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Complex Numbers and Quadratic Equations

JEE Main 2026 (Online) 23rd January Evening ShiftMCQEasy📈 Topic in 16% papers+4−1
Ques. 1JEE Main 2026, 23 January (Evening Shift)

If z = 32 + i2, i = -1, then (z201 - i )8 is equal to

  1. 1
  2. 0
  3. -1
  4. 256

Answer

(4) 256

Solution

z = 6 + i 6 = ei /6

z201 = ei 201 /6 = ei(32 + 3 /2)

= 3 2 + i 3 2 = -i

z201 - i = -2i

(z201 - i)8 = (-2i)8 = 28 i8 = 256

Answer: (4)

Keep practising — Complex Numbers and Quadratic Equations

Ques. 2JEE Main 2026, 22 January (Morning Shift)

The number of distinct real solutions of the equation x|x+4| + 3|x+2| + 10 = 0 is

  1. 2
  2. 3
  3. 0
  4. 1
Ques. 3JEE Main 2026, 22 January (Morning Shift)

Let = -1 + i 32 and = -1 - i 32, i = -1. If (7 - 7 + 9 )20 + (9 + 7 - 7 )20 + (-7 + 9 + 7 )20 + (14 + 7 + 7 )20 = m10, then m is

Ques. 4JEE Main 2026, 22 January (Evening Shift)

Let , be the roots of the quadratic equation 12x2 - 20x + 3 = 0, Z. If 12 | - | 32, then the sum of all possible values of is :

  1. 3
  2. 6
  3. 1
  4. 4
Ques. 5JEE Main 2026, 22 January (Evening Shift)

Let S = \z C : 4z2 + z = 0\. Then z S |z|2 is equal to:

  1. 564
  2. 116
  3. 764
  4. 316
Ques. 6JEE Main 2026, 21 January (Morning Shift)

If x2 + x + 1 = 0, then the value of (x + 1x )4 + (x2 + 1x2 )4 + (x3 + 1x3 )4 + + (x25 + 1x25 )4 is:

  1. 162
  2. 145
  3. 128
  4. 175
Ques. 7JEE Main 2026, 21 January (Morning Shift)

The sum of all the roots of the equation (x - 1)2 - 5|x - 1| + 6 = 0, is :

  1. 3
  2. 1
  3. 4
  4. 5
Ques. 8JEE Main 2026, 21 January (Evening Shift)

Let and be the roots of the equation x2 + 2ax + (3a + 10) = 0 such that < 1 < . Then the set of all possible values of a is :

  1. (- , -115 ) (5, )
  2. (- , -115 )
  3. (- , -3)
  4. (- , -2) (5, )
Ques. 9JEE Main 2026, 21 January (Evening Shift)

Let z be the complex number satisfying |z - 5| 3 and having maximum positive principal argument. Then 34 | 5z - 125iz + 16 |2 is equal to:

  1. 20
  2. 26
  3. 12
  4. 16
Ques. 10JEE Main 2025, 8 April (Evening Shift)

Let be a solution of x2 + x + 1 = 0, and for some a and b in R, bmatrix 4 & a & b bmatrix bmatrix 1 & 16 & 13 \ -1 & -1 & 2 \ -2 & -14 & -8 bmatrix = bmatrix 0 & 0 & 0 bmatrix. If 4 4 + m a + n b = 3, then m + n is equal to

  1. 11
  2. 3
  3. 8
  4. 7
Ques. 11JEE Main 2025, 8 April (Evening Shift)

The sum of the squares of the roots of |x-2|2 + |x-2| - 2 = 0 and the squares of the roots of x2 - 2|x-3| - 5 = 0, is

  1. 24
  2. 26
  3. 36
  4. 30

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