$ omega^k + omega^2k$\omega^k + \omega^{2k}$ will always be omega + omega^2$\omega + \omega^2$ or omega^2 + omega$\omega^2 + \omega$. Since 1 + omega + omega^2 = 0$1 + \omega + \omega^2 = 0$, we have omega + omega^2 = -1$\omega + \omega^2 = -1$. T_k neq 3m = (-1)^4 = 1$$T_{k \neq 3m} = (-1)^4 = 1$$ There are 25 - 8 = 17$25 - 8 = 17$ such non-multiples up to 25. ### Step 2: Total Sum textSum = 17 times 1 + 8 times 16$$\text{Sum} = 17 \times 1 + 8 \times 16$$ = 17 + 128 = 145$$= 17 + 128 = 145$$ ### Pattern Recognition Powers of x+1/x$x+1/x$ when x$x$ solves x^2 pm 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 omega$\omega$ identities). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations

### Related Formula textFor maximum argument of z text on a circle, the ray from origin is tangent to the circle.$$\text{For maximum argument of } z \text{ on a circle, the ray from origin is tangent to the circle.}$$ textIf |z-a| leq r, text maximum argument implies sintheta = fracr|a| text and coordinates are x=acos^2theta, y=asinthetacostheta$$\text{If } |z-a| \leq r, \text{ maximum argument implies } \sin\theta = \frac{r}{|a|} \text{ and coordinates are } x=a\cos^2\theta, y=a\sin\theta\cos\theta$$ ### Core Logic

The condition |z-5|leq 3$|z-5|\leq 3$ represents a disk centered at (5,0)$(5,0)$ with radius 3$3$. To maximize the principal argument theta$\theta$, the ray from origin must touch the circle in the first quadrant. The tangent, origin, and center form a right-angled triangle. ### 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 sqrt5^2 - 3^2 = 4$\sqrt{5^2 - 3^2} = 4$. The angle of tangency theta$\theta$ satisfies sintheta = frac35$\sin\theta = \frac{3}{5}$ and costheta = frac45$\cos\theta = \frac{4}{5}$. The point P(z)$P(z)$ lies on the circle and the tangent ray: z equiv (4costheta, 4sintheta) = left(4left(frac45right), 4left(frac35right)right) = left(frac165, frac125right)$$z \equiv (4\cos\theta, 4\sin\theta) = \left(4\left(\frac{4}{5}\right), 4\left(\frac{3}{5}\right)\right) = \left(\frac{16}{5}, \frac{12}{5}\right)$$ z = frac165 + frac125i$$z = \frac{16}{5} + \frac{12}{5}i$$ ### Step 2: Evaluate the Target Expression We need 34left|frac5z - 125iz + 16right|^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: 34 left|frac4 + 12i4 + 16iright|^2 = 34 frac|4 + 12i|^2|4 + 16i|^2 = 34 frac4^2 + 12^24^2 + 16^2$$34 \left|\frac{4 + 12i}{4 + 16i}\right|^2 = 34 \frac{|4 + 12i|^2}{|4 + 16i|^2} = 34 \frac{4^2 + 12^2}{4^2 + 16^2}$$ = 34 left(frac16 + 14416 + 256right) = 34 left(frac160272right)$$= 34 \left(\frac{16 + 144}{16 + 256}\right) = 34 \left(\frac{160}{272}\right)$$ = frac34 times 160272 = frac5440272 = 20$$= \frac{34 \times 160}{272} = \frac{5440}{272} = 20$$ ### Pattern Recognition For maximum arg(z)$\arg(z)$ on |z-c| = r$|z-c| = r$ where c$c$ is real, z$z$ coordinates are given by geometric projection: z = sqrtc^2-r^2(costheta + isintheta)$z = \sqrt{c^2-r^2}(\cos\theta + i\sin\theta)$ where sintheta = r/c$\sin\theta = r/c$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers

### Related Formula For z = x + iy$z = x + iy$, overlinez = x - iy$\overline{z} = x - iy$ and |z|^2 = x^2 + y^2$|z|^2 = x^2 + y^2$. ### Core Logic Substitute z = x + iy$z = x + iy$ into 4z^2 + overlinez = 0$4z^2 + \overline{z} = 0$: 4(x^2 - y^2 + 2ixy) + x - iy = 0$$4(x^2 - y^2 + 2ixy) + x - iy = 0$$ Separating real and imaginary parts: 4x^2 - 4y^2 + x = 0 quad textand quad y(8x - 1) = 0$$4x^2 - 4y^2 + x = 0 \quad \text{and} \quad y(8x - 1) = 0$$ Case 1: y = 0 implies 4x^2 + x = 0 implies x = 0$y = 0 \implies 4x^2 + x = 0 \implies x = 0$ or x = -1/4$x = -1/4$. Solutions: z_1 = 0 implies |z_1|^2 = 0$z_1 = 0 \implies |z_1|^2 = 0$, z_2 = -1/4 implies |z_2|^2 = 1/16$z_2 = -1/4 \implies |z_2|^2 = 1/16$. Case 2: x = 1/8 implies 4(1/64) - 4y^2 + 1/8 = 0 implies 4y^2 = 3/16 implies y = pm fracsqrt38$x = 1/8 \implies 4(1/64) - 4y^2 + 1/8 = 0 \implies 4y^2 = 3/16 \implies y = \pm \frac{\sqrt{3}}{8}$. Solutions: z_3, z_4 = frac18 pm i fracsqrt38 implies |z_3|^2 = |z_4|^2 = frac164 + frac364 = frac116$z_3, z_4 = \frac{1}{8} \pm i \frac{\sqrt{3}}{8} \implies |z_3|^2 = |z_4|^2 = \frac{1}{64} + \frac{3}{64} = \frac{1}{16}$. ### Step 1: Sum of Modulus Squared sum_z in S |z|^2 = 0 + frac116 + frac116 + frac116 = frac316$$\sum_{z \in S} |z|^2 = 0 + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{3}{16}$$ ### Pattern Recognition Separate complex equation into real and imaginary components to systematically find all roots. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers

