Jarvis OS
PYQ papersPricingSign inGet started
  1. Home
  2. /JEE Mathematics PYQs
  3. /Complex Numbers

Complex Numbers — JEE Previous Year Questions

Every Complex Numbers question asked in JEE Main and JEE Advanced across the last 186 papers — 191 questions, each with its correct answer. Free to read, no account needed.

Questions

191

Papers it appeared in

165/186

Appearance rate

89%

All 191 Complex Numbers questions

Most recent papers first.

Q1·MathematicsSingle correctJEE Main 2026
The number of values of z∈Cz \in \mathbb{C}z∈C, satisfying the equations ∣z−(4+8i)∣=10|z - (4 + 8i)| = \sqrt{10}∣z−(4+8i)∣=10​ and ∣z−(3+5i)∣+∣z−(5+11i)∣=45|z - (3 + 5i)| + |z - (5 + 11i)| = 4\sqrt{5}∣z−(3+5i)∣+∣z−(5+11i)∣=45​, is:
  1. (A)0
  2. (B)2
  3. (C)1
  4. (D)4

Correct answer: (B)

Step-by-step solution →
Q2·MathematicsSingle correctJEE Main 2026
Let the set of all values of k ∈ ℝ such that the equation z(zˉ+2+i)+k(2+3i)=0z(\bar{z} + 2 + i) + k(2 + 3i) = 0z(zˉ+2+i)+k(2+3i)=0, z ∈ ℂ, has at least one solution, be the interval [α, β]. Then 9(α + β) is equal to:
  1. (A)−10
  2. (B)−8
  3. (C)10√13
  4. (D)8√13

Correct answer: (A)

Step-by-step solution →
Q3·MathematicsSingle correctJEE Main 2026
Let S={z∈C:z2+6 iz−3=0}S = \{z \in \mathbb{C} : z^{2} + \sqrt{6}\,iz - 3 = 0\}S={z∈C:z2+6​iz−3=0}. Then ∑z∈Sz8\sum_{z \in S} z^{8}∑z∈S​z8 is equal to :
  1. (A)162
  2. (B)184
  3. (C)262
  4. (D)324

Correct answer: (A)

Step-by-step solution →
Q4·MathematicsSingle correctJEE Main 2026
Let z1,z2z_1, z_2z1​,z2​ ∈ ℂ be the distinct solutions of the equation z2+4z−(1+12i)=0z^2 + 4z - (1 + 12i) = 0z2+4z−(1+12i)=0. Then ∣z1∣2+∣z2∣2|z_1|^2 + |z_2|^2∣z1​∣2+∣z2​∣2 is equal to :
  1. (A)18
  2. (B)22
  3. (C)29
  4. (D)34

Correct answer: (D)

Step-by-step solution →
Q5·MathematicsSingle correctJEE Main 2026
Let zzz be a complex number such that ∣z+2∣=∣z−2∣|z + 2| = |z - 2|∣z+2∣=∣z−2∣ and arg⁡(z+3z−i)=π4\arg\left(\frac{z + 3}{z - i}\right) = \frac{\pi}{4}arg(z−iz+3​)=4π​. Then ∣z∣2|z|^2∣z∣2 is equal to:
  1. (A)999
  2. (B)444
  3. (C)555
  4. (D)111

Correct answer: (A)

Step-by-step solution →
Q6·MathematicsSingle correctJEE Main 2026
Let S={z∈C:z2+4z+16=0}S = \{z \in \mathbb{C} : z^{2} + 4z + 16 = 0\}S={z∈C:z2+4z+16=0}. Then ∑z∈S∣z+3i∣2\sum_{z \in S} |z + \sqrt{3}i|^{2}∑z∈S​∣z+3​i∣2 is equal to:
  1. (A)42
  2. (B)23
  3. (C)27
  4. (D)38

Correct answer: (D)

Step-by-step solution →
Q7·MathematicsSingle correctJEE Main 2026
If α=1\alpha = 1α=1 and β=1+i2\beta = 1 + i\sqrt{2}β=1+i2​, where i=−1i = \sqrt{-1}i=−1​ are two roots of the equation x3+ax2+bx+c=0x^{3} + ax^{2} + bx + c = 0x3+ax2+bx+c=0, a,b,c∈Ra, b, c \in \mathbb{R}a,b,c∈R, then ∫−11(x3+ax2+bx+c)dx\int_{-1}^{1}(x^{3} + ax^{2} + bx + c)dx∫−11​(x3+ax2+bx+c)dx is equal to:
  1. (A)−2
  2. (B)−4
  3. (C)−8
  4. (D)−10

Correct answer: (C)

Step-by-step solution →
Q8·MathematicsSingle correctJEE Main 2026
Let the circles C1:∣z∣=rC_1 : |z| = rC1​:∣z∣=r and C2:∣z−3−4i∣=5C_2 : |z - 3 - 4i| = 5C2​:∣z−3−4i∣=5, z∈Cz \in \mathbb{C}z∈C, be such that C2C_2C2​ lies within C1C_1C1​. If z1z_1z1​ moves on C1C_1C1​, z2z_2z2​ moves on C2C_2C2​ and min⁡∣z1−z2∣=2\min |z_1 - z_2| = 2min∣z1​−z2​∣=2, then max⁡∣z1−z2∣\max |z_1 - z_2|max∣z1​−z2​∣ is equal to:
  1. (A)121212
  2. (B)171717
  3. (C)222222
  4. (D)242424

Correct answer: (C)

Step-by-step solution →
Q9·MathematicsSingle correctJEE Main 2026
Let x and y be real numbers such that 50(2x1+3i−y1−2i)=31+17i50\left(\frac{2x}{1+3i}-\frac{y}{1-2i}\right)=31+17i50(1+3i2x​−1−2iy​)=31+17i, i=−1i=\sqrt{-1}i=−1​. Then the value of 10(x−3y)10(x-3y)10(x−3y) is :
  1. (A)20
  2. (B)31
  3. (C)35
  4. (D)75

Correct answer: (D)

Step-by-step solution →
Q10·MathematicsSingle correctJEE Main 2026
Let A={z∈C:∣z−2∣≤4}A = \left\{z \in \mathbb{C} : |z - 2| \leq 4\right\}A={z∈C:∣z−2∣≤4} and B={z∈C:∣z−2∣+∣z+2∣=5}B = \left\{z \in \mathbb{C} : |z - 2| + |z + 2| = 5\right\}B={z∈C:∣z−2∣+∣z+2∣=5}. Then the max {∣z1−z2∣:z1∈A and z2∈B}\left\{|z_1 - z_2| : z_1 \in A \text{ and } z_2 \in B\right\}{∣z1​−z2​∣:z1​∈A and z2​∈B} is
  1. (A)152\frac{15}{2}215​
  2. (B)8
  3. (C)172\frac{17}{2}217​
  4. (D)9

Correct answer: (C)

Step-by-step solution →
Q11·MathematicsSingle correctJEE Main 2026
Let zzz be a complex number such that ∣z−6∣=5|z - 6| = 5∣z−6∣=5 and ∣z+2−6i∣=5|z + 2 - 6i| = 5∣z+2−6i∣=5. Then the value of z3+3z2−15z+141z^3 + 3z^2 - 15z + 141z3+3z2−15z+141 is equal to
  1. (A)42
  2. (B)37
  3. (C)50
  4. (D)61

Correct answer: (C)

Step-by-step solution →
Q12·MathematicsNumericalJEE Main 2026
Let z=(1+i)(1+2i)(1+3i)…(1+ni)z = (1 + i)(1 + 2i)(1 + 3i) \ldots (1 + ni)z=(1+i)(1+2i)(1+3i)…(1+ni), where i=−1i = \sqrt{-1}i=−1​. If ∣z∣2=44200|z|^{2} = 44200∣z∣2=44200, then n is equal to −-−

Correct answer: 5

Step-by-step solution →
Q13·MathematicsSingle correctJEE Main 2026
Let S={z∈C:∣z−6iz−2i∣=1 and ∣z−8+2iz+2i∣=35}S=\left\{z\in\mathbb{C}:\left|\frac{z-6i}{z-2i}\right|=1\ \text{and}\ \left|\frac{z-8+2i}{z+2i}\right|=\frac{3}{5}\right\}S={z∈C:​z−2iz−6i​​=1 and ​z+2iz−8+2i​​=53​} . Then ∑z∈s∣z∣2\sum_{z\in s}|z|^{2}∑z∈s​∣z∣2 is equal to
  1. (A)398398398
  2. (B)413413413
  3. (C)423423423
  4. (D)385385385

Correct answer: (D)

Step-by-step solution →
Q14·MathematicsSingle correctJEE Main 2026
If z=32+i2,i=−1z = \frac{\sqrt{3}}{2} + \frac{i}{2}, i = \sqrt{-1}z=23​​+2i​,i=−1​, then (z201−i)8(z^{201} - i)^{8}(z201−i)8 is equal to
  1. (A)–1
  2. (B)0
  3. (C)1
  4. (D)256

Correct answer: (D)

Step-by-step solution →
Q15·MathematicsSingle correctJEE Main 2026
Let S={z:3≤∣2z−3(1+i)∣≤7}S = \{z : 3 \le | 2z - 3(1 + i)| \le 7\}S={z:3≤∣2z−3(1+i)∣≤7} be a set of complex numbers. Then Minz∈S ∣(z+12(5+3i))∣\underset{z \in S}{\mathrm{Min}}\ \left|\left(z + \frac{1}{2}(5 + 3i)\right)\right|z∈SMin​ ​(z+21​(5+3i))​ is equal to :
  1. (A)12\frac{1}{2}21​
  2. (B)32\frac{3}{2}23​
  3. (C)2
  4. (D)52\frac{5}{2}25​

Correct answer: (B)

Step-by-step solution →
Q16·MathematicsNumericalJEE Main 2026
Let α=−1+i32\alpha = \frac{-1+i\sqrt{3}}{2}α=2−1+i3​​ and β=−1−i32\beta = \frac{-1-i\sqrt{3}}{2}β=2−1−i3​​, i=−1i = \sqrt{-1}i=−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}(7−7α+9β)20+(9+7α−7β)20+(−7+9α+7β)20+(14+7α+7β)20=m10, then m is _________.

Correct answer: 49

Step-by-step solution →
Q17·MathematicsSingle correctJEE Main 2026
Let S={z∈C:4z2+zˉ=0}S = \left\{ z \in \mathbb{C} : 4z^{2} + \bar{z} = 0 \right\}S={z∈C:4z2+zˉ=0}. Then ∑z∈S∣z∣2\sum\limits_{z \in S} |z|^{2}z∈S∑​∣z∣2 is equal to :
  1. (A)316\frac{3}{16}163​
  2. (B)764\frac{7}{64}647​
  3. (C)116\frac{1}{16}161​
  4. (D)564\frac{5}{64}645​

Correct answer: (A)

Step-by-step solution →
Q18·MathematicsSingle correctJEE Main 2026
If x2+x+1=0x^{2}+x+1=0x2+x+1=0, then the value of (x+1x)4+(x2+1x2)4+(x3+1x3)4+...+(x25+1x25)4\left(x+\frac{1}{x}\right)^{4}+\left(x^{2}+\frac{1}{x^{2}}\right)^{4}+\left(x^{3}+\frac{1}{x^{3}}\right)^{4}+...+\left(x^{25}+\frac{1}{x^{25}}\right)^{4}(x+x1​)4+(x2+x21​)4+(x3+x31​)4+...+(x25+x251​)4 is :
  1. (A)128
  2. (B)162
  3. (C)175
  4. (D)145

Correct answer: (D)

Step-by-step solution →
Q19·MathematicsSingle correctJEE Main 2026
Let z be the complex number satisfying ∣z−5∣≤3|z - 5| \le 3∣z−5∣≤3 and having maximum positive principal argument. Then 34∣5z−125iz+16∣234\left|\frac{5z - 12}{5iz + 16}\right|^{2}34​5iz+165z−12​​2 is equal to:
  1. (A)16
  2. (B)12
  3. (C)26
  4. (D)20

Correct answer: (D)

Step-by-step solution →
Q20·MathematicsNumericalJEE Advanced 2025
For a non-zero complex number z, let arg(z) denote the principal argument of z, with −π < arg(z) ≤ π. Let ω be the cube root of unity for which 0 < arg(ω) < π. Let α=arg⁡(∑n=12025(−ω)n)\alpha = \arg\left( \sum_{n=1}^{2025} (-\omega)^n \right)α=arg(∑n=12025​(−ω)n). Then the value of 3απ\frac{3\alpha}{\pi}π3α​ is ______

Correct answer: -2

Step-by-step solution →
Q21·MathematicsMultiple correctJEE Advanced 2025
Let R denote the set of all real numbers. Let z1=1+2iz_1 = 1 + 2iz1​=1+2i and z2=3iz_2 = 3iz2​=3i be two complex numbers, where i=−1i = \sqrt{-1}i=−1​. Let S={(x,y)∈R×R:∣x+iy−z1∣=2∣x+iy−z2∣}S = \{(x, y) \in R \times R : |x + iy - z_1| = 2|x + iy - z_2|\}S={(x,y)∈R×R:∣x+iy−z1​∣=2∣x+iy−z2​∣}. Then which of the following statements is(are) TRUE ?
  1. (A)S is a circle with centre (−13,103)\left(-\frac{1}{3}, \frac{10}{3}\right)(−31​,310​)
  2. (B)S is a circle with centre (13,83)\left(\frac{1}{3}, \frac{8}{3}\right)(31​,38​)
  3. (C)S is a circle with radius 23\frac{\sqrt{2}}{3}32​​
  4. (D)S is a circle with radius 223\frac{2\sqrt{2}}{3}322​​

Correct answer: (A), (D)

Step-by-step solution →
Q22·MathematicsSingle correctJEE Main 2025
Let A={θ∈[0,2π]:1+10 Re(2cos⁡θ+isin⁡θcos⁡θ−3isin⁡θ)=0}A=\left\{\theta\in[0,2\pi]:1+10\,\mathrm{Re}\left(\dfrac{2\cos\theta+i\sin\theta}{\cos\theta-3i\sin\theta}\right)=0\right\}A={θ∈[0,2π]:1+10Re(cosθ−3isinθ2cosθ+isinθ​)=0}. Then ∑θ∈Aθ2\sum_{\theta\in A}\theta^2∑θ∈A​θ2 is equal to:
  1. (A)214π2\dfrac{21}{4}\pi^2421​π2
  2. (B)8π28\pi^28π2
  3. (C)274π2\dfrac{27}{4}\pi^2427​π2
  4. (D)6π26\pi^26π2

Correct answer: (A)

Step-by-step solution →
Q23·MathematicsSingle correctJEE Main 2025
Let the locus of z∈Cz\in\mathbb{C}z∈C, such that Re(z−12z+i)+Re(zˉ−12zˉ−i)=2\mathrm{Re}\left(\dfrac{z-1}{2z+i}\right)+\mathrm{Re}\left(\dfrac{\bar{z}-1}{2\bar{z}-i}\right)=2Re(2z+iz−1​)+Re(2zˉ−izˉ−1​)=2, is a circle of radius rrr and center (a,b)(a,b)(a,b), then 15abr2\dfrac{15ab}{r^2}r215ab​ is equal to:
  1. (A)24
  2. (B)12
  3. (C)18
  4. (D)16

Correct answer: (C)

Step-by-step solution →
Q24·MathematicsSingle correctJEE Main 2025
Among the statements (S1): The set {z∈C−{−i}:∣z∣=1\{z\in\mathbb{C}-\{-i\}:|z|=1{z∈C−{−i}:∣z∣=1 and z−iz+i\dfrac{z-i}{z+i}z+iz−i​ is purely real}\}} contains exactly two elements, and (S2): The set {z∈C−{−1}:∣z∣=1\{z\in\mathbb{C}-\{-1\}:|z|=1{z∈C−{−1}:∣z∣=1 and z−1z+1\dfrac{z-1}{z+1}z+1z−1​ is purely imaginary}\}} contains infinitely many elements.
  1. (A)both are incorrect
  2. (B)only (S1) is correct
  3. (C)only (S2) is correct
  4. (D)both are correct

Correct answer: (C)

Step-by-step solution →
Q25·MathematicsIntegerJEE Main 2025
Let A={z∈C:∣z−2−i∣=3}A=\{z\in\mathbb{C}:|z-2-i|=3\}A={z∈C:∣z−2−i∣=3}, B={z∈C:Re(z−iz)=2}B=\{z\in\mathbb{C}:\mathrm{Re}(z-iz)=2\}B={z∈C:Re(z−iz)=2} and S=A∩BS=A\cap BS=A∩B. Then ∑z∈S∣z∣2\displaystyle\sum_{z\in S}|z|^2z∈S∑​∣z∣2 is equal to ______.

Correct answer: 22

Step-by-step solution →
Q26·MathematicsIntegerJEE Main 2025
If α\alphaα is a root of the equation x2+x+1=0x^2+x+1=0x2+x+1=0 and ∑k=1n(αk+1αk)2=20\displaystyle\sum_{k=1}^{n}\left(\alpha^k+\dfrac{1}{\alpha^k}\right)^2=20k=1∑n​(αk+αk1​)2=20, then nnn is equal to ______.

