If x^2 + x + 1 = 0 , then the value of left(mathrmx+frac1mathrmxright)^4+left(mathrmx^2+frac1mathrmx^2right)^4+left(mathrmx^3+frac1mathrmx^3right)^4+ldots+left(mathrmx^25+frac1mathrmx^25right)^4 is :

Solution & Explanation

### Related Formula x^2 + x + 1 = 0 Rightarrow x = omega, omega^2 Properties of cube roots of unity: omega^3 = 1 and 1 + omega + omega^2 = 0. ### Core Logic Let alpha = omega. Then frac1x = frac1omega = omega^2. The series is sum_k=1^25 (omega^k + omega^2k)^4. Evaluate the term T_k = (omega^k + omega^2k)^4 based on the modulo of k with 3. ### Step 1: Cyclic Evaluation Case 1: k = 3m (multiples of 3) T_3m = (omega^3m + omega^6m)^4 = (1 + 1)^4 = 2^4 = 16 There are 8 such multiples up to 25 (3, 6, 9, dots, 24). Case 2: k neq 3m (non-multiples of 3) For k = 1, 2, 4, 5, dots omega^k + omega^2k will always be omega + omega^2 or omega^2 + omega. Since 1 + omega + omega^2 = 0, we have omega + omega^2 = -1. T_k neq 3m = (-1)^4 = 1 There are 25 - 8 = 17 such non-multiples up to 25. ### Step 2: Total Sum textSum = 17 times 1 + 8 times 16 = 17 + 128 = 145 ### Pattern Recognition Powers of x+1/x when x solves x^2 pm x + 1 = 0 perfectly orbit around periods of 3 or 6. Isolate the 3m resonant beats (which hit pure scalars like 1+1=2) versus the out-of-phase beats (which collapse to -1 or 1 via basic omega identities). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations

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Complex Numbers And Quadratic Equations

More Complex Numbers and Quadratic Equations Previous-Year Questions — Page 6

Q23 jee_main_2024_29_jan_morning Higher Powers of Roots
Let alpha, beta be the roots of the equation x^2-x+2=0 with Im(alpha) gt Im(beta). Then alpha^6+alpha^4+beta^4-5alpha^2 is equal to
Numerical Answer. Answer: 13 to 13

Solution

### Related Formula Since alpha is a root of x^2 - x + 2 = 0, it must satisfy the equation exactly: 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. Use the substitution alpha^2 = alpha - 2 to iteratively depress the powers of alpha. alpha^4 = (alpha^2)^2 = (alpha - 2)^2 = alpha^2 - 4alpha + 4 Substitute alpha^2 = alpha - 2 again into the result: alpha^4 = (alpha - 2) - 4alpha + 4 = -3alpha + 2 Now, generate alpha^6 using alpha^4: alpha^6 = alpha^4 cdot alpha^2 = (-3alpha + 2)(alpha - 2) = -3alpha^2 + 6alpha + 2alpha - 4 = -3alpha^2 + 8alpha - 4 Substitute alpha^2 = alpha - 2 into the result again: alpha^6 = -3(alpha - 2) + 8alpha - 4 = -3alpha + 6 + 8alpha - 4 = 5alpha + 2 ### Step 1: Simplify the Full Expression The symmetry of the roots dictates that beta^4 behaves identically to alpha^4. Thus: beta^4 = -3beta + 2 Substitute all depressed linear forms back into E = alpha^6 + alpha^4 + beta^4 - 5alpha^2: E = (5alpha + 2) + (-3alpha + 2) + (-3beta + 2) - 5(alpha - 2) E = 5alpha - 3alpha - 5alpha - 3beta + 2 + 2 + 2 + 10 E = -3alpha - 3beta + 16 E = -3(alpha + beta) + 16 ### Step 2: Apply Sum of Roots From the original quadratic equation x^2 - x + 2 = 0, the sum of roots is: alpha + beta = -frac-11 = 1 Substitute this back: E = -3(1) + 16 = 13 (Note: The condition Im(alpha) gt Im(beta) was a distractor since the expression simplified perfectly symmetrically into 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 or i. Always use the characteristic quadratic relation alpha^2 = palpha + 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
Q16 jee_main_2024_30_january_evening Roots of Unity
If z is a complex number, then the number of common roots of the equation z^1985 + z^100 + 1 = 0 and z^3 + 2z^2 + 2z + 1 = 0 , is equal to:
  • A. 1
  • B. 2
  • C. 0
  • D. 3