### Related Formula textFor a quadratic equation A x^2 + B x + C = 0 text to have two positive real roots:$$\text{For a quadratic equation } A x^2 + B x + C = 0 \text{ to have two positive real roots:}$$ text1. Real roots: D = B^2 - 4AC ge 0$$\text{1. Real roots: } D = B^2 - 4AC \ge 0$$ text2. Sum of roots: -fracBA > 0$$\text{2. Sum of roots: } -\frac{B}{A} > 0$$ text3. Product of roots: fracCA > 0$$\text{3. Product of roots: } \frac{C}{A} > 0$$ ### Core Logic We write down three systems of inequalities based on real and positive root conditions, find their intersection, and map the boundaries to solve for the parameters. ### Step 1: Apply the discriminant condition (Real roots) For real roots, the discriminant D ge 0$D \ge 0$: D = left[ 2(a - 3) right]^2 - 4(1 - a)(9) ge 0$$D = \left[ 2(a - 3) \right]^2 - 4(1 - a)(9) \ge 0$$ 4(a^2 - 6a + 9) - 36(1 - a) ge 0$$4(a^2 - 6a + 9) - 36(1 - a) \ge 0$$ (a^2 - 6a + 9) - 9(1 - a) ge 0$$(a^2 - 6a + 9) - 9(1 - a) \ge 0$$ a^2 - 6a + 9 - 9 + 9 a ge 0$$a^2 - 6a + 9 - 9 + 9 a \ge 0$$ a^2 + 3a ge 0 implies a(a + 3) ge 0$$a^2 + 3a \ge 0 \implies a(a + 3) \ge 0$$ Thus, the interval is: a in (-infty, -3] cup [0, infty) quad text--- (1)$$a \in (-\infty, -3] \cup [0, \infty) \quad \text{--- (1)}$$ ### Step 2: Apply the sum of roots condition (Positive sum) For positive roots, the sum of roots must be positive: -fracBA = frac-2(a - 3)1 - a = frac2(a - 3)a - 1 > 0$$-\frac{B}{A} = \frac{-2(a - 3)}{1 - a} = \frac{2(a - 3)}{a - 1} > 0$$ Using the wavy curve method for fraca-3a-1 > 0$\frac{a-3}{a-1} > 0$: a in (-infty, 1) cup (3, infty) quad text--- (2)$$a \in (-\infty, 1) \cup (3, \infty) \quad \text{--- (2)}$$ ### Step 3: Apply the product of roots condition (Positive product) For positive roots, the product of roots must be positive: fracCA = frac91 - a > 0 implies 1 - a > 0 implies a < 1$$\frac{C}{A} = \frac{9}{1 - a} > 0 \implies 1 - a > 0 \implies a < 1$$ Thus, the interval is: a in (-infty, 1) quad text--- (3)$$a \in (-\infty, 1) \quad \text{--- (3)}$$ ### Step 4: Find the intersection of all conditions Intersecting equations (1), (2), and (3): - First, intersect (2) and (3): ( (-infty, 1) cup (3, infty) ) cap (-infty, 1) = (-infty, 1)$$( (-\infty, 1) \cup (3, \infty) ) \cap (-\infty, 1) = (-\infty, 1)$$ - Next, intersect with (1): ( (-infty, -3] cup [0, infty) ) cap (-infty, 1) = (-infty, -3] cup [0, 1)$$( (-\infty, -3] \cup [0, \infty) ) \cap (-\infty, 1) = (-\infty, -3] \cup [0, 1)$$ Comparing this with (-infty, -alpha] cup [beta, gamma)$(-\infty, -\alpha] \cup [\beta, \gamma)$: - alpha = 3$\alpha = 3$ - beta = 0$\beta = 0$ - gamma = 1$\gamma = 1$ Now calculate the target sum: 2alpha + beta + gamma = 2(3) + 0 + 1 = 7$$2\alpha + \beta + \gamma = 2(3) + 0 + 1 = 7$$ ### Pattern Recognition Location of roots: When both roots are positive, checking sum and product signs along with D ge 0$D \ge 0$ is the standard and fastest set of inequalities, avoiding complex vertex projections. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Practical Chemistry