Correct answer: 11

Step-by-step solution →
Q27·MathematicsSingle correctJEE Main 2025
If z1,z2,z3∈Cz_1,z_2,z_3\in\mathbb{C}z1​,z2​,z3​∈C are the vertices of an equilateral triangle, whose centroid is z0z_0z0​, then ∑k=13(zk−z0)2\displaystyle\sum_{k=1}^{3}(z_k-z_0)^2k=1∑3​(zk​−z0​)2 is equal to:
  1. (A)0
  2. (B)1
  3. (C)iii
  4. (D)−i-i−i

Correct answer: (A)

Step-by-step solution →
Q28·MathematicsSingle correctJEE Main 2025
Let z∈Cz\in\mathbb{C}z∈C be such that z2+3iz−2+i=2+3i\dfrac{z^2+3i}{z-2+i}=2+3iz−2+iz2+3i​=2+3i. Then the sum of all possible values of z2z^2z2 is:
  1. (A)19−2i19-2i19−2i
  2. (B)−19−2i-19-2i−19−2i
  3. (C)19+2i19+2i19+2i
  4. (D)−19+2i-19+2i−19+2i

Correct answer: (B)

Step-by-step solution →
Q29·MathematicsSingle correctJEE Main 2025
Let zzz be a complex number such that ∣z∣=1|z|=1∣z∣=1. If 2+k2zk+zˉ=kz\dfrac{2+k^2z}{k+\bar{z}}=kzk+zˉ2+k2z​=kz, k∈Rk\in Rk∈R, then the maximum distance of k+ik2k+ik^2k+ik2 from the circle ∣z−(1+2i)∣=1|z-(1+2i)|=1∣z−(1+2i)∣=1 is:
  1. (A)5+1\sqrt{5}+15​+1
  2. (B)2
  3. (C)3
  4. (D)3+1\sqrt{3}+13​+1

Correct answer: (A)

Step-by-step solution →
Q30·MathematicsIntegerJEE Main 2025
Let integers a,b∈[−3,3]a, b\in[-3, 3]a,b∈[−3,3] be such that a+b≠0a+b\ne 0a+b=0. Then the number of all possible ordered pairs (a,b)(a, b)(a,b), for which ∣z−az+b∣=1\left|\dfrac{z-a}{z+b}\right|=1​z+bz−a​​=1 and ∣z+1ωω2ωz+ω21ω21z+ω∣=1\begin{vmatrix}z+1 & \omega & \omega^2\\ \omega & z+\omega^2 & 1\\ \omega^2 & 1 & z+\omega\end{vmatrix}=1​z+1ωω2​ωz+ω21​ω21z+ω​​=1, z∈Cz\in Cz∈C, where ω\omegaω and ω2\omega^2ω2 are the roots of x2+x+1=0x^2+x+1=0x2+x+1=0, is equal to ______.

Correct answer: 10

Step-by-step solution →
Q31·MathematicsSingle correctJEE Main 2025
Let ∣z1−8−2i∣≤1|z_1-8-2i|\le 1∣z1​−8−2i∣≤1 and ∣z2−2+6i∣≤2|z_2-2+6i|\le 2∣z2​−2+6i∣≤2, z1,z2∈Cz_1,z_2\in Cz1​,z2​∈C. Then the minimum value of ∣z1−z2∣|z_1-z_2|∣z1​−z2​∣ is:
  1. (A)3
  2. (B)7
  3. (C)13
  4. (D)10

Correct answer: (B)

Step-by-step solution →
Q32·MathematicsSingle correctJEE Main 2025
Let O be the origin, the point A be z1=3+22iz_1=\sqrt{3}+2\sqrt{2}iz1​=3​+22​i, the point B(z2)B(z_2)B(z2​) be such that 3∣z1∣=∣z2∣\sqrt{3}|z_1|=|z_2|3​∣z1​∣=∣z2​∣ and arg⁡(z2)=arg⁡(z1)+π6\arg(z_2)=\arg(z_1)+\frac{\pi}{6}arg(z2​)=arg(z1​)+6π​. Then
  1. (A)area of triangle ABO is 113\frac{11}{\sqrt{3}}3​11​
  2. (B)ABO is a scalene triangle
  3. (C)area of triangle ABO is 114\frac{11}{4}411​
  4. (D)ABO is an obtuse angled isosceles triangle

Correct answer: (D)

Step-by-step solution →
Q33·MathematicsSingle correctJEE Main 2025
If α\alphaα and β\betaβ are the roots of the equation 2z2−3z−2i=02z^2-3z-2i=02z2−3z−2i=0, where i=−1i=\sqrt{-1}i=−1​, then 16⋅Re(α19+β19+α11+β11α15+β15)⋅Im(α19+β19+α11+β11α15+β15)16\cdot\mathrm{Re}\left(\dfrac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)\cdot\mathrm{Im}\left(\dfrac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)16⋅Re(α15+β15α19+β19+α11+β11​)⋅Im(α15+β15α19+β19+α11+β11​) is equal to
  1. (A)398398398
  2. (B)312312312
  3. (C)409409409
  4. (D)441441441

Correct answer: (D)

Step-by-step solution →
Q34·MathematicsSingle correctJEE Main 2025
The number of complex numbers zzz, satisfying ∣z∣=1|z|=1∣z∣=1 and ∣zzˉ+zˉz∣=1\left|\dfrac{z}{\bar z}+\dfrac{\bar z}{z}\right|=1​zˉz​+zzˉ​​=1, is :
  1. (A)6
  2. (B)4
  3. (C)10
  4. (D)8

Correct answer: (D)

Step-by-step solution →
Q35·MathematicsSingle correctJEE Main 2025
Let ∣zˉ−i2zˉ+i∣=13\left|\dfrac{\bar z-i}{2\bar z+i}\right|=\dfrac{1}{3}​2zˉ+izˉ−i​​=31​, z∈Cz\in\mathbb{C}z∈C, be the equation of a circle with center at CCC. If the area of the triangle, whose vertices are at the points (0,0)(0,0)(0,0), CCC and (α,0)(\alpha,0)(α,0) is 11 square units, then α2\alpha^2α2 equals
  1. (A)100100100
  2. (B)505050
  3. (C)12125\dfrac{121}{25}25121​
  4. (D)8125\dfrac{81}{25}2581​

Correct answer: (A)

Step-by-step solution →
Q36·MathematicsSingle correctJEE Main 2025
Let z1,z2,z3z_1,z_2,z_3z1​,z2​,z3​ lie on ∣z∣=1|z|=1∣z∣=1 with arg⁡z1=−π4, arg⁡z2=0, arg⁡z3=π4\arg z_1=-\tfrac{\pi}{4},\ \arg z_2=0,\ \arg z_3=\tfrac{\pi}{4}argz1​=−4π​, argz2​=0, argz3​=4π​. If ∣z1zˉ2+z2zˉ3+z3zˉ1∣2=α+β2 (α,β∈Z)|z_1\bar z_2+z_2\bar z_3+z_3\bar z_1|^2=\alpha+\beta\sqrt2\ (\alpha,\beta\in\mathbb{Z})∣z1​zˉ2​+z2​zˉ3​+z3​zˉ1​∣2=α+β2​ (α,β∈Z), then α2+β2\alpha^2+\beta^2α2+β2 is:
  1. (A)24
  2. (B)41
  3. (C)33
  4. (D)29

Correct answer: (D)

Step-by-step solution →
Q37·MathematicsSingle correctJEE Main 2025
Let the curve z(1+i)+zˉ(1−i)=4z(1+i)+\bar{z}(1-i)=4z(1+i)+zˉ(1−i)=4, z∈Cz\in\mathbb{C}z∈C, divide the region ∣z−3∣≤1|z-3|\le1∣z−3∣≤1 into two parts of areas α\alphaα and β\betaβ. Then ∣α−β∣|\alpha-\beta|∣α−β∣ equals:
  1. (A)1+π21+\dfrac{\pi}{2}1+2π​
  2. (B)1+π31+\dfrac{\pi}{3}1+3π​
  3. (C)1+π41+\dfrac{\pi}{4}1+4π​
  4. (D)1+π61+\dfrac{\pi}{6}1+6π​

Correct answer: (A)

Step-by-step solution →
Q38·MathematicsNumericalJEE Main 2024
The sum of the square of the modulus of the elements in the set {z=a+ib:a,b∈Z,z∈C,∣z−1∣≤1,∣z−5∣≤∣z−5i∣}\{z = a + ib : a, b \in \mathbb{Z}, z \in \mathbb{C}, |z-1| \le 1, |z-5| \le |z-5i|\}{z=a+ib:a,b∈Z,z∈C,∣z−1∣≤1,∣z−5∣≤∣z−5i∣} is ________.

Correct answer: 9

Step-by-step solution →
Q39·MathematicsSingle correctJEE Main 2024
Let z be a complex number such that the real part of z−2iz+2i\dfrac{z-2i}{z+2i}z+2iz−2i​ is zero. Then, the maximum value of ∣z−(6+8i)∣|z-(6+8i)|∣z−(6+8i)∣ is equal to:
  1. (A)121212
  2. (B)∞\infty∞
  3. (C)101010
  4. (D)888

Correct answer: (A)

Step-by-step solution →
Q40·MathematicsSingle correctJEE Main 2024
Let zzz be a complex number such that ∣z+2∣=1|z+2|=1∣z+2∣=1 and Im(z+1z+2)=15\mathrm{Im}\left(\dfrac{z+1}{z+2}\right)=\dfrac15Im(z+2z+1​)=51​. Then the value of ∣Re(z+2‾)∣\left|\mathrm{Re}\left(\overline{z+2}\right)\right|​Re(z+2​)​ is:
  1. (A)65\dfrac{\sqrt{6}}{5}56​​
  2. (B)1+65\dfrac{1+\sqrt{6}}{5}51+6​​
  3. (C)245\dfrac{24}{5}524​
  4. (D)265\dfrac{2\sqrt{6}}{5}526​​

Correct answer: (D)

Step-by-step solution →
Q41·MathematicsSingle correctJEE Main 2024
The sum of all possible values of θ∈[−π,2π]\theta\in[-\pi,2\pi]θ∈[−π,2π], for which 1+icos⁡θ1−2icos⁡θ\dfrac{1+i\cos\theta}{1-2i\cos\theta}1−2icosθ1+icosθ​ is purely imaginary, is equal to
  1. (A)2π2\pi2π
  2. (B)3π3\pi3π
  3. (C)5π5\pi5π
  4. (D)4π4\pi4π

Correct answer: (B)

Step-by-step solution →
Q42·MathematicsSingle correctJEE Main 2024
If the set R={(a,b):a+5b=42, a,b∈N}R=\{(a,b):a+5b=42,\ a,b\in\mathbb{N}\}R={(a,b):a+5b=42, a,b∈N} has mmm elements and ∑n=1min!=x+iy\displaystyle\sum_{n=1}^{m}i^{n!}=x+iyn=1∑m​in!=x+iy, where i=−1i=\sqrt{-1}i=−1​, then the value of m+x+ym+x+ym+x+y is:
  1. (A)888
  2. (B)121212
  3. (C)444
  4. (D)555

Correct answer: (B)

Step-by-step solution →
Q43·MathematicsSingle correctJEE Main 2024
If z1,z2z_1, z_2z1​,z2​ are two distinct complex number such that ∣z1−2z212−z1z2ˉ∣=2\left|\frac{z_1 - 2z_2}{\frac{1}{2} - z_1\bar{z_2}}\right| = 2​21​−z1​z2​ˉ​z1​−2z2​​​=2, then
  1. (A)either z1z_1z1​ lies on a circle of radius 1 or z2z_2z2​ lies on a circle of radius 12\tfrac{1}{2}21​
  2. (B)either z1z_1z1​ lies on a circle of radius 12\tfrac{1}{2}21​ or z2z_2z2​ lies on a circle of radius 1.
  3. (C)z1z_1z1​ lies on a circle of radius 12\tfrac{1}{2}21​ and z2z_2z2​ lies on a circle of radius 1.
  4. (D)both z1z_1z1​ and z2z_2z2​ lie on the same circle.

Correct answer: (A)

Step-by-step solution →
Q44·MathematicsSingle correctJEE Main 2024
Consider the following two statements: Statement I: For any two non-zero complex numbers z1,z2z_1, z_2z1​,z2​, (∣z1∣+∣z2∣)∣z1∣z1∣+z2∣z2∣∣≤2(∣z1∣+∣z2∣)(|z_1| + |z_2|)\left|\dfrac{z_1}{|z_1|} + \dfrac{z_2}{|z_2|}\right| \le 2(|z_1| + |z_2|)(∣z1​∣+∣z2​∣)​∣z1​∣z1​​+∣z2​∣z2​​​≤2(∣z1​∣+∣z2​∣) and Statement II: If x,y,zx, y, zx,y,z are three distinct complex numbers and a,b,ca, b, ca,b,c are three positive real numbers such that a∣y−z∣=b∣z−x∣=c∣x−y∣\dfrac{a}{|y-z|} = \dfrac{b}{|z-x|} = \dfrac{c}{|x-y|}∣y−z∣a​=∣z−x∣b​=∣x−y∣c​, then a2y−z+b2z−x+c2x−y=1\dfrac{a^2}{y-z} + \dfrac{b^2}{z-x} + \dfrac{c^2}{x-y} = 1y−za2​+z−xb2​+x−yc2​=1. Between the above two statements,
  1. (A)both Statement I and Statement II are incorrect.
  2. (B)Statement I is incorrect but Statement II is correct.
  3. (C)Statement I is correct but Statement II is incorrect.
  4. (D)both Statement I and Statement II are correct.

Correct answer: (C)

Step-by-step solution →
Q45·MathematicsSingle correctJEE Main 2024
Let S1={z∈C:∣z∣≤5}S_1=\{z\in\mathbb{C}:|z|\le 5\}S1​={z∈C:∣z∣≤5}, S2={z∈C:Im⁡(z+1−3 i1−3 i)≥0}S_2=\left\{z\in\mathbb{C}:\operatorname{Im}\left(\dfrac{z+1-\sqrt{3}\,i}{1-\sqrt{3}\,i}\right)\ge 0\right\}S2​={z∈C:Im(1−3​iz+1−3​i​)≥0} and S3={z∈C:Re⁡(z)≥0}S_3=\{z\in\mathbb{C}:\operatorname{Re}(z)\ge 0\}S3​={z∈C:Re(z)≥0}. Then the area of S1∩S2∩S3S_1\cap S_2\cap S_3S1​∩S2​∩S3​ is:
  1. (A)125π6\dfrac{125\pi}{6}6125π​
  2. (B)125π24\dfrac{125\pi}{24}24125π​
  3. (C)125π4\dfrac{125\pi}{4}4125π​
  4. (D)125π12\dfrac{125\pi}{12}12125π​

Correct answer: (D)

Step-by-step solution →
Q46·MathematicsSingle correctJEE Main 2024
Let α\alphaα and β\betaβ be the sum and the product of all the non-zero solutions of the equation (zˉ)2+∣z∣=0(\bar z)^2+|z|=0(zˉ)2+∣z∣=0, z∈Cz\in\mathbb{C}z∈C. Then 4(α2+β2)4(\alpha^2+\beta^2)4(α2+β2) is equal to:
  1. (A)666
  2. (B)444
  3. (C)888
  4. (D)222

Correct answer: (B)

Step-by-step solution →
Q47·MathematicsSingle correctJEE Main 2024
Let S={z∈C:∣z−1∣=1S=\{z\in\mathbb{C}:|z-1|=1S={z∈C:∣z−1∣=1 and (2−1)(z+zˉ)−i(z−zˉ)=22}(\sqrt2-1)(z+\bar z)-i(z-\bar z)=2\sqrt2\}(2​−1)(z+zˉ)−i(z−zˉ)=22​}. Let z1,z2∈Sz_1,z_2\in Sz1​,z2​∈S be such that ∣z1∣=max⁡z∈S∣z∣|z_1|=\max\limits_{z\in S}|z|∣z1​∣=z∈Smax​∣z∣ and ∣z2∣=min⁡z∈S∣z∣|z_2|=\min\limits_{z\in S}|z|∣z2​∣=z∈Smin​∣z∣. Then ∣2 z1−z2∣2\left|\sqrt2\,z_1-z_2\right|^2​2​z1​−z2​​2 equals:
  1. (A)111
  2. (B)444
  3. (C)333
  4. (D)222

Correct answer: (D)

Step-by-step solution →
Q48·MathematicsSingle correctJEE Main 2024
If zzz is a complex number such that ∣z∣≥1|z|\ge 1∣z∣≥1, then the minimum value of ∣z+12(3+4i)∣\left|z+\dfrac{1}{2}(3+4i)\right|​z+21​(3+4i)​ is:
  1. (A)52\dfrac{5}{2}25​
  2. (B)2
  3. (C)3
  4. (D)32\dfrac{3}{2}23​

Correct answer: (D)

Step-by-step solution →
Q49·MathematicsSingle correctJEE Main 2024
Let z1z_1z1​ and z2z_2z2​ be two complex numbers such that z1+z2=5z_1+z_2=5z1​+z2​=5 and z13+z23=20+15iz_1^3+z_2^3=20+15iz13​+z23​=20+15i. Then ∣z14+z24∣\left|z_1^4+z_2^4\right|​z14​+z24​​ equals
  1. (A)30330\sqrt{3}303​
  2. (B)757575
  3. (C)151515\sqrt{15}1515​
  4. (D)25325\sqrt{3}253​

Correct answer: (B)

Step-by-step solution →
Q50·MathematicsNumericalJEE Main 2024
If α\alphaα denotes the number of solutions of ∣1−i∣x=2x|1-i|^x=2^x∣1−i∣x=2x and β=∣z∣arg⁡(z)\beta=\dfrac{|z|}{\arg(z)}β=arg(z)∣z∣​, where z=π4(1+i)4[1−πiπ+i+π−i1+πi]z=\dfrac{\pi}{4}(1+i)^4\left[\dfrac{1-\sqrt{\pi}i}{\sqrt{\pi}+i}+\dfrac{\sqrt{\pi}-i}{1+\sqrt{\pi}i}\right]z=4π​(1+i)4[π​+i1−π​i​+1+π​iπ​−i​], i=−1i=\sqrt{-1}i=−1​, then the distance of the point (α,β)(\alpha,\beta)(α,β) from the line 4x−3y=74x-3y=74x−3y=7 is ______.

