PYQSide ExamsMy practice

Complex Numbers and Quadratic Equations · Question 32

JEE Main 2026 (Online) 22nd January Evening ShiftMCQMedium📈 Topic in 13% papers+4−1
1

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

  1. 564
  2. 116
  3. 764
  4. 316

Answer

(4) 316

Solution

4z2 = - z 4|z|2 = |z|

|z| = 0 or |z| = 14

|z| = 0: z = 0 is a solution.

|z| = 14: let z = 14ei ; then 4 116e2i = - 14e-i

e3i = -1 = 3, , 5 3 (three distinct values of z)

z S |z|2 = 0 + 3 116 = 316

Answer: (4)

JEE Main 2026 (Online) 22nd January Evening Shift

Keep practising — Complex Numbers and Quadratic Equations

2

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
JEE Main 2026 (Online) 21st January Morning Shift
3

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
JEE Main 2026 (Online) 21st January Morning Shift
4

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, )
JEE Main 2026 (Online) 21st January Evening Shift
5

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
JEE Main 2026 (Online) 21st January Evening Shift
6

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
JEE Main 2025 (Online) 8th April Evening Shift
7

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
JEE Main 2025 (Online) 8th April Evening Shift
8

Let A = \ [0, 2 ] : 1 + 10 Re ( 2 + i - 3i ) = 0 \. Then A 2 is equal to

  1. 214 2
  2. 6 2
  3. 274 2
  4. 8 2
JEE Main 2025 (Online) 8th April Evening Shift
9

Let the set of all values of p R, for which both the roots of the equation x2 - (p + 2)x + (2p + 9) = 0 are negative real numbers, be the interval ( , ]. Then - 2 is equal to

  1. 5
  2. 0
  3. 20
  4. 9
JEE Main 2025 (Online) 7th April Morning Shift
10

Among the statements (S1) : The set \z C - \-i\ : |z| = 1 and z - iz + i is purely real \ contains exactly two elements, and (S2) : The set \z C - \-1\ : |z| = 1 and z - 1z + 1 is purely imaginary \ contains infinitely many elements.

  1. both are incorrect
  2. both are correct
  3. only (S2) is correct
  4. only (S1) is correct
JEE Main 2025 (Online) 7th April Morning Shift
11

If the locus of z C, such that Re ( z-12z+i ) + Re ( z-12 z-i ) = 2, is a circle of radius r and center (a, b), then 15abr2 is equal to :

  1. 16
  2. 24
  3. 12
  4. 18
JEE Main 2025 (Online) 7th April Evening Shift

All Complex Numbers and Quadratic Equations PYQs →

PDFDownload Complex Numbers and Quadratic Equations PYQs as PDF Free, with answer key and a QR code to every solution — print it or study offline⬇
All JEE Main Mathematics chapters