AktiraAktira JEE
Maths chapters

Complex Numbers

18 JEE Main questions · Maths

  1. MathsJEE Main 21 Jan 2026, Shift 1·Q17·MCQ·Moderate
    Open

    Cube roots of unity

    If x2+x+1=0x^2 + x + 1 = 0, then the value of (x+1x)4+(x2+1x2)4+(x3+1x3)4+…+(x25+1x25)4\left(x + \frac1x\right)^4 + \left(x^2 + \frac{1}{x^2}\right)^4 + \left(x^3 + \frac{1}{x^3}\right)^4 + \ldots + \left(x^{25} + \frac{1}{x^{25}}\right)^4 is:
    1. (1)128
    2. (2)162
    3. (3)175
    4. (4)145
  2. MathsJEE Main 21 Jan 2026, Shift 2·Q19·MCQ·Moderate
    Open

    Maximum argument in a disc

    Let zz be the complex number satisfying ∣z−5∣≤3\lvert z - 5\rvert \le 3 and having maximum positive principal argument. Then 34∣5z−125iz+16∣234\left\lvert\dfrac{5z - 12}{5iz + 16}\right\rvert^2 is equal to:
    1. (1)16
    2. (2)12
    3. (3)26
    4. (4)20
  3. MathsJEE Main 22 Jan 2026, Shift 1·Q21·Numerical·Moderate
    Open

    Cube roots of unity

    Let α=−1+i32\alpha = \dfrac{-1 + i\sqrt3}{2} and β=−1−i32\beta = \dfrac{-1 - i\sqrt3}{2}, i=−1i = \sqrt{-1}. If (7−7α+9β)20+(9+7α−7β)20+(−7+9α+7β)20+(14+7α+7β)20=m10,(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, then mm is .
  4. MathsJEE Main 22 Jan 2026, Shift 2·Q4·MCQ·Moderate
    Open

    Solving equations in conjugate

    Let S={z∈C:4z2+zˉ=0}S = \left\{z \in \mathbb{C} : 4z^2 + \bar{z} = 0\right\}. Then ∑z∈S∣z∣2\displaystyle\sum_{z\in S}\lvert z\rvert^2 is equal to:
    1. (1)316\dfrac{3}{16}
    2. (2)764\dfrac{7}{64}
    3. (3)116\dfrac{1}{16}
    4. (4)564\dfrac{5}{64}
  5. MathsJEE Main 23 Jan 2026, Shift 1·Q17·MCQ·Moderate
    Open

    Locus and minimum modulus

    Let S={z:3≤∣2z−3(1+i)∣≤7}S = \{z : 3 \le \lvert 2z - 3(1 + i)\rvert \le 7\} be a set of complex numbers. Then min⁡z∈S∣(z+12(5+3i))∣\displaystyle\min_{z\in S}\left\lvert\left(z + \frac12(5 + 3i)\right)\right\rvert is equal to:
    1. (1)12\dfrac12
    2. (2)32\dfrac32
    3. (3)2
    4. (4)52\dfrac52
  6. MathsJEE Main 23 Jan 2026, Shift 2·Q17·MCQ·Easy–moderate
    Open

    Powers in polar form

    If z=32+i2z = \dfrac{\sqrt3}{2} + \dfrac{i}{2}, i=−1i = \sqrt{-1}, then (z201−i)8(z^{201} - i)^8 is equal to
    1. (1)−1-1
    2. (2)0
    3. (3)1
    4. (4)256
  7. MathsJEE Main 24 Jan 2026, Shift 1·Q9·MCQ·Moderate
    Open

    Locus of complex numbers

    Let S={z∈C:∣z−6iz−2i∣=1 and ∣z−8+2iz+2i∣=35}S = \left\{z \in \mathbb{C} : \left\lvert\dfrac{z - 6i}{z - 2i}\right\rvert = 1 \text{ and } \left\lvert\dfrac{z - 8 + 2i}{z + 2i}\right\rvert = \dfrac35\right\}. Then ∑z∈S∣z∣2\displaystyle\sum_{z \in S}\lvert z\rvert^2 is equal to
    1. (1)398
    2. (2)413
    3. (3)423
    4. (4)385
    We corrected a printing error in the original paper

    Source printed the summation index as “z∈sz \in s” (lower-case ss; the set is SS).

  8. MathsJEE Main 24 Jan 2026, Shift 2·Q23·Numerical·Easy–moderate
    Open

    Modulus of a product

    Let z=(1+i)(1+2i)(1+3i)…(1+ni)z = (1 + i)(1 + 2i)(1 + 3i)\ldots(1 + ni), where i=−1i = \sqrt{-1}. If ∣z∣2=44200\lvert z\rvert^2 = 44200, then nn is equal to .
  9. MathsJEE Main 28 Jan 2026, Shift 1·Q5·MCQ·Moderate
    Open

    Modulus and touching circles

    Let zz be a complex number such that ∣z−6∣=5\lvert z - 6\rvert = 5 and ∣z+2−6i∣=5\lvert z + 2 - 6i\rvert = 5. Then the value of z3+3z2−15z+141z^3 + 3z^2 - 15z + 141 is equal to
    1. (1)42
    2. (2)37
    3. (3)50
    4. (4)61
  10. MathsJEE Main 28 Jan 2026, Shift 2·Q18·MCQ·Moderate
    Open