Correct answer: 3

Step-by-step solution →
Q51·MathematicsSingle correctJEE Main 2024
If z=x+iyz=x+iyz=x+iy, x≠0x\ne0x=0, satisfies the equation z2+izˉ=0z^2+i\bar z=0z2+izˉ=0, then ∣z∣2|z|^2∣z∣2 is equal to:
  1. (A)999
  2. (B)111
  3. (C)444
  4. (D)14\dfrac1441​

Correct answer: (B)

Step-by-step solution →
Q52·MathematicsSingle correctJEE Main 2024
If zzz is a complex number, then the number of common roots of the equation z1985+z100+1=0z^{1985}+z^{100}+1=0z1985+z100+1=0 and z3+2z2+2z+1=0z^3+2z^2+2z+1=0z3+2z2+2z+1=0, is equal to:
  1. (A)111
  2. (B)222
  3. (C)000
  4. (D)333

Correct answer: (B)

Step-by-step solution →
Q53·MathematicsSingle correctJEE Main 2024
If z=12−2iz=\dfrac{1}{2}-2iz=21​−2i, is such that ∣z+1∣=αz+β(1+i)|z+1|=\alpha z+\beta(1+i)∣z+1∣=αz+β(1+i), i=−1i=\sqrt{-1}i=−1​ and α,β∈R\alpha,\beta\in Rα,β∈R, then α+β\alpha+\betaα+β is equal to
  1. (A)−4-4−4
  2. (B)333
  3. (C)222
  4. (D)−1-1−1

Correct answer: (B)

Step-by-step solution →
Q54·MathematicsSingle correctJEE Main 2024
Let rrr and θ\thetaθ respectively be the modulus and amplitude of the complex number z=2−i(2tan⁡5π8)z=2-i\left(2\tan\dfrac{5\pi}{8}\right)z=2−i(2tan85π​), then (r,θ)(r,\theta)(r,θ) is equal to:
  1. (A)(2sec⁡3π8,3π8)\left(2\sec\dfrac{3\pi}{8},\dfrac{3\pi}{8}\right)(2sec83π​,83π​)
  2. (B)(2sec⁡3π8,5π8)\left(2\sec\dfrac{3\pi}{8},\dfrac{5\pi}{8}\right)(2sec83π​,85π​)
  3. (C)(2sec⁡5π8,3π8)\left(2\sec\dfrac{5\pi}{8},\dfrac{3\pi}{8}\right)(2sec85π​,83π​)
  4. (D)(2sec⁡11π8,11π8)\left(2\sec\dfrac{11\pi}{8},\dfrac{11\pi}{8}\right)(2sec811π​,811π​)

Correct answer: (A)

Step-by-step solution →
Q55·MathematicsSingle correctJEE Main 2024
If S={z∈C:∣z−i∣=∣z+i∣=∣z−1∣}S=\{z\in\mathbb C:|z-i|=|z+i|=|z-1|\}S={z∈C:∣z−i∣=∣z+i∣=∣z−1∣}, then n(S)n(S)n(S) is:
  1. (A)1
  2. (B)0
  3. (C)3
  4. (D)2

Correct answer: (A)

Step-by-step solution →
Q56·MathematicsNumericalJEE Main 2024
Let the complex numbers α\alphaα and 1αˉ\dfrac{1}{\bar\alpha}αˉ1​ lie on the circles ∣z−z0∣2=4|z-z_0|^2=4∣z−z0​∣2=4 and ∣z−z0∣2=16|z-z_0|^2=16∣z−z0​∣2=16 respectively, where z0=1+iz_0=1+iz0​=1+i. Then, the value of 100∣α∣2100|\alpha|^2100∣α∣2 is __________.

Correct answer: 20

Step-by-step solution →
Q57·MathematicsNumericalJEE Main 2024
If α\alphaα satisfies the equation x2+x+1=0x^2+x+1=0x2+x+1=0 and (1+α)7=A+Bα+Cα2(1+\alpha)^7=A+B\alpha+C\alpha^2(1+α)7=A+Bα+Cα2, A,B,C≥0A,B,C\ge 0A,B,C≥0, then 5(3A−2B−C)5(3A-2B-C)5(3A−2B−C) is equal to __________.

Correct answer: 5

Step-by-step solution →
Q58·MathematicsIntegerJEE Advanced 2023
Let A1A_{1}A1​, A2A_{2}A2​, A3A_{3}A3​, ....., A8A_{8}A8​ be the vertices of a regular octagon that lie on a circle of radius 2. Let P be a point on the circle and let PAiPA_{i}PAi​ denote the distance between the points PPP and AiA_{i}Ai​ for i=1,2,.....,8i = 1, 2, ....., 8i=1,2,.....,8. If PPP varies over the circle, then the maximum value of the product PA1⋅PA2.....PA8PA_{1} \cdot PA_{2} ..... PA_{8}PA1​⋅PA2​.....PA8​, is

Correct answer: 512

Step-by-step solution →
Q59·MathematicsSingle correctJEE Advanced 2023
Let z be a complex number satisfying ∣z∣3+2z2+4zˉ−8=0|z|^3 + 2z^2 + 4\bar{z} - 8 = 0∣z∣3+2z2+4zˉ−8=0, where zˉ\bar{z}zˉ denotes the complex conjugate of z . Let the imaginary part of z be nonzero. Match each entry in List-I to the correct entries in List-II. The correct option is:
List – IList – II
P.∣z∣2|z|^2∣z∣2 is equal to1.12
Q.∣z−zˉ∣2\left|z - \bar{z}\right|^2∣z−zˉ∣2 is equal to2.4
R.∣z∣2+∣z+zˉ∣2|z|^2 + \left|z + \bar{z}\right|^2∣z∣2+∣z+zˉ∣2 is equal to3.8
S.∣z+1∣2\left|z + 1\right|^2∣z+1∣2 is equal to4.10
5.7
  1. (A)(P) → (1) (Q) → (3) (R) → (5) (S) → (4)
  2. (B)(P) → (2) (Q) → (1) (R) → (3) (S) → (5)
  3. (C)(P) → (2) (Q) → (4) (R) → (5) (S) → (1)
  4. (D)(P) → (2) (Q) → (3) (R) → (5) (S) → (4)

Correct answer: (B)

Step-by-step solution →
Q60·MathematicsNumericalJEE Advanced 2023
Let A={1967+1686 isin⁡θ7−3icos⁡θ:θ∈R}A = \left\{\frac{1967 + 1686\,i\sin\theta}{7 - 3i\cos\theta} : \theta \in R\right\}A={7−3icosθ1967+1686isinθ​:θ∈R}. If A contains exactly one positive integer n, then the value of n is

Correct answer: 281

Step-by-step solution →
Q61·MathematicsSingle correctJEE Main 2023
If the set {Re(z−zˉ+zzˉ2−3z+5zˉ):z∈C,Re(z)=3}\left\{\mathrm{Re}\left(\frac{z-\bar{z}+z\bar{z}}{2-3z+5\bar{z}}\right):z\in\mathbb{C},\mathrm{Re}(z)=3\right\}{Re(2−3z+5zˉz−zˉ+zzˉ​):z∈C,Re(z)=3} is equal to the interval (α,β](\alpha,\beta](α,β], then 24(β−α)24(\beta-\alpha)24(β−α) is equal to
  1. (A)36
  2. (B)42
  3. (C)27
  4. (D)30

Correct answer: (D)

Step-by-step solution →
Q62·MathematicsSingle correctJEE Main 2023
Let α,β\alpha, \betaα,β be the roots of the equation x2−2x+2=0x^2 - \sqrt{2}x + 2 = 0x2−2​x+2=0. Then α14+β14\alpha^{14} + \beta^{14}α14+β14 is equal to
  1. (A)−642-64\sqrt{2}−642​
  2. (B)−1282-128\sqrt{2}−1282​
  3. (C)−64-64−64
  4. (D)−128-128−128

Correct answer: (D)

Step-by-step solution →
Q63·MathematicsNumericalJEE Main 2023
Let ω=zzˉ+k1z+kˉ1zˉ+λ(1+i), k1∈C\omega=z\bar z+k_{1}z+\bar k_{1}\bar z+\lambda(1+i),\,k_{1}\in\mathbb{C}ω=zzˉ+k1​z+kˉ1​zˉ+λ(1+i),k1​∈C. Re⁡(ω)=0\operatorname{Re}(\omega)=0Re(ω)=0 be the circle CCC of radius 111 in the first quadrant touching the line y=1y=1y=1 and the yyy-axis. If the curve Im⁡(ω)=0\operatorname{Im}(\omega)=0Im(ω)=0 intersects CCC at AAA and BBB, then 30(AB)230(AB)^{2}30(AB)2 is equal to _____.

Correct answer: 24

Step-by-step solution →
Q64·MathematicsSingle correctJEE Main 2023
Let S={z∈C:zˉ=i(z2+Re(zˉ))}S = \left\{ z \in \mathbb{C} : \bar{z} = i(z^2 + \text{Re}(\bar{z})) \right\}S={z∈C:zˉ=i(z2+Re(zˉ))}. Then ∑z∈S∣z∣2\sum\limits_{z \in S} |z|^2z∈S∑​∣z∣2 is equal to
  1. (A)72\dfrac{7}{2}27​
  2. (B)444
  3. (C)52\dfrac{5}{2}25​
  4. (D)333

Correct answer: (B)

Step-by-step solution →
Q65·MathematicsSingle correctJEE Main 2023
Let C be the circle in the complex plane with centre z0=12(1+3i)z_0=\dfrac12(1+3i)z0​=21​(1+3i) and radius r=1r=1r=1. Let z1=1+iz_1=1+iz1​=1+i and the complex number z2z_2z2​ be outside the circle C such that ∣z1−z0∣ ∣z2−z0∣=1|z_1-z_0|\,|z_2-z_0|=1∣z1​−z0​∣∣z2​−z0​∣=1. If z0z_0z0​, z1z_1z1​ and z2z_2z2​ are collinear, then the smaller value of ∣z2∣2|z_2|^2∣z2​∣2 is equal to
  1. (A)132\dfrac{13}{2}213​
  2. (B)52\dfrac5225​
  3. (C)32\dfrac3223​
  4. (D)72\dfrac7227​

Correct answer: (B)

Step-by-step solution →
Q66·MathematicsNumericalJEE Main 2023
Let S={z∈C−{i,2i}:z2+8iz−15z2−3iz−2∈R}S=\left\{z\in\mathbb{C}-\{i,2i\}:\dfrac{z^2+8iz-15}{z^2-3iz-2}\in\mathbb{R}\right\}S={z∈C−{i,2i}:z2−3iz−2z2+8iz−15​∈R}. If α−1311i∈S\alpha-\dfrac{13}{11}i\in Sα−1113​i∈S, α∈R−{0}\alpha\in\mathbb{R}-\{0\}α∈R−{0}, then 242α2242\alpha^2242α2 is equal to

Correct answer: 1680

Step-by-step solution →
Q67·MathematicsSingle correctJEE Main 2023
For a∈Ca\in\mathbb{C}a∈C, let A={z∈C:Re⁡(a+zˉ)>Im⁡(aˉ+z)}A=\{z\in\mathbb{C}:\operatorname{Re}(a+\bar{z})>\operatorname{Im}(\bar{a}+z)\}A={z∈C:Re(a+zˉ)>Im(aˉ+z)} and B={z∈C:Re⁡(a+zˉ)<Im⁡(aˉ+z)}B=\{z\in\mathbb{C}:\operatorname{Re}(a+\bar{z})<\operatorname{Im}(\bar{a}+z)\}B={z∈C:Re(a+zˉ)<Im(aˉ+z)}. Then among the two statements: (S1): If Re⁡(a),Im⁡(a)>0\operatorname{Re}(a),\operatorname{Im}(a)>0Re(a),Im(a)>0, then the set AAA contains all the real numbers; (S2): If Re⁡(a),Im⁡(a)<0\operatorname{Re}(a),\operatorname{Im}(a)<0Re(a),Im(a)<0, then the set BBB contains all the real numbers,
  1. (A)Only (S1) is true
  2. (B)Both are false
  3. (C)Only (S2) is true
  4. (D)Both are true

Correct answer: (B)

Step-by-step solution →
Q68·MathematicsSingle correctJEE Main 2023
Let w1w_1w1​ be the point obtained by the rotation of z1=5+4iz_1 = 5 + 4iz1​=5+4i about the origin through a right angle in the anticlockwise direction, and w2w_2w2​ be the point obtained by the rotation of z2=3+5iz_2 = 3 + 5iz2​=3+5i about the origin through a right angle in the clockwise direction. Then the principal argument of w1−w2w_1 - w_2w1​−w2​ is equal to
  1. (A)−π+tan⁡−1335-\pi + \tan^{-1} \frac{33}{5}−π+tan−1533​
  2. (B)−π−tan⁡−1335-\pi - \tan^{-1} \frac{33}{5}−π−tan−1533​
  3. (C)−π+tan⁡−189-\pi + \tan^{-1} \frac{8}{9}−π+tan−198​
  4. (D)π−tan⁡−189\pi - \tan^{-1} \frac{8}{9}π−tan−198​

Correct answer: (D)

Step-by-step solution →
Q69·MathematicsSingle correctJEE Main 2023
Let S={z=x+iy:2z−3i4z+2i is a real number}S=\left\{z=x+iy:\frac{2z-3i}{4z+2i}\text{ is a real number}\right\}S={z=x+iy:4z+2i2z−3i​ is a real number}. Then which of the following is NOT correct?
  1. (A)y+x2+y2≠−14y+x^2+y^2\ne-\frac{1}{4}y+x2+y2=−41​
  2. (B)x=0x=0x=0
  3. (C)(x,y)=(0,−12)(x,y)=\left(0,-\frac{1}{2}\right)(x,y)=(0,−21​)
  4. (D)y∈(−∞,−12)∪(−12,∞)y\in\left(-\infty,-\frac{1}{2}\right)\cup\left(-\frac{1}{2},\infty\right)y∈(−∞,−21​)∪(−21​,∞)

Correct answer: (C)

Step-by-step solution →
Q70·MathematicsSingle correctJEE Main 2023
Let the complex number z=x+iyz=x+iyz=x+iy be such that 2z−3i2z+i\dfrac{2z-3i}{2z+i}2z+i2z−3i​ is purely imaginary. If x+y2=0x+y^2=0x+y2=0, then y4+y2−yy^4+y^2-yy4+y2−y is equal to:
  1. (A)32\dfrac3223​
  2. (B)43\dfrac4334​
  3. (C)23\dfrac2332​
  4. (D)34\dfrac3443​

Correct answer: (D)

Step-by-step solution →
Q71·MathematicsSingle correctJEE Main 2023
Let A={θ∈(0,2π):1+2isin⁡θ1−isin⁡θ is purely imaginary}A=\left\{\theta\in(0,2\pi):\dfrac{1+2i\sin\theta}{1-i\sin\theta}\text{ is purely imaginary}\right\}A={θ∈(0,2π):1−isinθ1+2isinθ​ is purely imaginary}. Then the sum of the elements in A is
  1. (A)π\piπ
  2. (B)2π2\pi2π
  3. (C)4π4\pi4π
  4. (D)3π3\pi3π

Correct answer: (C)

Step-by-step solution →
Q72·MathematicsSingle correctJEE Main 2023
If for z=α+iβz=\alpha+i\betaz=α+iβ, ∣z+2∣=z+4(1+i)|z+2|=z+4(1+i)∣z+2∣=z+4(1+i), then α+β\alpha+\betaα+β and αβ\alpha\betaαβ are the roots of the equation
  1. (A)x2+7x+12=0x^{2}+7x+12=0x2+7x+12=0
  2. (B)x2+3x−4=0x^{2}+3x-4=0x2+3x−4=0
  3. (C)x2+2x−3=0x^{2}+2x-3=0x2+2x−3=0
  4. (D)x2+x−12=0x^{2}+x-12=0x2+x−12=0

Correct answer: (A)

Step-by-step solution →
Q73·MathematicsNumericalJEE Main 2023
For α,β,z∈C\alpha,\beta,z\in \mathbb{C}α,β,z∈C and λ>1\lambda>1λ>1, if λ−1\sqrt{\lambda-1}λ−1​ is the radius of the circle ∣z−α∣2+∣z−β∣2=2λ|z-\alpha|^2+|z-\beta|^2=2\lambda∣z−α∣2+∣z−β∣2=2λ, then ∣α−β∣|\alpha-\beta|∣α−β∣ is equal to

Correct answer: 2

Step-by-step solution →
Q74·MathematicsSingle correctJEE Main 2023
Let a≠ba\neq ba=b be two non-zero real numbers. Then the number of elements in the set X={z∈C:Re⁡(az2+bz)=a and Re⁡(bz2+az)=b}X=\{z\in \mathbb{C}:\operatorname{Re}(az^2+bz)=a \text{ and } \operatorname{Re}(bz^2+az)=b\}X={z∈C:Re(az2+bz)=a and Re(bz2+az)=b} is equal to
  1. (A)1
  2. (B)3
  3. (C)0
  4. (D)2

