### Related Formula z^3-1 = (z-1)(z^2+z+1) quad textCube roots of unity: 1, omega, omega^2$z^3-1 = (z-1)(z^2+z+1) \quad \text{Cube roots of unity: } 1, \omega, \omega^2$ omega^3 = 1 quad textand quad 1+omega+omega^2 = 0$\omega^3 = 1 \quad \text{and} \quad 1+\omega+\omega^2 = 0$ ### Core Logic Let's first find the roots of the lower degree polynomial: z^3 + 2z^2 + 2z + 1 = 0$z^3 + 2z^2 + 2z + 1 = 0$ Group the terms: (z^3 + 1) + 2z(z + 1) = 0$(z^3 + 1) + 2z(z + 1) = 0$ (z + 1)(z^2 - z + 1) + 2z(z + 1) = 0$(z + 1)(z^2 - z + 1) + 2z(z + 1) = 0$ (z + 1)(z^2 - z + 1 + 2z) = 0$(z + 1)(z^2 - z + 1 + 2z) = 0$ (z + 1)(z^2 + z + 1) = 0$(z + 1)(z^2 + z + 1) = 0$ Thus, the roots are z = -1$z = -1$, and the roots of z^2 + z + 1 = 0$z^2 + z + 1 = 0$ which are z = omega, omega^2$z = \omega, \omega^2$. ### Step 1: Check z = -1 Substitute z = -1$z = -1$ into the first equation z^1985 + z^100 + 1 = 0$z^{1985} + z^{100} + 1 = 0$: (-1)^1985 + (-1)^100 + 1 = -1 + 1 + 1 = 1 neq 0$(-1)^{1985} + (-1)^{100} + 1 = -1 + 1 + 1 = 1 \neq 0$ So, z = -1$z = -1$ is not a common root. ### Step 2: Check z = ω and ω² Substitute z = omega$z = \omega$: omega^1985 + omega^100 + 1$\omega^{1985} + \omega^{100} + 1$ Reduce the powers modulo 3 (since omega^3 = 1$\omega^3 = 1$): 1985 = 3 times 661 + 2 Rightarrow omega^1985 = omega^2$1985 = 3 \times 661 + 2 \Rightarrow \omega^{1985} = \omega^2$ 100 = 3 times 33 + 1 Rightarrow omega^100 = omega^1 = omega$100 = 3 \times 33 + 1 \Rightarrow \omega^{100} = \omega^1 = \omega$ So, omega^1985 + omega^100 + 1 = omega^2 + omega + 1 = 0$\omega^{1985} + \omega^{100} + 1 = \omega^2 + \omega + 1 = 0$. z = omega$z = \omega$ is a common root. Substitute z = omega^2$z = \omega^2$: (omega^2)^1985 + (omega^2)^100 + 1 = omega^3970 + omega^200 + 1$(\omega^2)^{1985} + (\omega^2)^{100} + 1 = \omega^{3970} + \omega^{200} + 1$ 3970 = 3 times 1323 + 1 Rightarrow omega^3970 = omega$3970 = 3 \times 1323 + 1 \Rightarrow \omega^{3970} = \omega$ 200 = 3 times 66 + 2 Rightarrow omega^200 = omega^2$200 = 3 \times 66 + 2 \Rightarrow \omega^{200} = \omega^2$ So, omega^3970 + omega^200 + 1 = omega + omega^2 + 1 = 0$\omega^{3970} + \omega^{200} + 1 = \omega + \omega^2 + 1 = 0$. z = omega^2$z = \omega^2$ is also a common root. ### Step 3: Conclusion There are exactly 2 common roots: omega$\omega$ and omega^2$\omega^2$. ### Pattern Recognition Whenever z^2+z+1$z^2+z+1$ emerges as a factor, its roots omega$\omega$ and omega^2$\omega^2$ can be directly tested in massive degree polynomials using exponent reduction modulo 3. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers

### Related Formula |z^n| = |z|^n$|z^n| = |z|^n$ |barz| = |z|$|\bar{z}| = |z|$ ### Core Logic Given the equation: z^2 = -ibarz$z^2 = -i\bar{z}$ Taking the modulus on both sides: |z^2| = |-ibarz|$|z^2| = |-i\bar{z}|$ Using properties of modulus: |z|^2 = |-i| cdot |barz|$|z|^2 = |-i| \cdot |\bar{z}|$ Since |-i| = 1$|-i| = 1$ and |barz| = |z|$|\bar{z}| = |z|$: |z|^2 = |z|$|z|^2 = |z|$ |z|^2 - |z| = 0$|z|^2 - |z| = 0$ |z|(|z| - 1) = 0$|z|(|z| - 1) = 0$ ### Step 1: Applying the non-zero condition This gives two possibilities: |z| = 0$|z| = 0$ or |z| = 1$|z| = 1$. Since z = x + iy$z = x + iy$ and xy neq 0$xy \neq 