    Locus and maximum modulus

    Let A={z∈C:∣z−2∣≤4}andB={z∈C:∣z−2∣+∣z+2∣=5}.A = \{z \in \mathbb{C} : \lvert z - 2\rvert \le 4\} \quad\text{and}\quad B = \{z \in \mathbb{C} : \lvert z - 2\rvert + \lvert z + 2\rvert = 5\}. Then the max⁡{∣z1−z2∣:z1∈A and z2∈B}\max\{\lvert z_1 - z_2\rvert : z_1 \in A \text{ and } z_2 \in B\} is
    1. (1)152\dfrac{15}{2}
    2. (2)8
    3. (3)172\dfrac{17}{2}
    4. (4)9
  11. MathsJEE Main 2 Apr 2026, Shift 1·Q2·MCQ·Easy–moderate
    Open

    Equality of complex numbers

    Let xx and yy be real numbers such that 50(2x1+3i−y1−2i)=31+17i,i=−1.50\left(\dfrac{2x}{1+3i} - \dfrac{y}{1-2i}\right) = 31 + 17i,\quad i = \sqrt{-1}. Then the value of 10(x−3y)10(x - 3y) is:
    1. (1)20
    2. (2)31
    3. (3)35
    4. (4)75
    We corrected a printing error in the original paper

    Source printed “Let xx and yy be real number”.

  12. MathsJEE Main 2 Apr 2026, Shift 2·Q2·MCQ·Moderate
    Open

    Circles in the complex plane

    Let the circles C1:∣z∣=rC_1 : \lvert z\rvert = r and C2:∣z−3−4i∣=5C_2 : \lvert z - 3 - 4i\rvert = 5, z∈Cz \in \mathbb{C}, be such that C2C_2 lies within C1C_1. If z1z_1 moves on C1C_1, z2z_2 moves on C2C_2 and min⁡∣z1−z2∣=2\min\lvert z_1 - z_2\rvert = 2, then max⁡∣z1−z2∣\max\lvert z_1 - z_2\rvert is equal to:
    1. (1)12
    2. (2)17
    3. (3)22
    4. (4)24
    We corrected a printing error in the original paper

    Source printed “Let the circle C1C_1 … and C2C_2 …” (singular for two circles).

  13. MathsJEE Main 4 Apr 2026, Shift 1·Q3·MCQ·Moderate
    Open

    Locus and argument of complex numbers

    Let zz be a complex number such that ∣z+2∣=∣z−2∣\lvert z + 2\rvert = \lvert z - 2\rvert and arg⁡(z+3z−i)=π4\arg\left(\dfrac{z + 3}{z - i}\right) = \dfrac{\pi}{4}. Then ∣z∣2\lvert z\rvert^2 is equal to:
    1. (1)9
    2. (2)4
    3. (3)5
    4. (4)1
  14. MathsJEE Main 4 Apr 2026, Shift 2·Q2·MCQ·Easy–moderate
    Open

    Modulus of complex numbers

    Let S={z∈C:z2+4z+16=0}S = \{z \in \mathbb{C} : z^2 + 4z + 16 = 0\}. Then ∑z∈S∣z+3 i∣2\displaystyle\sum_{z\in S}\left\lvert z + \sqrt3\,i\right\rvert^2 is equal to:
    1. (1)42
    2. (2)23
    3. (3)27
    4. (4)38
  15. MathsJEE Main 5 Apr 2026, Shift 2·Q2·MCQ·Easy–moderate
    Open

    Quadratic with complex coefficients

    Let z1,z2∈Cz_1, z_2 \in \mathbb{C} be the distinct solutions of the equation z2+4z−(1+12i)=0z^2 + 4z - (1 + 12i) = 0. Then ∣z1∣2+∣z2∣2\lvert z_1\rvert^2 + \lvert z_2\rvert^2 is equal to:
    1. (1)18
    2. (2)22
    3. (3)29
    4. (4)34
  16. MathsJEE Main 6 Apr 2026, Shift 1·Q4·MCQ·Moderate
    Open

    Complex equation with real parameter

    Let the set of all values of k∈Rk \in \mathbb{R} such that the equation z(zˉ+2+i)+k(2+3i)=0z(\bar{z} + 2 + i) + k(2 + 3i) = 0, z∈Cz \in \mathbb{C}, has at least one solution, be the interval [α,β][\alpha, \beta]. Then 9(α+β)9(\alpha + \beta) is equal to:
    1. (1)−10-10
    2. (2)−8-8
    3. (3)101310\sqrt{13}
    4. (4)8138\sqrt{13}
  17. MathsJEE Main 6 Apr 2026, Shift 2·Q3·MCQ·Easy–moderate
    Open

    Roots and powers of complex numbers

    Let S={z∈C:z2+6 iz−3=0}S = \left\{z \in \mathbb{C} : z^2 + \sqrt6\,iz - 3 = 0\right\}. Then ∑z∈Sz8\displaystyle\sum_{z\in S} z^8 is equal to:
    1. (1)162
    2. (2)184
    3. (3)262
    4. (4)324
  18. MathsJEE Main 8 Apr 2026, Shift 2·Q2·MCQ·Moderate
    Open

    Locus of complex numbers

    The number of values of z∈Cz \in \mathbb{C}, satisfying the equations ∣z−(4+8i)∣=10\lvert z - (4 + 8i)\rvert = \sqrt{10} and ∣z−(3+5i)∣+∣z−(5+11i)∣=45\lvert z - (3 + 5i)\rvert + \lvert z - (5 + 11i)\rvert = 4\sqrt5, is:
    1. (1)0
    2. (2)2
    3. (3)1
    4. (4)4