Correct answer: (C)

Step-by-step solution →
Q75·MathematicsSingle correctJEE Main 2023
If the center and radius of the circle ∣z−2z−3∣=2\left|\frac{z-2}{z-3}\right|=2​z−3z−2​​=2 are respectively (α,β)(\alpha,\beta)(α,β) and γ\gammaγ, then 3(α+β+γ)3(\alpha+\beta+\gamma)3(α+β+γ) is equal to
  1. (A)111111
  2. (B)121212
  3. (C)999
  4. (D)101010

Correct answer: (B)

Step-by-step solution →
Q76·MathematicsSingle correctJEE Main 2023
For all z∈Cz\in Cz∈C on the curve C1:∣z∣=4C_1:|z|=4C1​:∣z∣=4, let the locus of the point z+1zz+\frac{1}{z}z+z1​ be the curve C2C_2C2​. Then :
  1. (A)the curve C1C_1C1​ lies inside C2C_2C2​
  2. (B)the curve C2C_2C2​ lies inside C1C_1C1​
  3. (C)the curves C1C_1C1​ and C2C_2C2​ intersect at 4 points
  4. (D)the curves C1C_1C1​ and C2C_2C2​ intersect at 2 points

Correct answer: (C)

Step-by-step solution →
Q77·MathematicsSingle correctJEE Main 2023
The complex number z=i−1cos⁡π3+isin⁡π3z=\dfrac{i-1}{\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}}z=cos3π​+isin3π​i−1​ is equal to:
  1. (A)2i(cos⁡5π12−isin⁡5π12)\sqrt2 i\left(\cos\frac{5\pi}{12}-i\sin\frac{5\pi}{12}\right)2​i(cos125π​−isin125π​)
  2. (B)2(cos⁡π12+isin⁡π12)\sqrt2\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)2​(cos12π​+isin12π​)
  3. (C)2(cos⁡5π12+isin⁡5π12)\sqrt2\left(\cos\frac{5\pi}{12}+i\sin\frac{5\pi}{12}\right)2​(cos125π​+isin125π​)
  4. (D)cos⁡π12−isin⁡π12\cos\frac{\pi}{12}-i\sin\frac{\pi}{12}cos12π​−isin12π​

Correct answer: (C)

Step-by-step solution →
Q78·MathematicsNumericalJEE Main 2023
Let z=1+iz=1+iz=1+i and z1=1+izˉzˉ(1−z)+1zz_1=\dfrac{1+i\bar z}{\bar z(1-z)+\dfrac1z}z1​=zˉ(1−z)+z1​1+izˉ​. Then 12πarg⁡(z1)\dfrac{12}{\pi}\arg(z_1)π12​arg(z1​) is equal to

Correct answer: 9

Step-by-step solution →
Q79·MathematicsNumericalJEE Main 2023
Let α=8−14i, A={z∈C:αz−αˉzˉz2−(zˉ)2−112i=1}\alpha=8-14i,\,A=\left\{z\in\mathbb{C}:\dfrac{\alpha z-\bar\alpha\bar z}{z^{2}-(\bar z)^{2}-112i}=1\right\}α=8−14i,A={z∈C:z2−(zˉ)2−112iαz−αˉzˉ​=1} and B={z∈C: ∣z+3i∣=4}B=\{z\in\mathbb{C}:\,|z+3i|=4\}B={z∈C:∣z+3i∣=4}. Then ∑z∈A∩B(Re⁡z−Im⁡z)\displaystyle\sum_{z\in A\cap B}(\operatorname{Re}z-\operatorname{Im}z)z∈A∩B∑​(Rez−Imz) is equal to _____.

Correct answer: 14

Step-by-step solution →
Q80·MathematicsSingle correctJEE Main 2023
For two non-zero complex numbers z1z_1z1​ and z2z_2z2​, if Re(z1z2ˉ)=0\text{Re}(z_1 \bar{z_2}) = 0Re(z1​z2​ˉ​)=0 and Re(z1+z2)=0\text{Re}(z_1 + z_2) = 0Re(z1​+z2​)=0, then which of the following are possible? A. Im(z1)>0\text{Im}(z_1) > 0Im(z1​)>0 and Im(z2)>0\text{Im}(z_2) > 0Im(z2​)>0 B. Im(z1)<0\text{Im}(z_1) < 0Im(z1​)<0 and Im(z2)>0\text{Im}(z_2) > 0Im(z2​)>0 C. Im(z1)>0\text{Im}(z_1) > 0Im(z1​)>0 and Im(z2)<0\text{Im}(z_2) < 0Im(z2​)<0 D. Im(z1)<0\text{Im}(z_1) < 0Im(z1​)<0 and Im(z2)<0\text{Im}(z_2) < 0Im(z2​)<0. Choose the correct answer from the options given below:
  1. (A)B and D
  2. (B)A and B
  3. (C)B and C
  4. (D)A and C

Correct answer: (C)

Step-by-step solution →
Q81·MathematicsNumericalJEE Main 2023
Let α1,α2,…,α7\alpha_1,\alpha_2,\dots,\alpha_7α1​,α2​,…,α7​ be the roots of the equation x7+3x5−13x3−15x=0x^7+3x^5-13x^3-15x=0x7+3x5−13x3−15x=0 and ∣α1∣≥∣α2∣≥⋯≥∣α7∣|\alpha_1|\ge|\alpha_2|\ge\dots\ge|\alpha_7|∣α1​∣≥∣α2​∣≥⋯≥∣α7​∣. Then α1α2−α3α4+α5α6\alpha_1\alpha_2-\alpha_3\alpha_4+\alpha_5\alpha_6α1​α2​−α3​α4​+α5​α6​ is equal to _____.

Correct answer: 3

Step-by-step solution →
Q82·MathematicsSingle correctJEE Main 2023
Let z1=2+3iz_1=2+3iz1​=2+3i and z2=3+4iz_2=3+4iz2​=3+4i. The set S={z∈C:∣z−z1∣2−∣z−z2∣2=∣z1−z2∣2}S=\{z\in\mathbb{C}:|z-z_1|^2-|z-z_2|^2=|z_1-z_2|^2\}S={z∈C:∣z−z1​∣2−∣z−z2​∣2=∣z1​−z2​∣2} represents a:
  1. (A)hyperbola with the length of the transverse axis 777
  2. (B)hyperbola with eccentricity 222
  3. (C)straight line with the sum of its intercepts on the coordinate axes equals −18-18−18
  4. (D)straight line with the sum of its intercepts on the coordinate axes equals 141414

Correct answer: (D)

Step-by-step solution →
Q83·MathematicsSingle correctJEE Main 2023
Let zzz be a complex number such that ∣z−2iz+i∣=2\left|\dfrac{z-2i}{z+i}\right|=2​z+iz−2i​​=2, z≠−iz\neq -iz=−i. Then zzz lies on the circle of radius 222 and centre
  1. (A)(2,0)(2,0)(2,0)
  2. (B)(0,2)(0,2)(0,2)
  3. (C)(0,−2)(0,-2)(0,−2)
  4. (D)(0,0)(0,0)(0,0)

Correct answer: (C)

Step-by-step solution →
Q84·MathematicsSingle correctJEE Main 2023
The value of (1+sin⁡2π9+icos⁡2π91+sin⁡2π9−icos⁡2π9)3\left(\dfrac{1+\sin\frac{2\pi}{9}+i\cos\frac{2\pi}{9}}{1+\sin\frac{2\pi}{9}-i\cos\frac{2\pi}{9}}\right)^3(1+sin92π​−icos92π​1+sin92π​+icos92π​​)3 is
  1. (A)−12(3−i)-\dfrac12(\sqrt3-i)−21​(3​−i)
  2. (B)−12(1−i3)-\dfrac12(1-i\sqrt3)−21​(1−i3​)
  3. (C)12(1−i3)\dfrac12(1-i\sqrt3)21​(1−i3​)
  4. (D)12(3+i)\dfrac12(\sqrt3+i)21​(3​+i)

Correct answer: (A)

Step-by-step solution →
Q85·MathematicsSingle correctJEE Main 2023
Let p,q∈Rp,q\in\mathbb{R}p,q∈R and (1−3 i)200=2199(p+iq)(1-\sqrt{3}\,i)^{200}=2^{199}(p+iq)(1−3​i)200=2199(p+iq), i=−1i=\sqrt{-1}i=−1​. Then p+q+q2p+q+q^2p+q+q2 and p−q+q2p-q+q^2p−q+q2 are roots of the equation
  1. (A)x2−4x−1=0x^2-4x-1=0x2−4x−1=0
  2. (B)x2−4x+1=0x^2-4x+1=0x2−4x+1=0
  3. (C)x2+4x−1=0x^2+4x-1=0x2+4x−1=0
  4. (D)x2+4x+1=0x^2+4x+1=0x2+4x+1=0

Correct answer: (B)

Step-by-step solution →
Q86·MathematicsMultiple correctJEE Advanced 2022
Let z̄ denote the complex conjugate of a complex number z. If z is a non-zero complex number for which both real and imaginary parts of (zˉ)2\left(\bar{z}\right)^{2}(zˉ)2 + 1z2\frac{1}{z^{2}}z21​ are integers, than which of the following is/are possible value(s) of |z|?
  1. (A)(43+32052)14\left(\frac{43+3\sqrt{205}}{2}\right)^{\frac{1}{4}}(243+3205​​)41​
  2. (B)(7+334)14\left(\frac{7+\sqrt{33}}{4}\right)^{\frac{1}{4}}(47+33​​)41​
  3. (C)(9+654)14\left(\frac{9+\sqrt{65}}{4}\right)^{\frac{1}{4}}(49+65​​)41​
  4. (D)(7+136)14\left(\frac{7+\sqrt{13}}{6}\right)^{\frac{1}{4}}(67+13​​)41​

Correct answer: (A)

Step-by-step solution →
Q87·MathematicsNumericalJEE Advanced 2022
Let zˉ\bar{z}zˉ denote the complex conjugate of a complex number z and let i=−1i = \sqrt{-1}i=−1​. In the set of complex numbers, the number of distinct roots of the equation zˉ−z2=i(zˉ+z2)\bar{z} - z^{2} = i\left(\bar{z} + z^{2}\right)zˉ−z2=i(zˉ+z2) is __________.

Correct answer: 4

Step-by-step solution →
Q88·MathematicsNumericalJEE Advanced 2022
Let z be a complex number with non-zero imaginary part. If 2+3z+4z22−3z+4z2\frac{2+3z+4z^{2}}{2-3z+4z^{2}}2−3z+4z22+3z+4z2​ is a real number, then the value of ∣z∣2|z|^{2}∣z∣2 is __________.

Correct answer: 0.50

Step-by-step solution →
Q89·MathematicsSingle correctJEE Main 2022
If z≠0z \neq 0z=0 be a complex number such that ∣z−1z∣=2\left| z - \frac{1}{z} \right| = 2​z−z1​​=2, then the maximum value of ∣z∣|z|∣z∣ is:
  1. (A)2\sqrt{2}2​
  2. (B)111
  3. (C)2−1\sqrt{2} - 12​−1
  4. (D)2+1\sqrt{2} + 12​+1

Correct answer: (D)

Step-by-step solution →
Q90·MathematicsSingle correctJEE Main 2022
If z=2+3iz = 2 + 3iz=2+3i, then z5+(zˉ)5z^{5} + (\bar{z})^{5}z5+(zˉ)5 is equal to :
  1. (A)244
  2. (B)224
  3. (C)245
  4. (D)265

Correct answer: (A)

Step-by-step solution →
Q91·MathematicsSingle correctJEE Main 2022
Let S={z=x+iy:∣z−1+i∣≥∣z∣,∣z∣<2,∣z+i∣=∣z−1∣}S = \{ z = x + iy : |z - 1 + i| \ge |z|, |z| < 2, |z + i| = |z - 1| \}S={z=x+iy:∣z−1+i∣≥∣z∣,∣z∣<2,∣z+i∣=∣z−1∣}. Then the set of all values of x, for which w=2x+iy∈Sw = 2x + iy \in Sw=2x+iy∈S for some y∈Ry \in \mathbb{R}y∈R, is
  1. (A)(−2,122]\left(-\sqrt{2}, \frac{1}{2\sqrt{2}}\right](−2​,22​1​]
  2. (B)(−12,14]\left(-\frac{1}{\sqrt{2}}, \frac{1}{4}\right](−2​1​,41​]
  3. (C)(−2,12]\left(-\sqrt{2}, \frac{1}{2}\right](−2​,21​]
  4. (D)(−12,122]\left(-\frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}\right](−2​1​,22​1​]

Correct answer: (B)

Step-by-step solution →
Q92·MathematicsSingle correctJEE Main 2022
Let S1={z1∈C:∣z1−3∣=12}S_{1} = \left\{ z_{1} \in C : \left| z_{1} - 3 \right| = \frac{1}{2} \right\}S1​={z1​∈C:∣z1​−3∣=21​} and S2={z2∈C:∣z2−∣z2+1∣∣=∣z2+∣z2−1∣∣}S_{2} = \left\{ z_{2} \in C : \left| z_{2} - \left| z_{2} + 1 \right| \right| = \left| z_{2} + \left| z_{2} - 1 \right| \right| \right\}S2​={z2​∈C:∣z2​−∣z2​+1∣∣=∣z2​+∣z2​−1∣∣}. Then, for z1∈S1z_{1} \in S_{1}z1​∈S1​ and z2∈S2z_{2} \in S_{2}z2​∈S2​, the least value of ∣z2−z1∣\left| z_{2} - z_{1} \right|∣z2​−z1​∣ is :
  1. (A)0
  2. (B)12\frac{1}{2}21​
  3. (C)32\frac{3}{2}23​
  4. (D)52\frac{5}{2}25​

Correct answer: (C)

Step-by-step solution →
Q93·MathematicsNumericalJEE Main 2022
Let z=a+ibz = a + ibz=a+ib, b≠0b \neq 0b=0 be complex numbers satisfying z2=zˉ⋅21−∣z∣z^{2}=\bar{z}\cdot 2^{1-|z|}z2=zˉ⋅21−∣z∣. Then the least value of n∈Nn \in Nn∈N, such that zn=(z+1)nz^{n} = (z+1)^{n}zn=(z+1)n, is equal to _____

Correct answer: 6

Step-by-step solution →
Q94·MathematicsSingle correctJEE Main 2022
Let the minimum value v0v_0v0​ of v=∣z∣2+∣z−3∣2+∣z−6i∣2v = |z|^2 + |z-3|^2 + |z-6i|^2v=∣z∣2+∣z−3∣2+∣z−6i∣2, z∈Cz \in \mathbb{C}z∈C is attained at z=z0z = z_0z=z0​. Then ∣2z02−zˉ03+3∣2+v02\left| 2z_0^2 - \bar{z}_0^3 + 3 \right|^2 + v_0^2​2z02​−zˉ03​+3​2+v02​ is equal to
  1. (A)1000
  2. (B)1024
  3. (C)1105
  4. (D)1196

Correct answer: (A)

Step-by-step solution →
Q95·MathematicsNumericalJEE Main 2022
Let S={z∈C:z2+zˉ=0}S=\left\{z\in\mathbb{C}:z^{2}+\bar{z}=0\right\}S={z∈C:z2+zˉ=0}. Then ∑z∈S(Re(z)+Im(z))\sum_{z\in S}\left(\mathrm{Re}\left(z\right)+\mathrm{Im}\left(z\right)\right)∑z∈S​(Re(z)+Im(z)) is equal to ______.