0$, neither x$x$ nor y$y$ is zero, which implies |z| neq 0$|z| \neq 0$. Therefore, |z| = 1$|z| = 1$. We are asked for |z^2|$|z^2|$: |z^2| = |z|^2 = 1^2 = 1$|z^2| = |z|^2 = 1^2 = 1$ ### Pattern Recognition When dealing with equations involving z$z$ and barz$\bar{z}$, applying modulus to both sides rapidly simplifies the problem, turning complex equations into simple real algebraic equations in |z|$|z|$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations

### Related Formula textSum of roots (alpha + beta) = -fracba$\text{Sum of roots } (\alpha + \beta) = -\frac{b}{a}$ textProduct of roots (alpha beta) = fracca$\text{Product of roots } (\alpha \beta) = \frac{c}{a}$ ### Core Logic For the equation x^2 - 70x + lambda = 0$x^2 - 70x + \lambda = 0$: alpha + beta = 70$\alpha + \beta = 70$ alpha beta = lambda$\alpha \beta = \lambda$ Since alpha, beta in mathbbN$\alpha, \beta \in \mathbb{N}$, their sum is 70. This gives alpha(70 - alpha) = lambda$\alpha(70 - \alpha) = \lambda$. We are given lambda / 2 notin mathbbN$\lambda / 2 \notin \mathbb{N}$ and lambda / 3 notin mathbbN$\lambda / 3 \notin \mathbb{N}$. This means lambda$\lambda$ is not divisible by 2 or 3. So lambda$\lambda$ must be an odd number and not a multiple of 3. Thus, neither alpha$\alpha$ nor beta$\beta$ can be a multiple of 2 or 3. ### Step 1: Finding minimum lambda We need to find the minimum value of lambda$\lambda$. lambda = alpha(70 - alpha)$\lambda = \alpha(70 - \alpha)$. This parabola opens downwards, so the minimum product occurs when the integers alpha$\alpha$ and beta$\beta$ are as far apart as possible. Let's test small values for alpha$\alpha$: If alpha = 1$\alpha = 1$, beta = 69 Rightarrow lambda = 69$\beta = 69 \Rightarrow \lambda = 69$, but 69 = 3 times 23$69 = 3 \times 23$, which is divisible by 3. Rejected. If alpha = 2$\alpha = 2$, divisible by 2. Rejected. If alpha = 3$\alpha = 3$, divisible by 3. Rejected. If alpha = 4$\alpha = 4$, divisible by 2. Rejected. If alpha = 5$\alpha = 5$, beta = 65 Rightarrow lambda = 325$\beta = 65 \Rightarrow \lambda = 325$. Check divisibility: 325 is odd (not div by 2). 3+2+5 = 10$3+2+5 = 10$ (not div by 3). So minimum lambda = 325$\lambda = 325$ with roots alpha = 5, beta = 65$\alpha = 5, \beta = 65$. ### Step 2: Evaluating the target expression We need to compute: E = frac(sqrtalpha - 1 + sqrtbeta - 1)(lambda + 35)|alpha - beta|$E = \frac{(\sqrt{\alpha - 1} + \sqrt{\beta - 1})(\lambda + 35)}{|\alpha - \beta|}$ Substitute the values alpha = 5$\alpha = 5$, beta = 65$\beta = 65$, lambda = 325$\lambda = 325$: |alpha - beta| = |5 - 65| = 60$|\alpha - \beta| = |5 - 65| = 60$ sqrtalpha - 1 = sqrt4 = 2$\sqrt{\alpha - 1} = \sqrt{4} = 2$ sqrtbeta - 1 = sqrt64 = 8$\sqrt{\beta - 1} = \sqrt{64} = 8$ E = frac(2 + 8)(325 + 35)60 = frac(10)(360)60 = 10 times 6 = 60$E = \frac{(2 + 8)(325 + 35)}{60} = \frac{(10)(360)}{60} = 10 \times 6 = 60$ The result is exactly 60. ### Pattern Recognition Minimizing the product of two numbers with a fixed sum requires them to be as far apart as possible. Divisibility constraints are quickly verified using prime modulus filters (%2 neq 0$%2 \neq 0$, %3 neq 0$%3 \neq 0$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations

### Related Formula a^3+b^3 = (a+b)^3 - 3ab(a+b)$a^3+b^3 = (a+b)^3 - 3ab(a+b)$ a^2+b^2 = (a+b)^2 - 2ab$a^2+b^2 = (a+b)^2 - 2ab$ a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2$a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2$ ### Core Logic Given z_1+z_2=5$z_1+z_2=5$ and z_1^3+z_2^3=20+15i$z_1^3+z_2^3=20+15i$. z_1^3+z_2^3 = (z_1+z_2)^3 - 3z_1z_2(z_1+z_2)$z_1^3+z_2^3 = (z_1+z_2)^3 - 3z_1z_2(z_1+z_2)$ 20+15i = 125 - 15z_1z_2$20+15i = 125 - 15z_1z_2$ 15z_1z_2 = 105 - 15i implies z_1z_2 = 7-i$15z_1z_2 = 105 - 15i \implies z_1z_2 = 7-i$ Now, compute z_1^2+z_2^2$z_1^2+z_2^2$: z_1^2+z_2^2 = (z_1+z_2)^2 - 2z_1z_2 = 25 - 2(7-i) = 11+2i$z_1^2+z_2^2 = (z_1+z_2)^2 - 2z_1z_2 = 25 - 2(7-i) = 11+2i$ Now, compute z_1^4+z_2^4$z_1^4+z_2^4$: z_1^4+z_2^4 = (z_1^2+z_2^2)^2 - 2(z_1z_2)^2$z_1^4+z_2^4 = (z_1^2+z_2^2)^2 - 2(z_1z_2)^2$ = (11+2i)^2 - 2(7-i)^2$= (11+2i)^2 - 2(7-i)^2$ = (121 - 4 + 44i) - 2(49 - 1 - 14i)$= (121 - 4 + 44i) - 2(49 - 1 - 14i)$ = 117 + 44i - 2(48 - 14i)$= 117 + 44i - 2(48 - 14i)$ = 117 + 44i - 96 + 28i = 21 + 72i$= 117 + 44i - 96 + 28i = 21 + 72i$ Finally, find magnitude: |z_1^4+z_2^4| = |21+72i| = sqrt21^2 + 72^2 = sqrt441 + 5184 = sqrt5625 = 75$|z_1^4+z_2^4| = |21+72i| = \sqrt{21^2 + 72^2} = \sqrt{441 + 5184} = \sqrt{5625} = 75$ ### Pattern Recognition Standard algebraic identities recursively applied. Extract sum and product, then ladder up to higher powers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations

### Core Logic |1 - i|^x = 2^x implies (sqrt2)^x = 2^x implies 2^x/2 = 2^x$|1 - i|^x = 2^x \implies (\sqrt{2})^x = 2^x \implies 2^{x/2} = 2^x$ This implies fracx2 = x implies x = 0$\frac{x}{2} = x \implies x = 0$. There is exactly 1 solution, so alpha = 1$\alpha = 1$. ### Step 1: Simplify complex number z (1+i)^4 = ((1+i)^2)^2 = (1 + i^2 + 2i)^2 = (2i)^2 = -4$(1+i)^4 = ((1+i)^2)^2 = (1 + i^2 + 2i)^2 = (2i)^2 = -4$ Thus, z = -pi left( frac(1-sqrtpii)(sqrtpi-i)pi + 1 + frac(sqrtpi-i)(1-sqrtpii)1 + pi right)$z = -\pi \left( \frac{(1-\sqrt{\pi}i)(\sqrt{\pi}-i)}{\pi + 1} + \frac{(\sqrt{\pi}-i)(1-\sqrt{\pi}i)}{1 + \pi} \right)$ ### Step 2: Simplify Bracket Let's expand the terms directly: z = fracpi4(-4) left[ fracsqrtpi - pi i - i - sqrtpipi + 1 + fracsqrtpi - i - pi i - sqrtpi1 + pi right]$z = \frac{\pi}{4}(-4) \left[ \frac{\sqrt{\pi} - \pi i - i - \sqrt{\pi}}{\pi + 1} + \frac{\sqrt{\pi} - i - \pi i - \sqrt{\pi}}{1 + \pi} \right]$ = -pi left[ frac-i(pi+1)pi+1 + frac-i(pi+1)pi+1 right]$= -\pi \left[ \frac{-i(\pi+1)}{\pi+1} + \frac{-i(\pi+1)}{\pi+1} \right]$ = -pi [ -i - i ] = 2pi i$= -\pi [ -i - i ] = 2\pi i$ ### Step 3: Find beta For z = 2pi i$z = 2\pi i$: |z| = 2pi$|z| = 2\pi$ and arg(z) = fracpi2$\arg(z) = \frac{\pi}{2}$. beta = frac|z|arg(z) = frac2pipi/2 = 4$\beta = \frac{|z|}{\arg(z)} = \frac{2\pi}{\pi/2} = 4$ ### Step 4: Distance from Line Distance of point (alpha, beta) = (1, 4)$(\alpha, \beta) = (1, 4)$ from the line 4x - 3y - 7 = 0$4x - 3y - 7 = 0$: D = frac|4(1) - 3(4) - 7|sqrt4^2 + (-3)^2 = frac|4 - 12 - 7|5 = frac|-15|5 = 3$D = \frac{|4(1) - 3(4) - 7|}{\sqrt{4^2 + (-3)^2}} = \frac{|4 - 12 - 7|}{5} = \frac{|-15|}{5} = 3$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations Class 11 Maths: Straight Lines