Solution

### Related Formula z^3-1 = (z-1)(z^2+z+1) quad textCube roots of unity: 1, omega, omega^2 omega^3 = 1 quad textand 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 Group the terms: (z^3 + 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) = 0 Thus, the roots are z = -1, and the roots of z^2 + z + 1 = 0 which are z = omega, omega^2. ### Step 1: Check z = -1 Substitute z = -1 into the first equation z^1985 + z^100 + 1 = 0: (-1)^1985 + (-1)^100 + 1 = -1 + 1 + 1 = 1 neq 0 So, z = -1 is not a common root. ### Step 2: Check z = ω and ω² Substitute z = omega: omega^1985 + omega^100 + 1 Reduce the powers modulo 3 (since omega^3 = 1): 1985 = 3 times 661 + 2 Rightarrow omega^1985 = omega^2 100 = 3 times 33 + 1 Rightarrow omega^100 = omega^1 = omega So, omega^1985 + omega^100 + 1 = omega^2 + omega + 1 = 0. z = omega is a common root. Substitute z = omega^2: (omega^2)^1985 + (omega^2)^100 + 1 = omega^3970 + omega^200 + 1 3970 = 3 times 1323 + 1 Rightarrow omega^3970 = omega 200 = 3 times 66 + 2 Rightarrow omega^200 = omega^2 So, omega^3970 + omega^200 + 1 = omega + omega^2 + 1 = 0. z = omega^2 is also a common root. ### Step 3: Conclusion There are exactly 2 common roots: omega and omega^2. ### Pattern Recognition Whenever z^2+z+1 emerges as a factor, its roots omega and 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
Q3 jee_main_2024_30_jan_morning Modulus and Conjugate of a Complex Number
If z = x + iy, xy neq 0, satisfies the equation z^2 + ibarz = 0, then |z^2| is equal to:
  • A. 9
  • B. 1
  • C. 4
  • D. frac14

Solution

### Related Formula |z^n| = |z|^n |barz| = |z| ### Core Logic Given the equation: z^2 = -ibarz Taking the modulus on both sides: |z^2| = |-ibarz| Using properties of modulus: |z|^2 = |-i| cdot |barz| Since |-i| = 1 and |barz| = |z|: |z|^2 = |z| |z|^2 - |z| = 0 |z|(|z| - 1) = 0 ### Step 1: Applying the non-zero condition This gives two possibilities: |z| = 0 or |z| = 1. Since z = x + iy and xy neq 0, neither x nor y is zero, which implies |z| neq 0. Therefore, |z| = 1. We are asked for |z^2|: |z^2| = |z|^2 = 1^2 = 1 ### Pattern Recognition When dealing with equations involving z and barz, applying modulus to both sides rapidly simplifies the problem, turning complex equations into simple real algebraic equations in |z|. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations
Q28 jee_main_2024_30_jan_morning Quadratic Equations
Let alpha, beta in mathbbN be roots of equation x^2 - 70x + lambda = 0 where fraclambda2, fraclambda3 notin mathbbN. If lambda assumes the minimum possible value, then frac(sqrtalpha - 1 + sqrtbeta - 1)(lambda + 35)|alpha - beta| is equal to:
Numerical Answer. Answer: 60 to 60

Solution

### Related Formula textSum of roots (alpha + beta) = -fracba textProduct of roots (alpha beta) = fracca ### Core Logic For the equation x^2 - 70x + lambda = 0: alpha + beta = 70 alpha beta = lambda Since alpha, beta in mathbbN, their sum is 70. This gives alpha(70 - alpha) = lambda. We are given lambda / 2 notin mathbbN and lambda / 3 notin mathbbN. This means lambda is not divisible by 2 or 3. So lambda must be an odd number and not a multiple of 3. Thus, neither alpha nor beta can be a multiple of 2 or 3. ### Step 1: Finding minimum lambda We need to find the minimum value of lambda. lambda = alpha(70 - alpha). This parabola opens downwards, so the minimum product occurs when the integers alpha and beta are as far apart as possible. Let's test small values for alpha: If alpha = 1, beta = 69 Rightarrow lambda = 69, but 69 = 3 times 23, which is divisible by 3. Rejected. If alpha = 2, divisible by 2. Rejected. If alpha = 3, divisible by 3. Rejected. If alpha = 4, divisible by 2. Rejected. If alpha = 5, beta = 65 Rightarrow lambda = 325. Check divisibility: 325 is odd (not div by 2). 3+2+5 = 10 (not div by 3). So minimum lambda = 325 with roots alpha = 5, beta = 65. ### Step 2: Evaluating the target expression We need to compute: E = frac(sqrtalpha - 1 + sqrtbeta - 1)(lambda + 35)|alpha - beta| Substitute the values alpha = 5, beta = 65, lambda = 325: |alpha - beta| = |5 - 65| = 60 sqrtalpha - 1 = sqrt4 = 2 sqrtbeta - 1 = sqrt64 = 8 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, %3 neq 0). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations
Q3 jee_main_2024_31_jan_evening Algebra of Complex Numbers
Let z_1 and z_2 be two complex numbers such that z_1 + z_2 = 5 and z_1^3 + z_2^3 = 20 + 15i. Then |z_1^4 + z_2^4| equals-
  • A. 30sqrt3
  • B. 75
  • C. 15sqrt15
  • D. 25sqrt3

Solution

### Related Formula a^3+b^3 = (a+b)^3 - 3ab(a+b) a^2+b^2 = (a+b)^2 - 2ab a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2 ### Core Logic Given z_1+z_2=5 and 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) 20+15i = 125 - 15z_1z_2 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+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_1^2+z_2^2)^2 - 2(z_1z_2)^2 = (11+2i)^2 - 2(7-i)^2 = (121 - 4 + 44i) - 2(49 - 1 - 14i) = 117 + 44i - 2(48 - 14i) = 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 ### 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

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