Correct answer: 0

Step-by-step solution →
Q96·MathematicsSingle correctJEE Main 2022
Let S be the set of all (α,β)(\alpha, \beta)(α,β), π<α,β<2π\pi < \alpha, \beta < 2\piπ<α,β<2π, for which the complex number 1−isin⁡α1+2isin⁡α\frac{1-i\sin\alpha}{1+2i\sin\alpha}1+2isinα1−isinα​ is purely imaginary and 1+icos⁡β1−2icos⁡β\frac{1+i\cos\beta}{1-2i\cos\beta}1−2icosβ1+icosβ​ is purely real. Let Zαβ=sin⁡2α+icos⁡2β,(α,β)∈SZ_{\alpha\beta} = \sin 2\alpha + i\cos 2\beta, (\alpha,\beta) \in SZαβ​=sin2α+icos2β,(α,β)∈S. Then ∑(α,β)∈S(iZαβ+1i Z‾αβ)\sum_{(\alpha,\beta)\in S}\left(i Z_{\alpha\beta} + \frac{1}{i\,\overline{Z}_{\alpha\beta}}\right)∑(α,β)∈S​(iZαβ​+iZαβ​1​) is equal to :
  1. (A)333
  2. (B)3i3i3i
  3. (C)111
  4. (D)2−i2-i2−i

Correct answer: (C)

Step-by-step solution →
Q97·MathematicsSingle correctJEE Main 2022
If z=x+iyz = x + iyz=x+iy satisfies ∣z∣−2=0|z| - 2 = 0∣z∣−2=0 and ∣z−i∣−∣z+5i∣=0|z-i|-|z+5i|=0∣z−i∣−∣z+5i∣=0, then
  1. (A)x+2y−4=0x + 2y - 4 = 0x+2y−4=0
  2. (B)x2+y−4=0x^{2} + y - 4 = 0x2+y−4=0
  3. (C)x+2y+4=0x + 2y + 4 = 0x+2y+4=0
  4. (D)x2−y+3=0x^{2}- y + 3 = 0x2−y+3=0

Correct answer: (C)

Step-by-step solution →
Q98·MathematicsSingle correctJEE Main 2022
Let O be the origin and A be the point z1_{1}1​ = 1 + 2i. If B is the point z2_{2}2​, Re(z2_{2}2​) < 0, such that OAB is a right angled isosceles triangle with OB as hypotenuse, then which of the following is NOT true ?
  1. (A)arg z2_{2}2​ = π – tan−1^{-1}−1 3
  2. (B)arg (z1_{1}1​ – 2z2_{2}2​) = – tan⁡−143\tan^{-1}\dfrac{4}{3}tan−134​
  3. (C)∣z2∣=10\left|z_{2}\right| = \sqrt{10}∣z2​∣=10​
  4. (D)∣2z1−z2∣=5\left|2z_{1} - z_{2}\right| = 5∣2z1​−z2​∣=5

Correct answer: (D)

Step-by-step solution →
Q99·MathematicsSingle correctJEE Main 2022
If α,β,γ,δ\alpha, \beta, \gamma, \deltaα,β,γ,δ are the roots of the equation x4+x3+x2+x+1=0x^{4} + x^{3} + x^{2} + x + 1 = 0x4+x3+x2+x+1=0, then α2021+β2021+γ2021+δ2021\alpha^{2021} + \beta^{2021} + \gamma^{2021} + \delta^{2021}α2021+β2021+γ2021+δ2021 is equal to .
  1. (A)−4-4−4
  2. (B)−1-1−1
  3. (C)111
  4. (D)444

Correct answer: (B)

Step-by-step solution →
Q100·MathematicsNumericalJEE Main 2022
Let S={z∈C:∣z−2∣≤1,z(1+i)+zˉ(1−i)≤2}S = \{z \in C : |z - 2| \leq 1, z(1 + i) + \bar{z}(1 - i) \leq 2\}S={z∈C:∣z−2∣≤1,z(1+i)+zˉ(1−i)≤2}. Let ∣z−4i∣|z - 4i|∣z−4i∣ attains minimum and maximum values, respectively, at z1∈Sz_1 \in Sz1​∈S and z2∈Sz_2 \in Sz2​∈S. If 5(∣z1∣2+∣z2∣2)=α+β55(|z_1|^2 + |z_2|^2) = \alpha + \beta\sqrt{5}5(∣z1​∣2+∣z2​∣2)=α+β5​, where α and β are integers, then the value of α + β is equal to _____.

Correct answer: 26

Step-by-step solution →
Q101·MathematicsSingle correctJEE Main 2022
Let α and β be the roots of the equation x2+(2i−1)=0x^{2} + (2i - 1) = 0x2+(2i−1)=0. Then, the value of ∣α8+β8∣|\alpha^{8} + \beta^{8}|∣α8+β8∣ is equal to :
  1. (A)50
  2. (B)250
  3. (C)1250
  4. (D)1500

Correct answer: (A)

Step-by-step solution →
Q102·MathematicsSingle correctJEE Main 2022
Let arg (z) represent the principal argument of the complex number z. The, ∣z∣=3\left| z \right| = 3∣z∣=3 and arg⁡(z−1)−arg⁡(z+1)=π4\arg (z - 1) - \arg (z + 1) = \frac{\pi}{4}arg(z−1)−arg(z+1)=4π​ intersect:
  1. (A)Exactly at one point
  2. (B)Exactly at two points
  3. (C)Nowhere
  4. (D)At infinitely many points.

Correct answer: (C)

Step-by-step solution →
Q103·MathematicsSingle correctJEE Main 2022
Let α\alphaα be a root of the equation 1+x2+x4=01 + x^2 + x^4 = 01+x2+x4=0. Then the value of α1011+α2022−α3033\alpha^{1011} + \alpha^{2022} - \alpha^{3033}α1011+α2022−α3033 is equal to:
  1. (A)111
  2. (B)α\alphaα
  3. (C)1+α1 + \alpha1+α
  4. (D)1+2α1 + 2\alpha1+2α

Correct answer: (A)

Step-by-step solution →
Q104·MathematicsNumericalJEE Main 2022
The number of elements in the set {z=a+ib∈C:a,b∈Z and 1<∣z−3+2i∣<4}\{z = a + ib \in \mathbb{C} : a, b \in \mathbb{Z} \text{ and } 1 < |z - 3 + 2i| < 4\}{z=a+ib∈C:a,b∈Z and 1<∣z−3+2i∣<4} is ______.

Correct answer: 40

Step-by-step solution →
Q105·MathematicsNumericalJEE Main 2022
Sum of squares of modulus of all the complex numbers zzz satisfying zˉ=iz2+z2−z\bar{z} = iz^2 + z^2 - zzˉ=iz2+z2−z is equal to

Correct answer: 2

Step-by-step solution →
Q106·MathematicsSingle correctJEE Main 2022
The area of the polygon, whose vertices are the non-real roots of the equation z‾=iz2\overline{z} = iz^{2}z=iz2 is :
  1. (A)334\frac{3\sqrt{3}}{4}433​​
  2. (B)332\frac{3\sqrt{3}}{2}233​​
  3. (C)32\frac{3}{2}23​
  4. (D)34\frac{3}{4}43​

Correct answer: (A)

Step-by-step solution →
Q107·MathematicsSingle correctJEE Main 2022
Let for some real numbers α\alphaα and β\betaβ, a=α−iβa=\alpha-i\betaa=α−iβ. If the system of equations 4ix+(1+i)y=04ix+(1+i)y=04ix+(1+i)y=0 and 8(cos⁡2π3+isin⁡2π3)x+aˉy=08\left(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right)x+\bar{a}y=08(cos32π​+isin32π​)x+aˉy=0 has more than one solution then αβ\dfrac{\alpha}{\beta}βα​ is equal to :
  1. (A)−2+3-2+\sqrt{3}−2+3​
  2. (B)2−32-\sqrt{3}2−3​
  3. (C)2+32+\sqrt{3}2+3​
  4. (D)−2−3-2-\sqrt{3}−2−3​

Correct answer: (B)

Step-by-step solution →
Q108·MathematicsSingle correctJEE Main 2022
The number of points of intersection of ∣z−(4+3i)∣=2\left|z-(4+3i)\right|=2∣z−(4+3i)∣=2 and ∣z∣+∣z−4∣=6\left|z\right|+\left|z-4\right|=6∣z∣+∣z−4∣=6, z∈Cz \in \mathbb{C}z∈C is :
  1. (A)000
  2. (B)111
  3. (C)222
  4. (D)333

Correct answer: (C)

Step-by-step solution →
Q109·MathematicsNumericalJEE Main 2022
If z2+z+1=0z^{2} + z + 1 = 0z2+z+1=0, z∈Cz \in \mathbb{C}z∈C, then ∣∑n=115(Zn+(−1)n1Zn)2∣\left|\sum_{n=1}^{15}\left(Z^{n} + \left(-1\right)^{n}\frac{1}{Z^{n}}\right)^{2}\right|​∑n=115​(Zn+(−1)nZn1​)2​ is equal to ______ .

Correct answer: 2

Step-by-step solution →
Q110·MathematicsSingle correctJEE Main 2022
Let A={z∈C:∣z+1z−1∣<1}A = \left\{ z \in C : \left| \frac{z+1}{z-1} \right| < 1 \right\}A={z∈C:​z−1z+1​​<1} and B={z∈C:arg⁡(z−1z+1)=2π3}B = \left\{ z \in C : \arg\left( \frac{z-1}{z+1} \right) = \frac{2\pi}{3} \right\}B={z∈C:arg(z+1z−1​)=32π​}. Then A∩BA \cap BA∩B is :
  1. (A)a portion of a circle centred at (0,−13)\left(0, -\frac{1}{\sqrt{3}}\right)(0,−3​1​) that lies in the second and third quadrants only
  2. (B)a portion of a circle centred at (0,−13)\left(0, -\frac{1}{\sqrt{3}}\right)(0,−3​1​) that lies in the second quadrant only
  3. (C)an empty set
  4. (D)a portion of a circle of radius 23\frac{2}{\sqrt{3}}3​2​ that lies in the third quadrant only

Correct answer: (B)

Step-by-step solution →
Q111·MathematicsSingle correctJEE Main 2022
Let z1z_1z1​ and z2z_2z2​ be two complex numbers such that z‾1=iz‾2\overline{z}_1 = i\overline{z}_2z1​=iz2​ and arg⁡(z1z‾2)=π\arg\left(\frac{z_1}{\overline{z}_2}\right) = \piarg(z2​z1​​)=π. Then
  1. (A)arg⁡z2=π4\arg z_2 = \frac{\pi}{4}argz2​=4π​
  2. (B)arg⁡z2=−3π4\arg z_2 = -\frac{3\pi}{4}argz2​=−43π​
  3. (C)arg⁡z1=π4\arg z_1 = \frac{\pi}{4}argz1​=4π​
  4. (D)arg⁡z1=−3π4\arg z_1 = -\frac{3\pi}{4}argz1​=−43π​

Correct answer: (C)

Step-by-step solution →
Q112·MathematicsNumericalJEE Main 2022
Let S={z∈C:∣z−3∣≤1 and z(4+3i)+zˉ(4−3i)≤24}S = \left\{z \in \mathbb{C} : |z-3| \le 1 \text{ and } z(4+3i) + \bar{z}(4-3i) \le 24\right\}S={z∈C:∣z−3∣≤1 and z(4+3i)+zˉ(4−3i)≤24}. If α+iβ\alpha + i\betaα+iβ is the point in S which is closest to 4i4i4i, then 25(α+β)25(\alpha + \beta)25(α+β) is equal to ______.

Correct answer: 80

Step-by-step solution →
Q113·MathematicsSingle correctJEE Main 2022
Let A={z∈C:1≤∣z−(1+i)∣≤2}A=\{z\in\mathbb{C}:1\le |z-(1+i)|\le 2\}A={z∈C:1≤∣z−(1+i)∣≤2} and B={z∈A:∣z−(1−i)∣=1}B=\{z\in A:|z-(1-i)|=1\}B={z∈A:∣z−(1−i)∣=1}. Then, B :
  1. (A)is an empty set
  2. (B)contains exactly two elements
  3. (C)contains exactly three elements
  4. (D)is an infinite set

Correct answer: (D)

Step-by-step solution →
Q114·MathematicsSingle correctJEE Advanced 2021
Let θ1,θ2,...,θ10\theta_1, \theta_2, ..., \theta_{10}θ1​,θ2​,...,θ10​ be positive valued angles (in radian) such that θ1+θ2+...+θ10=2π\theta_1 + \theta_2 + ... + \theta_{10} = 2\piθ1​+θ2​+...+θ10​=2π. Define the complex numbers z1=eiθ1z_1 = e^{i\theta_1}z1​=eiθ1​, zk=zk−1eiθkz_k = z_{k-1}e^{i\theta_k}zk​=zk−1​eiθk​ for k=2,3,...,10k = 2, 3, ..., 10k=2,3,...,10, where i=−1i = \sqrt{-1}i=−1​. Consider the statements P and Q given below : P : ∣z2−z1∣+∣z3−z2∣+...+∣z10−z9∣+∣z1−z10∣≤2π\left|z_2 - z_1\right| + \left|z_3 - z_2\right| + ... + \left|z_{10} - z_9\right| + \left|z_1 - z_{10}\right| \le 2\pi∣z2​−z1​∣+∣z3​−z2​∣+...+∣z10​−z9​∣+∣z1​−z10​∣≤2π Q : ∣z22−z12∣+∣z32−z22∣+....+∣z102−z92∣+∣z12−z102∣≤4π\left|z_2^2 - z_1^2\right| + \left|z_3^2 - z_2^2\right| + .... + \left|z_{10}^2 - z_9^2\right| + \left|z_1^2 - z_{10}^2\right| \le 4\pi​z22​−z12​​+​z32​−z22​​+....+​z102​−z92​​+​z12​−z102​​≤4π Then,
  1. (A)P is TRUE and Q is FALSE
  2. (B)Q is TRUE and P is FALSE
  3. (C)both P and Q are TRUE
  4. (D)both P and Q are FALSE

Correct answer: (C)

Step-by-step solution →
Q115·MathematicsMultiple correctJEE Advanced 2021
For any complex number w=c+idw = c + idw=c+id, let arg⁡(w)∈(−π,π]\arg(w) \in (-\pi, \pi]arg(w)∈(−π,π], where i=−1i = \sqrt{-1}i=−1​. Let α\alphaα and β\betaβ be real numbers such that for all complex numbers z=x+iyz = x + iyz=x+iy satisfying arg⁡(z+αz+β)=π4\arg\left(\frac{z + \alpha}{z + \beta}\right) = \frac{\pi}{4}arg(z+βz+α​)=4π​, the ordered pair (x,y)(x, y)(x,y) lies on the circle x2+y2+5x−3y+4=0x^2 + y^2 + 5x - 3y + 4 = 0x2+y2+5x−3y+4=0. Then which of the following statements is (are) TRUE ?
  1. (A)α=−1\alpha = -1α=−1
  2. (B)αβ=4\alpha\beta = 4αβ=4
  3. (C)αβ=−4\alpha\beta = -4αβ=−4
  4. (D)β=4\beta = 4β=4

Correct answer: (B), (D)

Step-by-step solution →
Q116·MathematicsNumericalJEE Main 2021
If for the complex numbers z satisfying |z − 2 − 2i| ≤ 1, the maximum value of |3iz + 6| is attained at a + ib, then a + b is equal to ________ .

Correct answer: 5

Step-by-step solution →
Q117·MathematicsSingle correctJEE Main 2021
If ar=cos⁡2rπ9+isin⁡2rπ9a_{r} = \cos\frac{2r\pi}{9} + i \sin\frac{2r\pi}{9}ar​=cos92rπ​+isin92rπ​, r=1,2,3,…r = 1, 2, 3, \ldotsr=1,2,3,…, i=−1i = \sqrt{-1}i=−1​, then the determinant ∣a1a2a3a4a5a6a7a8a9∣\begin{vmatrix} a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9} \end{vmatrix}​a1​a4​a7​​a2​a5​a8​​a3​a6​a9​​​ is equal to :
  1. (A)a2a6−a4a8a_{2}a_{6} - a_{4}a_{8}a2​a6​−a4​a8​
  2. (B)a9a_{9}a9​
  3. (C)a1a9−a3a7a_{1}a_{9} - a_{3}a_{7}a1​a9​−a3​a7​
  4. (D)a5a_{5}a5​

Correct answer: (C)

Step-by-step solution →
Q118·MathematicsNumericalJEE Main 2021
A point z moves in the complex plane such that arg⁡(z−2z+2)=π4\arg\left( \frac{z-2}{z+2} \right) = \frac{\pi}{4}arg(z+2z−2​)=4π​, then the minimum value of ∣z−92−2i∣2\left| z - 9\sqrt{2} - 2i \right|^{2}​z−92​−2i​2 is equal to ______.

Correct answer: 98

Step-by-step solution →
Q119·MathematicsSingle correctJEE Main 2021
If zzz is a complex number such that z−iz−1\frac{z - i}{z - 1}z−1z−i​ is purely imaginary, then the minimum value of ∣z−(3+3i)∣|z - (3 + 3i)|∣z−(3+3i)∣ is :
  1. (A)22−12\sqrt{2} - 122​−1
  2. (B)323\sqrt{2}32​
  3. (C)626\sqrt{2}62​
  4. (D)222\sqrt{2}22​

Correct answer: (D)

Step-by-step solution →
Q120·MathematicsNumericalJEE Main 2021
Let z1z_{1}z1​ and z2z_{2}z2​ be two complex numbers such that arg⁡ (z1−z2)=π4\arg\,(z_{1} - z_{2}) = \frac{\pi}{4}arg(z1​−z2​)=4π​ and z1z_{1}z1​, z2z_{2}z2​ satisfy the equation ∣z−3∣=Re(z)|z - 3| = \mathrm{Re}(z)∣z−3∣=Re(z). Then the imaginary part of z1+z2z_{1} + z_{2}z1​+z2​ is equal to ___________ .

Correct answer: 6

Step-by-step solution →
Q121·MathematicsSingle correctJEE Main 2021
If S={z∈C:z−iz+2i∈R}S = \left\{ z \in \mathbb{C} : \frac{z-i}{z+2i} \in \mathbb{R} \right\}S={z∈C:z+2iz−i​∈R}, then :
  1. (A)S contains exactly two elements
  2. (B)S contains only one element
  3. (C)S is a circle in the complex plane
  4. (D)S is a straight line in the complex plane

Correct answer: (D)

Step-by-step solution →
Q122·MathematicsSingle correctJEE Main 2021
The equation arg⁡(z−1z+1)=π4\arg\left(\frac{z-1}{z+1}\right) = \frac{\pi}{4}arg(z+1z−1​)=4π​ represents a circle with:
  1. (A)centre at (0,−1)(0, -1)(0,−1) and radius 2\sqrt{2}2​
  2. (B)centre at (0,1)(0, 1)(0,1) and radius 2\sqrt{2}2​
  3. (C)centre at (0,0)(0,0)(0,0) and radius 2\sqrt{2}2​
  4. (D)centre at (0,1)(0,1)(0,1) and radius 222

Correct answer: (B)

Step-by-step solution →
Q123·MathematicsNumericalJEE Main 2021
The least positive integer n such that (2i)n(1−i)n−2,i=−1\frac{(2i)^{n}}{(1-i)^{n-2}}, i = \sqrt{-1}(1−i)n−2(2i)n​,i=−1​ is a positive integer, is ________.

Correct answer: 6

Step-by-step solution →
Q124·MathematicsNumericalJEE Main 2021
Let z=1−i32z = \frac{1-i\sqrt{3}}{2}z=21−i3​​, i=−1i = \sqrt{-1}i=−1​. Then the value of 21+(z+1z)3+(z2+1z2)3+(z3+1z3)3+...+(z21+1z21)321 + \left(z + \frac{1}{z}\right)^{3} + \left(z^{2} + \frac{1}{z^{2}}\right)^{3} + \left(z^{3} + \frac{1}{z^{3}}\right)^{3} + ... + \left(z^{21} + \frac{1}{z^{21}}\right)^{3}21+(z+z1​)3+(z2+z21​)3+(z3+z31​)3+...+(z21+z211​)3 is ________.

Correct answer: 13

Step-by-step solution →
Q125·MathematicsSingle correctJEE Main 2021
If (3+i)100=299(p+iq)\left(\sqrt{3} + i\right)^{100} = 2^{99}\left(p + iq\right)(3​+i)100=299(p+iq), then p and q are roots of the equation :
  1. (A)x2−(3−1)x−3=0x^{2} - \left(\sqrt{3} - 1\right)x - \sqrt{3} = 0x2−(3​−1)x−3​=0
  2. (B)x2+(3+1)x+3=0x^{2} + \left(\sqrt{3} + 1\right)x + \sqrt{3} = 0x2+(3​+1)x+3​=0
  3. (C)x2+(3−1)x−3=0x^{2} + \left(\sqrt{3} - 1\right)x - \sqrt{3} = 0x2+(3​−1)x−3​=0
  4. (D)x2−(3+1)x+3=0x^{2} - \left(\sqrt{3} + 1\right)x + \sqrt{3} = 0x2−(3​+1)x+3​=0

Correct answer: (A)

Step-by-step solution →
Q126·MathematicsSingle correctJEE Main 2021
Let c be the set of all complex numbers. Let S1_11​ = {z∈C:∣z−2∣≤1}\left\{ z \in C : |z - 2| \le 1 \right\}{z∈C:∣z−2∣≤1} and S2_22​ = {z∈C:z(1+i)+zˉ(1−i)≥4}\left\{ z \in C : z(1+i) + \bar{z}(1-i) \ge 4 \right\}{z∈C:z(1+i)+zˉ(1−i)≥4}. Then, the maximum value of ∣z−52∣2\left| z - \frac{5}{2} \right|^2​z−25​​2 for z∈S1∩S2z \in S_1 \cap S_2z∈S1​∩S2​ is equal to :
  1. (A)5+224\frac{5 + 2\sqrt{2}}{4}45+22​​
  2. (B)5+222\frac{5 + 2\sqrt{2}}{2}25+22​​
  3. (C)3+224\frac{3 + 2\sqrt{2}}{4}43+22​​
  4. (D)3+222\frac{3 + 2\sqrt{2}}{2}23+22​​

Correct answer: (A)

Step-by-step solution →
Q127·MathematicsNumericalJEE Main 2021
If the real part of the complex number z = 3+2icos⁡θ1−3icos⁡θ\frac{3+2i\cos\theta}{1-3i\cos\theta}1−3icosθ3+2icosθ​ , θ ∈ (0,π2)\left(0,\frac{\pi}{2}\right)(0,2π​) is zero, then the value of sin⁡23θ\sin^{2}3\thetasin23θ + cos⁡2θ\cos^{2}\thetacos2θ is equal to………

Correct answer: 1

Step-by-step solution →
Q128·MathematicsSingle correctJEE Main 2021
Let C be the set of all complex numbers. Let S1={z∈C | ∣z−3−2i∣2=8}S_1 = \left\{ z \in C \,\middle|\, \left| z - 3 - 2i \right|^2 = 8 \right\}S1​={z∈C​∣z−3−2i∣2=8}, S2={z∈C | Re⁡(z)≥5}S_2 = \left\{ z \in C \,\middle|\, \operatorname{Re}(z) \ge 5 \right\}S2​={z∈C∣Re(z)≥5} and S3={z∈C | ∣z−zˉ∣≥8}S_3 = \left\{ z \in C \,\middle|\, \left| z - \bar{z} \right| \ge 8 \right\}S3​={z∈C∣∣z−zˉ∣≥8}. Then the number of element in S1∩S2∩S3S_1 \cap S_2 \cap S_3S1​∩S2​∩S3​ is equal to:
  1. (A)0
  2. (B)2
  3. (C)1
  4. (D)infinite

Correct answer: (C)

Step-by-step solution →
Q129·MathematicsNumericalJEE Main 2021
The equation of a circle is Re(z2)+2(Im(z))2+2Re(z)=0Re\left(z^{2}\right) + 2\left(Im(z)\right)^{2} + 2Re(z) = 0Re(z2)+2(Im(z))2+2Re(z)=0, where z=x+iyz = x + iyz=x+iy. A line which passes through the center of the given circle and the vertex of the parabola, x2−6x−y+13=0x^{2} - 6x - y + 13 = 0x2−6x−y+13=0, has y-intercept equal to.........

Correct answer: 1

Step-by-step solution →
Q130·MathematicsSingle correctJEE Main 2021
Let n denote the number of solutions of the equation z2+3zˉ=0z^2 + 3\bar{z} = 0z2+3zˉ=0, where z is a complex number. Then the value of ∑k=0∞1nk\sum_{k=0}^{\infty} \frac{1}{n^k}∑k=0∞​nk1​ is equal to :
  1. (A)1
  2. (B)2
  3. (C)32\frac{3}{2}23​
  4. (D)43\frac{4}{3}34​

Correct answer: (D)

Step-by-step solution →
Q131·MathematicsSingle correctJEE Main 2021
If the real part of the complex number (1−cos⁡θ+2isin⁡θ)−1\left(1-\cos\theta+2i\sin\theta\right)^{-1}(1−cosθ+2isinθ)−1 is 15\frac{1}{5}51​ for θ∈(0,π)\theta\in(0,\pi)θ∈(0,π), then the value of the integral ∫0θsin⁡x dx\int_{0}^{\theta}\sin x\,dx∫0θ​sinxdx is equal to :
  1. (A)−1-1−1
  2. (B)000
  3. (C)222
  4. (D)111

Correct answer: (D)

Step-by-step solution →
Q132·MathematicsSingle correctJEE Main 2021
If z and ω are two complex numbers such that ∣zω∣=1|z\omega| = 1∣zω∣=1 and arg⁡(z)−arg⁡(w)=3π2\arg(z) - \arg(w) = \frac{3\pi}{2}arg(z)−arg(w)=23π​, then arg⁡(1−2zˉω1+3zˉω)\arg\left(\frac{1 - 2\bar{z}\omega}{1 + 3\bar{z}\omega}\right)arg(1+3zˉω1−2zˉω​) is : (Here arg(z) denotes the principal argument of complex number z)
  1. (A)π4\frac{\pi}{4}4π​
  2. (B)−3π4-\frac{3\pi}{4}−43π​
  3. (C)−π4-\frac{\pi}{4}−4π​
  4. (D)3π4\frac{3\pi}{4}43π​

Correct answer: (B)

Step-by-step solution →
Q133·MathematicsNumericalJEE Main 2021
If f(x)f(x)f(x) and g(x) are two polynomials such that the polynomial P(x)=f(x3)+xg(x3)P(x) = f(x^{3}) + xg(x^{3})P(x)=f(x3)+xg(x3) is divisible by x2+x+1x^{2} + x + 1x2+x+1, then P(1) is equal to ________ .

Correct answer: 0

Step-by-step solution →
Q134·MathematicsNumericalJEE Main 2021
Let z1,z2z_{1}, z_{2}z1​,z2​ be the roots of the equation z2+az+12=0z^{2} + az + 12 = 0z2+az+12=0 and z1,z2z_{1}, z_{2}z1​,z2​ form an equilateral triangle with origin. Then, the value of ∣a∣|a|∣a∣ is

Correct answer: 6

Step-by-step solution →
Q135·MathematicsSingle correctJEE Main 2021
Let a complex number be w = 1 − 3i\sqrt{3}i3​i. Let another complex number z be such that |zw| = 1 and arg(z) − arg(w) = π2\frac{\pi}{2}2π​. Then the area of the triangle with vertices origin, z and w is equal to :
  1. (A)4
  2. (B)12\frac{1}{2}21​
  3. (C)14\frac{1}{4}41​
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q136·MathematicsSingle correctJEE Main 2021
If the equation a∣z∣2+α‾z+αz‾+d=0a\left|z\right|^{2} + \overline{\alpha}z + \alpha\overline{z} + d = 0a∣z∣2+αz+αz+d=0 represents a circle where a,d are real constants then which of the following condition is correct ?
  1. (A)∣α∣2−ad≠0\left|\alpha\right|^{2} - ad \neq 0∣α∣2−ad=0
  2. (B)∣α∣2−ad>0|\alpha|^{2} - ad > 0∣α∣2−ad>0 and a∈R−{0}a \in R - \{0\}a∈R−{0}
  3. (C)∣α∣2−ad≥0|\alpha|^{2} - ad \geq 0∣α∣2−ad≥0 and a∈Ra \in Ra∈R
  4. (D)α=0\alpha = 0α=0, a,d∈R+a, d \in R^{+}a,d∈R+

Correct answer: (B)

Step-by-step solution →
Q137·MathematicsSingle correctJEE Main 2021
The area of the triangle with vertices A(z)A(z)A(z), B(iz)B(iz)B(iz) and C(z+iz)C(z + iz)C(z+iz) is :
  1. (A)1
  2. (B)12∣z∣2\frac{1}{2}|z|^221​∣z∣2
  3. (C)12\frac{1}{2}21​
  4. (D)12∣z+iz∣2\frac{1}{2}|z + iz|^221​∣z+iz∣2

Correct answer: (B)

Step-by-step solution →
Q138·MathematicsSingle correctJEE Main 2021
Let S1S_{1}S1​, S2S_{2}S2​ and S3S_{3}S3​ be three sets defined as S1={z∈C:∣z−1∣≤2}S_{1} = \left\{z \in \mathbb{C} : |z-1| \le \sqrt{2}\right\}S1​={z∈C:∣z−1∣≤2​} S2={z∈C:Re((1−i)z)≥1}S_{2} = \left\{z \in \mathbb{C} : \mathrm{Re}\left((1-i)z\right) \ge 1\right\}S2​={z∈C:Re((1−i)z)≥1} S3={z∈C:Im(z)≤1}S_{3} = \left\{z \in \mathbb{C} : \mathrm{Im}(z) \le 1\right\}S3​={z∈C:Im(z)≤1} Then the set S1∩S2∩S3S_{1} \cap S_{2} \cap S_{3}S1​∩S2​∩S3​
  1. (A)is a singleton
  2. (B)has exactly two elements
  3. (C)has infinitely many elements
  4. (D)has exactly three elements

Correct answer: (C)

Step-by-step solution →
Q139·MathematicsNumericalJEE Main 2021
Let z and w be two complex numbers such that w=zzˉ−2z+2w = z\bar{z} - 2z + 2w=zzˉ−2z+2, ∣z+iz−3i∣=1\left|\frac{z+i}{z-3i}\right| = 1​z−3iz+i​​=1 and Re(w) has minimum value. Then, the minimum value of n ∈ ℕ for which wnw^{n}wn is real, is equal to _____.

Correct answer: 4

Step-by-step solution →
Q140·MathematicsSingle correctJEE Main 2021
The least value of ∣z∣|z|∣z∣ where z is complex number which satisfies the inequality exp⁡((∣z∣+3)(∣z∣−1)∣∣z∣+1∣log⁡e2)≥log⁡2∣57+9i∣,\exp\left(\frac{\left(|z|+3\right)\left(|z|-1\right)}{\left||z|+1\right|}\log_{e}2\right) \ge \log_{\sqrt{2}}\left|5\sqrt{7}+9i\right|,exp(∣∣z∣+1∣(∣z∣+3)(∣z∣−1)​loge​2)≥log2​​​57​+9i​, i=−1i=\sqrt{-1}i=−1​, is equal to :
  1. (A)333
  2. (B)5\sqrt{5}5​
  3. (C)222
  4. (D)888

Correct answer: (A)

Step-by-step solution →
Q141·MathematicsSingle correctJEE Main 2021
Let a complex number z, |z| ≠ 1, satisfy log⁡12(∣z∣+11(∣z∣−1)2)≤2\log_{\frac{1}{\sqrt{2}}}\left(\frac{|z| + 11}{(|z| - 1)^{2}}\right) \le 2log2​1​​((∣z∣−1)2∣z∣+11​)≤2. Then, the largest value of |z| is equal to _____ .
  1. (A)8
  2. (B)7
  3. (C)6
  4. (D)5

Correct answer: (B)

Step-by-step solution →
Q142·MathematicsNumericalJEE Main 2021
Let z be those complex number which satisfy ∣z+5∣≤4|z+5| \leq 4∣z+5∣≤4 and z(1+i)+zˉ(1−i)≥−10,i=−1z(1+i) + \bar{z}\left(1-i\right) \geq -10, i = \sqrt{-1}z(1+i)+zˉ(1−i)≥−10,i=−1​. If the maximum value of ∣z+1∣2|z+1|^{2}∣z+1∣2 is α+β2\alpha + \beta\sqrt{2}α+β2​, then the value of (α+β)\left(\alpha + \beta\right)(α+β) is ________________.

Correct answer: 48

Step-by-step solution →
Q143·MathematicsNumericalJEE Main 2021
The sum of 162th162^{\text{th}}162th power of the roots of the equation x3−2x2+2x−1=0x^{3}-2x^{2}+2x-1=0x3−2x2+2x−1=0 is _______.

Correct answer: 3

Step-by-step solution →
Q144·MathematicsSingle correctJEE Main 2021
Let the lines (2−i)z=(2+i)zˉ(2 - i)z = (2 + i)\bar{z}(2−i)z=(2+i)zˉ and (2+i)z+(i−2)zˉ−4i=0(2 + i)z + (i - 2)\bar{z} - 4i = 0(2+i)z+(i−2)zˉ−4i=0, (here i2=−1i^{2} = -1i2=−1) be normal to a circle C. If the line iz+zˉ+1+i=0iz + \bar{z} + 1 + i = 0iz+zˉ+1+i=0 is tangent to this circle C, then its radius is :
  1. (A)32\frac{3}{\sqrt{2}}2​3​
  2. (B)323\sqrt{2}32​
  3. (C)322\frac{3}{2\sqrt{2}}22​3​
  4. (D)122\frac{1}{2\sqrt{2}}22​1​

Correct answer: (C)

Step-by-step solution →
Q145·MathematicsSingle correctJEE Main 2021
If α, β∈ R are such that 1 – 2i (here i2^{2}2 = –1) is a root of z2^{2}2 + αz + β = 0, then (α – β) is equal to:
  1. (A)7
  2. (B)–3
  3. (C)3
  4. (D)–7

Correct answer: (D)

Step-by-step solution →
Q146·MathematicsNumericalJEE Main 2021
If the least and the largest real values of α\alphaα, for which the equation z+α∣z−1∣+2i=0z + \alpha|z - 1| + 2i = 0z+α∣z−1∣+2i=0 (z∈C and i=−1)\left(z \in C \text{ and } i = \sqrt{-1}\right)(z∈C and i=−1​) has a solution, are p and q respectively; then 4(p2+q2)4(p^2 + q^2)4(p2+q2) is equal to ____

Correct answer: 10

Step-by-step solution →
Q147·MathematicsNumericalJEE Main 2021
Let i=−1i=\sqrt{-1}i=−1​. If (−1+i3)21(1−i)24+(1+i3)21(1+i)24=k,\frac{(-1+i\sqrt{3})^{21}}{(1-i)^{24}}+\frac{(1+i\sqrt{3})^{21}}{(1+i)^{24}}=k,(1−i)24(−1+i3​)21​+(1+i)24(1+i3​)21​=k, and n=[∣k∣]\mathrm{n}=\left[\left|k\right|\right]n=[∣k∣] be the greatest integral part of ∣k∣|k|∣k∣. Then ∑j=0n+5(j+5)2−∑j=0n+5(j+5)\sum_{j=0}^{n+5}(j+5)^{2}-\sum_{j=0}^{n+5}(j+5)∑j=0n+5​(j+5)2−∑j=0n+5​(j+5) is equal to________.

Correct answer: 310

Step-by-step solution →
Q148·MathematicsIntegerJEE Advanced 2020
For a complex number zzz, let Re(z)\mathrm{Re}(z)Re(z) denote the real part of zzz. Let SSS be the set of all complex numbers zzz satisfying z4−∣z∣4=4iz2z^{4} - |z|^{4} = 4iz^{2}z4−∣z∣4=4iz2, where i=−1i = \sqrt{-1}i=−1​. Then the minimum possible value of ∣z1−z2∣2|z_{1} - z_{2}|^{2}∣z1​−z2​∣2, where z1,z2∈Sz_{1}, z_{2} \in Sz1​,z2​∈S with Re(z1)>0\mathrm{Re}(z_{1}) > 0Re(z1​)>0 and Re(z2)<0\mathrm{Re}(z_{2}) < 0Re(z2​)<0, is ________

Correct answer: 8

Step-by-step solution →
Q149·MathematicsMultiple correctJEE Advanced 2020
Let S be the set of all complex numbers z satisfying ∣z2+z+1∣=1|z^{2} + z + 1| = 1∣z2+z+1∣=1. Then which of the following statements is/are TRUE ?
  1. (A)∣z+12∣≤12\left|z + \frac{1}{2}\right| \leq \frac{1}{2}​z+21​​≤21​ for all z∈Sz \in Sz∈S
  2. (B)∣z∣≤2|z| \leq 2∣z∣≤2 for all z∈Sz \in Sz∈S
  3. (C)∣z+12∣≥12\left|z + \frac{1}{2}\right| \geq \frac{1}{2}​z+21​​≥21​ for all z∈Sz \in Sz∈S
  4. (D)The set S has exactly four elements

Correct answer: (B), (C)

Step-by-step solution →
Q150·MathematicsSingle correctJEE Main 2020
Let z=x+iyz = x + iyz=x+iy be a non-zero complex number such that z2=i∣z∣2z^2 = i|z|^2z2=i∣z∣2, where i=−1i = \sqrt{-1}i=−1​, then z lies on the:
  1. (A)line, y=−xy = -xy=−x
  2. (B)imaginary axis
  3. (C)line, y=xy = xy=x
  4. (D)real axis

Correct answer: (C)

Step-by-step solution →
Q151·MathematicsSingle correctJEE Main 2020
The region represented by {z=x+iy∈C:∣z∣−Re(z)≤1}\{z = x+iy \in C : |z| - \mathrm{Re}(z) \le 1\}{z=x+iy∈C:∣z∣−Re(z)≤1} is also given by the inequality:
  1. (A)y2≥2(x+1)y^2 \ge 2(x+1)y2≥2(x+1)
  2. (B)y2≤2(x+12)y^2 \le 2\left(x+\dfrac{1}{2}\right)y2≤2(x+21​)
  3. (C)y2≤x+12y^2 \le x + \dfrac{1}{2}y2≤x+21​
  4. (D)y2≥x+1y^2 \ge x+1y2≥x+1

Correct answer: (B)

Step-by-step solution →
Q152·MathematicsSingle correctJEE Main 2020
The value of (−1+i31−i)30\left(\dfrac{-1+i\sqrt{3}}{1-i}\right)^{30}(1−i−1+i3​​)30 is:
  1. (A)215i2^{15}i215i
  2. (B)−215-2^{15}−215
  3. (C)656^565
  4. (D)−215i-2^{15}i−215i

Correct answer: (D)

Step-by-step solution →
Q153·MathematicsSingle correctJEE Main 2020
If the four complex numbers, z,zˉ,zˉ−2Re(zˉ)z, \bar{z}, \bar{z} - 2\text{Re}(\bar{z})z,zˉ,zˉ−2Re(zˉ) and z−2Re(z)z - 2\text{Re}(z)z−2Re(z) represent the vertices of a square of side 4 units in the Argand plane, then ∣z∣|z|∣z∣ is equal to:
  1. (A)222\sqrt{2}22​
  2. (B)2
  3. (C)424\sqrt{2}42​
  4. (D)4

Correct answer: (A)

Step-by-step solution →
Q154·MathematicsSingle correctJEE Main 2020
If a and b are real numbers such that (2+α)4=a+bα(2 + \alpha)^{4} = a + b\alpha(2+α)4=a+bα, where α=−1+i32\alpha = \frac{-1 + i\sqrt{3}}{2}α=2−1+i3​​, then a + b is equal to:
  1. (A)9
  2. (B)33
  3. (C)57
  4. (D)24

Correct answer: (A)

Step-by-step solution →
Q155·MathematicsSingle correctJEE Main 2020
Let u=2z+iz−kiu = \frac{2z + i}{z - ki}u=z−ki2z+i​, z=x+iyz = x + iyz=x+iy and k>0k > 0k>0. If the curve represented by Re(u)+Im(u)=1\mathrm{Re}(u) + \mathrm{Im}(u) = 1Re(u)+Im(u)=1 intersects the y-axis at the points P and Q where PQ = 5, then the value of k is:
  1. (A)222
  2. (B)12\frac{1}{2}21​
  3. (C)32\frac{3}{2}23​
  4. (D)444

Correct answer: (A)

Step-by-step solution →
Q156·MathematicsNumericalJEE Main 2020
If (1+i1−i)m/2=(1+ii−1)n/3=1\left(\frac{1+i}{1-i}\right)^{m/2} = \left(\frac{1+i}{i-1}\right)^{n/3} = 1(1−i1+i​)m/2=(i−11+i​)n/3=1, (m,n∈N)(m, n \in N)(m,n∈N) Then the greatest common divisor of the least values of m and n is __________.

Correct answer: 4

Step-by-step solution →
Q157·MathematicsSingle correctJEE Main 2020
If z1,z2z_1, z_2z1​,z2​ are complex numbers such that Re(z1)=∣z1−1∣,Re(z2)=∣z2−1∣\mathrm{Re}(z_1) = |z_1 - 1|, \mathrm{Re}(z_2) = |z_2 - 1|Re(z1​)=∣z1​−1∣,Re(z2​)=∣z2​−1∣ and arg⁡(z1−z2)=π6\arg(z_1 - z_2) = \frac{\pi}{6}arg(z1​−z2​)=6π​, then Im(z1+z2)\mathrm{Im}(z_1 + z_2)Im(z1​+z2​) is equal to:
  1. (A)232\sqrt{3}23​
  2. (B)32\frac{\sqrt{3}}{2}23​​
  3. (C)13\frac{1}{\sqrt{3}}3​1​
  4. (D)23\frac{2}{\sqrt{3}}3​2​

Correct answer: (A)

Step-by-step solution →
Q158·MathematicsSingle correctJEE Main 2020
If z be a complex number satisfying ∣Re(z)∣+∣Im(z)∣=4\left|Re(z)\right| + \left|Im(z)\right| = 4∣Re(z)∣+∣Im(z)∣=4, then ∣z∣\left|z\right|∣z∣ cannot be:
  1. (A)172\sqrt{\dfrac{17}{2}}217​​
  2. (B)10\sqrt{10}10​
  3. (C)8\sqrt{8}8​
  4. (D)7\sqrt{7}7​

Correct answer: (D)

Step-by-step solution →
Q159·MathematicsSingle correctJEE Main 2020
Let z be a complex number such that ∣z−iz+2i∣=1\left|\dfrac{z-i}{z+2i}\right|=1​z+2iz−i​​=1 and ∣z∣=52|z|=\dfrac{5}{2}∣z∣=25​. Then the value of ∣z+3i∣|z+3i|∣z+3i∣ is:
  1. (A)154\dfrac{15}{4}415​
  2. (B)232\sqrt{3}23​
  3. (C)10\sqrt{10}10​
  4. (D)72\dfrac{7}{2}27​

Correct answer: (D)

Step-by-step solution →
Q160·MathematicsSingle correctJEE Main 2020
If the equation, x2+bx+45=0 (b∈R)x^{2}+bx+45=0\ (b\in R)x2+bx+45=0 (b∈R) has conjugate complex roots and they satisfy ∣z+1∣=210|z+1|=2\sqrt{10}∣z+1∣=210​, then:
  1. (A)b2+b=12b^{2}+b=12b2+b=12
  2. (B)b2−b=42b^{2}-b=42b2−b=42
  3. (C)b2−b=30b^{2}-b=30b2−b=30
  4. (D)b2+b=72b^{2}+b=72b2+b=72

Correct answer: (C)

Step-by-step solution →
Q161·MathematicsSingle correctJEE Main 2020
Let α=−1+i32\alpha=\dfrac{-1+i\sqrt{3}}{2}α=2−1+i3​​. If a=(1+α)∑k=0100α2ka=(1+\alpha)\displaystyle\sum_{k=0}^{100}\alpha^{2k}a=(1+α)k=0∑100​α2k and b=∑k=0100α3kb=\displaystyle\sum_{k=0}^{100}\alpha^{3k}b=k=0∑100​α3k, then a and b are the roots of the quadratic equation:
  1. (A)x2−101x+100=0x^{2}-101x+100=0x2−101x+100=0
  2. (B)x2−102x+101=0x^{2}-102x+101=0x2−102x+101=0
  3. (C)x2+102x+101=0x^{2}+102x+101=0x2+102x+101=0
  4. (D)x2+101x+100=0x^{2}+101x+100=0x2+101x+100=0

Correct answer: (B)

Step-by-step solution →
Q162·MathematicsSingle correctJEE Main 2020
If 3+isin⁡θ4−icos⁡θ,θ∈[0,2π]\dfrac{3+i\sin\theta}{4-i\cos\theta}, \theta \in [0,2\pi]4−icosθ3+isinθ​,θ∈[0,2π], is a real number, then an argument of sin⁡θ+icos⁡θ\sin\theta+i\cos\thetasinθ+icosθ is:
  1. (A)−tan⁡−1(34)-\tan^{-1}\left(\dfrac{3}{4}\right)−tan−1(43​)
  2. (B)π−tan⁡−1(43)\pi-\tan^{-1}\left(\dfrac{4}{3}\right)π−tan−1(34​)
  3. (C)π−tan⁡−1(34)\pi-\tan^{-1}\left(\dfrac{3}{4}\right)π−tan−1(43​)
  4. (D)tan⁡−1(43)\tan^{-1}\left(\dfrac{4}{3}\right)tan−1(34​)

Correct answer: (B)

Step-by-step solution →
Q163·MathematicsSingle correctJEE Main 2020
If Re(z−12z+i)=1Re\left(\frac{z-1}{2z+i}\right)=1Re(2z+iz−1​)=1, where z=x+iyz=x+iyz=x+iy, then the point (x, y) lies on a:
  1. (A)straight line whose slope is 32\frac{3}{2}23​
  2. (B)circle whose diameter is 52\frac{\sqrt{5}}{2}25​​
  3. (C)straight line whose slope is −23-\frac{2}{3}−32​
  4. (D)circle whose centre is at (−12,−32)\left(-\frac{1}{2},-\frac{3}{2}\right)(−21​,−23​)

Correct answer: (B)

Step-by-step solution →
Q164·MathematicsNumericalJEE Advanced 2019
Let ω≠1\omega \ne 1ω=1 be a cube root of unity. Then the minimum of the set {∣a+bω+cω2∣2:a,b,c\{|a + b\omega + c\omega^2|^2 : a, b, c{∣a+bω+cω2∣2:a,b,c distinct non-zero integers}\}} equals ____

Correct answer: 3.00

Step-by-step solution →
Q165·MathematicsSingle correctJEE Advanced 2019
Let S be the set of all complex numbers z satisfying ∣z−2+i∣≥5|z - 2 + i| \ge \sqrt{5}∣z−2+i∣≥5​. If the complex number z0z_0z0​ is such that 1∣z0−1∣\frac{1}{|z_0 - 1|}∣z0​−1∣1​ is the maximum of the set {1∣z−1∣:z∈S}\left\{ \frac{1}{|z - 1|} : z \in S \right\}{∣z−1∣1​:z∈S}, then the principal argument of 4−z0−z0‾z0−z0‾+2i\frac{4 - z_0 - \overline{z_0}}{z_0 - \overline{z_0} + 2i}z0​−z0​​+2i4−z0​−z0​​​ is
  1. (A)3π4\frac{3\pi}{4}43π​
  2. (B)π4\frac{\pi}{4}4π​
  3. (C)−π2-\frac{\pi}{2}−2π​
  4. (D)π2\frac{\pi}{2}2π​

Correct answer: (C)

Step-by-step solution →
Q166·MathematicsSingle correctJEE Main 2019
Let z∈Cz \in Cz∈C with Im(z)=10\text{Im}(z) = 10Im(z)=10 and it satisfies 2z−n2z+n=2i−1\dfrac{2z-n}{2z+n} = 2i - 12z+n2z−n​=2i−1 for some natural number n. Then :
  1. (A)n = 40 and Re(z) = 10
  2. (B)n = 20 and Re(z) = 10
  3. (C)n = 40 and Re(z) = −10-10−10
  4. (D)n = 20 and Re(z) = −10-10−10

Correct answer: (C)

Step-by-step solution →
Q167·MathematicsSingle correctJEE Main 2019
The equation ∣z−i∣=∣z−1∣,i=−1|z-i|=|z-1|, i=\sqrt{-1}∣z−i∣=∣z−1∣,i=−1​, represents :
  1. (A)a circle of radius 12\frac{1}{2}21​
  2. (B)the line through the origin with slope 1
  3. (C)a circle of radius 1
  4. (D)the line through the origin with slope −1-1−1

Correct answer: (B)

Step-by-step solution →
Q168·MathematicsSingle correctJEE Main 2019
If a > 0 and z = (1+i)2a−i\frac{(1+i)^{2}}{a-i}a−i(1+i)2​, has magnitude 25\sqrt{\frac{2}{5}}52​​, then zˉ\bar{z}zˉ is equal to:
  1. (A)−35−15i-\frac{3}{5}-\frac{1}{5}i−53​−51​i
  2. (B)−15−35i-\frac{1}{5}-\frac{3}{5}i−51​−53​i
  3. (C)−15+35i-\frac{1}{5}+\frac{3}{5}i−51​+53​i
  4. (D)15−35i\frac{1}{5}-\frac{3}{5}i51​−53​i

Correct answer: (B)

Step-by-step solution →
Q169·MathematicsSingle correctJEE Main 2019
If z and w are two complex numbers such that ∣zw∣=1|zw| = 1∣zw∣=1 and arg⁡(z)−arg⁡(w)=π2\arg(z) - \arg(w) = \dfrac{\pi}{2}arg(z)−arg(w)=2π​, then
  1. (A)zˉw=i\bar{z}w = izˉw=i
  2. (B)zwˉ=−1+i2z\bar{w} = \dfrac{-1 + i}{\sqrt{2}}zwˉ=2​−1+i​
  3. (C)zwˉ=1−i2z\bar{w} = \dfrac{1 - i}{\sqrt{2}}zwˉ=2​1−i​
  4. (D)zˉw=−i\bar{z}w = -izˉw=−i

Correct answer: (D)

Step-by-step solution →
Q170·MathematicsSingle correctJEE Main 2019
All the points in the set S={α+iα−i:α∈R}S = \left\{\dfrac{\alpha+i}{\alpha-i} : \alpha \in R\right\}S={α−iα+i​:α∈R} (i=−1)(i=\sqrt{-1})(i=−1​) lie on a:
  1. (A)straight line whose slope is 1
  2. (B)circle whose radius is 2\sqrt{2}2​
  3. (C)straight line whose slope is −1-1−1
  4. (D)circle whose radius is 1

Correct answer: (D)

Step-by-step solution →
Q171·MathematicsSingle correctJEE Main 2019
Let z∈Cz \in Cz∈C be such that ∣z∣<1|z|<1∣z∣<1. If w=5+3z5(1−z)w=\dfrac{5+3z}{5(1-z)}w=5(1−z)5+3z​, then
  1. (A)5 Im(w)<15\,\text{Im}(w)<15Im(w)<1
  2. (B)4 Im(w)>54\,\text{Im}(w)>54Im(w)>5
  3. (C)5 Re(w)>15\,\text{Re}(w)>15Re(w)>1
  4. (D)5 Re(w)>45\,\text{Re}(w)>45Re(w)>4

Correct answer: (C)

Step-by-step solution →
Q172·MathematicsSingle correctJEE Main 2019
If z=32+i2(i+−1)z=\frac{\sqrt{3}}{2}+\frac{i}{2}\left(i+\sqrt{-1}\right)z=23​​+2i​(i+−1​), then (1+iz+z5+iz8)9\left(1+iz+z^{5}+iz^{8}\right)^{9}(1+iz+z5+iz8)9 is equal to:
  1. (A)-1
  2. (B)1
  3. (C)(−1+2i)9(-1+2i)^{9}(−1+2i)9
  4. (D)0

Correct answer: (A)

Step-by-step solution →
Q173·MathematicsSingle correctJEE Main 2019
If α\alphaα and β\betaβ be the roots of the equation x2−2x+2=0x^{2}-2x+2=0x2−2x+2=0, then the least value of n for which (αβ)n=1\left(\frac{\alpha}{\beta}\right)^{n}=1(βα​)n=1 is:
  1. (A)444
  2. (B)222
  3. (C)555
  4. (D)333

Correct answer: (A)

Step-by-step solution →
Q174·MathematicsSingle correctJEE Main 2019
Let z1z_{1}z1​ and z2z_{2}z2​ be two complex numbers satisfying ∣z1∣=9|z_{1}| = 9∣z1​∣=9 and ∣z2−3−4i∣=4|z_{2} - 3 - 4i| = 4∣z2​−3−4i∣=4. Then the minimum value of ∣z1−z2∣|z_{1} - z_{2}|∣z1​−z2​∣ is :
  1. (A)000
  2. (B)2\sqrt{2}2​
  3. (C)111
  4. (D)222

Correct answer: (A)

Step-by-step solution →
Q175·MathematicsSingle correctJEE Main 2019
If z−αz+α(α∈R)\frac{z-\alpha}{z+\alpha}(\alpha \in R)z+αz−α​(α∈R) is a purely imaginary number and ∣z∣=2|z|=2∣z∣=2, then a value of α\alphaα is:
  1. (A)2
  2. (B)1
  3. (C)12\frac{1}{2}21​
  4. (D)2\sqrt{2}2​

Correct answer: (A)

Step-by-step solution →
Q176·MathematicsSingle correctJEE Main 2019
Let z be a complex number such that ∣z∣+z=3+i|z|+z=3+i∣z∣+z=3+i (where i=−1i=\sqrt{-1}i=−1​). Then ∣z∣|z|∣z∣ is equal to:
  1. (A)343\dfrac{\sqrt{34}}{3}334​​
  2. (B)53\dfrac{5}{3}35​
  3. (C)414\dfrac{\sqrt{41}}{4}441​​
  4. (D)54\dfrac{5}{4}45​

Correct answer: (B)

Step-by-step solution →
Q177·MathematicsSingle correctJEE Main 2019
Let (−2−13i)3=x+iy27\left(-2-\dfrac{1}{3}i\right)^{3}=\dfrac{x+iy}{27}(−2−31​i)3=27x+iy​ (i=−1)(i=\sqrt{-1})(i=−1​), where x and y are real numbers, then y−xy-xy−x equals:
  1. (A)91
  2. (B)-85
  3. (C)85
  4. (D)-91

Correct answer: (A)

Step-by-step solution →
Q178·MathematicsSingle correctJEE Main 2019
Let A={θ∈(−π2,π):3+2isin⁡θ1−2isin⁡θ purely imaginary}A=\left\{\theta\in\left(-\dfrac{\pi}{2},\pi\right):\dfrac{3+2i\sin\theta}{1-2i\sin\theta}\ \text{purely imaginary}\right\}A={θ∈(−2π​,π):1−2isinθ3+2isinθ​ purely imaginary}. Then the sum of the elements in A is:
  1. (A)5π6\dfrac{5\pi}{6}65π​
  2. (B)π\piπ
  3. (C)3π4\dfrac{3\pi}{4}43π​
  4. (D)2π3\dfrac{2\pi}{3}32π​

Correct answer: (D)

Step-by-step solution →
Q179·MathematicsSingle correctJEE Main 2019
Let α\alphaα and β\betaβ be two roots of the equation x2+2x+2=0x^{2}+2x+2=0x2+2x+2=0, then α15+β15\alpha^{15}+\beta^{15}α15+β15 is equal to:
  1. (A)−256-256−256
  2. (B)512512512
  3. (C)−512-512−512
  4. (D)256256256

Correct answer: (A)

Step-by-step solution →
Q180·MathematicsSingle correctJEE Main 2019
Let z0z_{0}z0​ be a root of the quadratic equation, x2+x+1=0x^{2}+x+1=0x2+x+1=0. If z=3+6iz081−3iz093z=3+6iz_{0}^{81}-3iz_{0}^{93}z=3+6iz081​−3iz093​, then arg⁡z\arg zargz is equal to
  1. (A)π4\frac{\pi}{4}4π​
  2. (B)π3\frac{\pi}{3}3π​
  3. (C)0
  4. (D)π6\frac{\pi}{6}6π​

Correct answer: (A)

Step-by-step solution →
Q181·MathematicsMultiple correctJEE Advanced 2018
Let sss, ttt, rrr be non-zero complex numbers and L be the set of solutions z=x+iyz = x + iyz=x+iy (xxx, y∈Ry \in Ry∈R, i=−1i = \sqrt{-1}i=−1​) of the equation sz+tzˉ+r=0sz + t\bar{z} + r = 0sz+tzˉ+r=0, where zˉ=x−iy\bar{z} = x - iyzˉ=x−iy. Then, which of the following statement(s) is (are) TRUE ?
  1. (A)If L has exactly one element, then ∣s∣≠∣t∣|s| \neq |t|∣s∣=∣t∣
  2. (B)If ∣s∣=∣t∣|s| = |t|∣s∣=∣t∣, then L has infinitely many elements
  3. (C)The number of elements in L∩{z:∣z−1+i∣=5}L \cap \{z : |z - 1 + i| = 5\}L∩{z:∣z−1+i∣=5} is at most 2
  4. (D)If L has more than one element, then L has infinitely many elements

Correct answer: (A), (C), (D)

Step-by-step solution →
Q182·MathematicsMultiple correctJEE Advanced 2018
For a non-zero complex number z, let arg(z) denote the principal argument with −π<arg⁡(z)≤π-\pi < \arg(z) \le \pi−π<arg(z)≤π. Then, which of the following statement(s) is (are) FALSE ?
  1. (A)arg⁡(−1−i)=π4\arg(-1-i) = \frac{\pi}{4}arg(−1−i)=4π​, where i=−1i = \sqrt{-1}i=−1​
  2. (B)The function f:R→(−π,π]f : R \to (-\pi, \pi]f:R→(−π,π], defined by f(t)=arg⁡(−1+it)f(t) = \arg(-1+it)f(t)=arg(−1+it) for all t∈Rt \in Rt∈R, is continuous at all points of R, where i=−1i = \sqrt{-1}i=−1​
  3. (C)For any two non-zero complex numbers z1z_{1}z1​ and z2z_{2}z2​, arg⁡(z1z2)−arg⁡(z1)+arg⁡(z2)\arg\left(\frac{z_{1}}{z_{2}}\right) - \arg(z_{1}) + \arg(z_{2})arg(z2​z1​​)−arg(z1​)+arg(z2​) is an integer multiple of 2π2\pi2π
  4. (D)For any three given distinct complex numbers z1z_{1}z1​, z2z_{2}z2​ and z3z_{3}z3​, the locus of the point z satisfying the condition arg⁡((z−z1)(z2−z3)(z−z3)(z2−z1))=π\arg\left(\frac{(z-z_{1})(z_{2}-z_{3})}{(z-z_{3})(z_{2}-z_{1})}\right) = \piarg((z−z3​)(z2​−z1​)(z−z1​)(z2​−z3​)​)=π, lies on a straight line

Correct answer: (A), (B), (D)

Step-by-step solution →
Q183·MathematicsMultiple correctJEE Advanced 2017
Let aaa, bbb, xxx and yyy be real numbers such that a−b=1a - b = 1a−b=1 and y≠0y \neq 0y=0. If the complex number z=x+iyz = x + iyz=x+iy satisfies Im(az+bz+1)=y\mathrm{Im}\left(\frac{az + b}{z + 1}\right) = yIm(z+1az+b​)=y, then which of the following is(are) possible value(s) of xxx ?
  1. (A)−1−1−y2-1 - \sqrt{1 - y^{2}}−1−1−y2​
  2. (B)1+1+y21 + \sqrt{1 + y^{2}}1+1+y2​
  3. (C)1−1+y21 - \sqrt{1 + y^{2}}1−1+y2​
  4. (D)−1+1−y2-1 + \sqrt{1 - y^{2}}−1+1−y2​

Correct answer: (A), (D)

Step-by-step solution →
Q184·MathematicsIntegerJEE Advanced 2016
Let z=−1+3i2z = \frac{-1 + \sqrt{3}i}{2}z=2−1+3​i​, where i=−1i = \sqrt{-1}i=−1​, and r,s∈{1,2,3}r, s \in \{1, 2, 3\}r,s∈{1,2,3}. Let P=[(−z)rz2sz2szr]P = \begin{bmatrix} (-z)^{r} & z^{2s} \\ z^{2s} & z^{r} \end{bmatrix}P=[(−z)rz2s​z2szr​] and III be the identity matrix of order 2. Then the total number of ordered pairs (r,s)(r, s)(r,s) for which P2=−IP^{2} = -IP2=−I is

Correct answer: 1

Step-by-step solution →
Q185·MathematicsMultiple correctJEE Advanced 2016
Let a,b∈Ra, b \in \mathbb{R}a,b∈R and a2+b2≠0a^{2} + b^{2} \neq 0a2+b2=0. Suppose S={z∈C:z=1a+ibt,t∈R,t≠0}S = \left\{ z \in \mathbb{C} : z = \frac{1}{a + ibt}, t \in \mathbb{R}, t \neq 0 \right\}S={z∈C:z=a+ibt1​,t∈R,t=0}, where i=−1i = \sqrt{-1}i=−1​. If z=x+iyz = x + iyz=x+iy and z∈Sz \in Sz∈S, then (x, y) lies on
  1. (A)the circle with radius 12a\frac{1}{2a}2a1​ and centre (12a,0)\left( \frac{1}{2a}, 0 \right)(2a1​,0) for a>0,b≠0a > 0, b \neq 0a>0,b=0
  2. (B)the circle with radius −12a-\frac{1}{2a}−2a1​ and centre (−12a,0)\left( -\frac{1}{2a}, 0 \right)(−2a1​,0) for a<0,b≠0a < 0, b \neq 0a<0,b=0
  3. (C)the x-axis for a≠0,b=0a \neq 0, b = 0a=0,b=0
  4. (D)the y-axis for a=0,b≠0a = 0, b \neq 0a=0,b=0

Correct answer: (A), (C), (D)

Step-by-step solution →
Q186·MathematicsIntegerJEE Advanced 2015
For any integer kkk, let αk=cos⁡(kπ7)+isin⁡(kπ7)\alpha_{k} = \cos\left(\dfrac{k\pi}{7}\right) + i\sin\left(\dfrac{k\pi}{7}\right)αk​=cos(7kπ​)+isin(7kπ​), where i=−1i = \sqrt{-1}i=−1​. The value of the expression ∑k=112∣αk+1−αk∣∑k=13∣α4k−1−α4k−2∣\dfrac{\displaystyle\sum_{k=1}^{12}\left|\alpha_{k+1} - \alpha_{k}\right|}{\displaystyle\sum_{k=1}^{3}\left|\alpha_{4k-1} - \alpha_{4k-2}\right|}k=1∑3​∣α4k−1​−α4k−2​∣k=1∑12​∣αk+1​−αk​∣​ is

Correct answer: 4

Step-by-step solution →
Q187·MathematicsSingle correctJEE Advanced 2014
Let zk=cos⁡(2kπ10)+isin⁡(2kπ10)z_{k} = \cos\left(\frac{2k\pi}{10}\right) + i\sin\left(\frac{2k\pi}{10}\right)zk​=cos(102kπ​)+isin(102kπ​) ; k=1,2,…,9k = 1, 2, \ldots, 9k=1,2,…,9.
List – IList – II
P.For each zkz_{k}zk​ there exists a zjz_{j}zj​ such that zk⋅zj=1z_{k} \cdot z_{j} = 1zk​⋅zj​=11.True
Q.There exists a k∈{1,2,…,9}k \in \{1, 2, \ldots, 9\}k∈{1,2,…,9} such that z1⋅z=zkz_{1} \cdot z = z_{k}z1​⋅z=zk​ has no solution zzz in the set of complex numbers2.False
R.∣1−z1∣∣1−z2∣…∣1−z9∣10\frac{|1-z_{1}||1-z_{2}| \ldots |1-z_{9}|}{10}10∣1−z1​∣∣1−z2​∣…∣1−z9​∣​ equals3.1
S.1−∑k=19cos⁡(2kπ10)1 - \sum_{k=1}^{9} \cos\left(\frac{2k\pi}{10}\right)1−∑k=19​cos(102kπ​) equals4.2
  1. (A)P-1, Q-2, R-4, S-3
  2. (B)P-2, Q-1, R-3, S-4
  3. (C)P-1, Q-2, R-3, S-4
  4. (D)P-2, Q-1, R-4, S-3

Correct answer: (C)

Step-by-step solution →
Q188·MathematicsMultiple correctJEE Advanced 2013
Let w=3+i2w = \frac{\sqrt{3} + i}{2}w=23​+i​ and P={wn:n=1,2,3,…}P = \{w^{n} : n = 1, 2, 3, \ldots\}P={wn:n=1,2,3,…}. Further H1={z∈C:Re⁡z>12}H_{1} = \left\{z \in \mathbb{C} : \operatorname{Re} z > \frac{1}{2}\right\}H1​={z∈C:Rez>21​} and H2={z∈C:Re⁡z<−12}H_{2} = \left\{z \in \mathbb{C} : \operatorname{Re} z < \frac{-1}{2}\right\}H2​={z∈C:Rez<2−1​}, where C\mathbb{C}C is the set of all complex numbers. If z1∈P∩H1z_{1} \in P \cap H_{1}z1​∈P∩H1​, z2∈P∩H2z_{2} \in P \cap H_{2}z2​∈P∩H2​ and O represents the origin, then ∠z1Oz2=\angle z_{1} O z_{2} =∠z1​Oz2​=
  1. (A)π2\frac{\pi}{2}2π​
  2. (B)π6\frac{\pi}{6}6π​
  3. (C)2π3\frac{2\pi}{3}32π​
  4. (D)5π6\frac{5\pi}{6}65π​

Correct answer: (C), (D)

Step-by-step solution →
Q189·MathematicsSingle correctJEE Advanced 2013
Let S=S1∩S2∩S3S = S_{1} \cap S_{2} \cap S_{3}S=S1​∩S2​∩S3​, where S1={z∈C:∣z∣<4}S_{1} = \{z \in \mathbb{C} : |z| < 4\}S1​={z∈C:∣z∣<4}, S2={z∈C:Im⁡[z−1+3i1−3i]>0}S_{2} = \left\{z \in \mathbb{C} : \operatorname{Im}\left[\frac{z - 1 + \sqrt{3}i}{1 - \sqrt{3}i}\right] > 0\right\}S2​={z∈C:Im[1−3​iz−1+3​i​]>0} and S3={z∈C:Re⁡z>0}S_{3} = \{z \in \mathbb{C} : \operatorname{Re} z > 0\}S3​={z∈C:Rez>0}. Area of S=S =S=
  1. (A)10π3\frac{10\pi}{3}310π​
  2. (B)20π3\frac{20\pi}{3}320π​
  3. (C)16π3\frac{16\pi}{3}316π​
  4. (D)32π3\frac{32\pi}{3}332π​

Correct answer: (B)

Step-by-step solution →
Q190·MathematicsSingle correctJEE Advanced 2013
Let complex numbers α\alphaα and 1αˉ\frac{1}{\bar{\alpha}}αˉ1​ lie on circles (x−x0)2+(y−y0)2=r2(x-x_0)^2+(y-y_0)^2=r^2(x−x0​)2+(y−y0​)2=r2 and (x−x0)2+(y−y0)2=4r2(x-x_0)^2+(y-y_0)^2=4r^2(x−x0​)2+(y−y0​)2=4r2, respectively. If z0=x0+iy0z_0=x_0+iy_0z0​=x0​+iy0​ satisfies the equation 2∣z0∣2=r2+22|z_0|^2=r^2+22∣z0​∣2=r2+2, then ∣α∣=|\alpha|=∣α∣=
  1. (A)12\frac{1}{\sqrt{2}}2​1​
  2. (B)12\frac{1}{2}21​
  3. (C)17\frac{1}{\sqrt{7}}7​1​
  4. (D)13\frac{1}{3}31​

Correct answer: (C)

Step-by-step solution →
Q191·MathematicsSingle correctJEE Advanced 2013
Let S=S1∩S2∩S3S = S_{1} \cap S_{2} \cap S_{3}S=S1​∩S2​∩S3​, where S1={z∈C:∣z∣<4}S_{1} = \{z \in \mathbb{C} : |z| < 4\}S1​={z∈C:∣z∣<4}, S2={z∈C:Im⁡[z−1+3i1−3i]>0}S_{2} = \left\{z \in \mathbb{C} : \operatorname{Im}\left[\frac{z - 1 + \sqrt{3}i}{1 - \sqrt{3}i}\right] > 0\right\}S2​={z∈C:Im[1−3​iz−1+3​i​]>0} and S3={z∈C:Re⁡z>0}S_{3} = \{z \in \mathbb{C} : \operatorname{Re} z > 0\}S3​={z∈C:Rez>0}. min⁡z∈S∣1−3i−z∣=\min_{z \in S} |1 - 3i - z| =minz∈S​∣1−3i−z∣=
  1. (A)2−32\frac{2 - \sqrt{3}}{2}22−3​​
  2. (B)2+32\frac{2 + \sqrt{3}}{2}22+3​​
  3. (C)3−32\frac{3 - \sqrt{3}}{2}23−3​​
  4. (D)3+32\frac{3 + \sqrt{3}}{2}23+3​​

Correct answer: (C)

Step-by-step solution →

Complex Numbers — frequently asked

How many questions from Complex Numbers appear in JEE?

Complex Numbers has appeared in 165 of the last 186 JEE Main and JEE Advanced papers — about 89% of them — contributing 191 questions in total across those papers.

Is Complex Numbers an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 89% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Complex Numbers questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Mathematics chapters

  • Three Dimensional Geometry 344
  • Matrices and Determinants 342
  • Sets, Relations and Functions 313
  • Sequence and Series 287
  • Definite Integration 267
  • Vector Algebra 245
  • Differential Equations 240
  • Probability 226

All 26 Mathematics chapters →

Practise Complex Numbers until it stops costing you marks.

Build a timed test from these 191 questions in one click. Jarvis marks it, names the specific misconception behind each wrong answer, and brings the ones you failed back at the right interval.

Practise Complex Numbers freeSee pricing

Free plan, no time limit · No credit card needed

Jarvis OS

AI-powered JEE preparation.

Question papers

JEE Main PYQsJEE Advanced PYQsPhysics PYQsChemistry PYQsMaths PYQs

Product

TourPricingSign inCreate account

Legal

Privacy PolicyTerms of ServiceRefund & CancellationShipping & DeliveryContact Us

Operated by

Venyou Craft Private Limited

info@jarvisos.net

© 2026 Jarvis OS