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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 961-979 of 979 questions
961 Medium
961 of 979
Application +4 | -1
If ∣z1∣=∣z2∣=∣z3∣=1\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=1, then value of ∣z2−z3∣2+\left|z_2-z_3\right|^2+ ∣z3−z1∣2+∣z1−z2∣2\left|z_3-z_1\right|^2+\left|z_1-z_2\right|^2 cannot exceed
View answer and explanation
Correct option: C
9
Option C (9) is correct because the maximum value of the expression is obtained when the real part of the sum of complex conjugates is minimized, leading to a maximum value of 9.
962 Easy
962 of 979
Application +4 | -1
Let (−2−13i)3=x+iy27(i=−1)\left(-2-\frac{1}{3} \mathrm{i}\right)^3=\frac{\mathrm{x}+\mathrm{iy}}{27}(\mathrm{i}=\sqrt{-1}), where x\mathrm{x} and yy are real numbers, then y−xy-x equals :
View answer and explanation
Correct option: D
91
Option D (91) is correct because expanding the complex number expression and then calculating y - x results in 91.
963 Easy
963 of 979
Application +4 | -1
If the cube roots of unity are 1,  ω,  ω21,\;\omega ,\;{\omega ^2}, then the roots of the equation (x−2)3+27=0{\left( {x - 2} \right)^3} + 27 = 0 are
View answer and explanation
Correct option: D
−1,  2−3ω,  2−3ω2- 1,\;2 - 3\omega ,\;2 - 3{\omega ^2}
The correct answer is D (−1,  2−3ω,  2−3ω2- 1,\;2 - 3\omega ,\;2 - 3{\omega ^2}) because it represents the solutions obtained by solving the given cubic equation using the properties of cube roots of unity.
964 Medium
964 of 979
Analyzing +4 | -1
If z1,z2{z_1},{z_2} are two complex numbers such that ∣z1−z2z1+z2∣=1\left| {\frac{{{z_1} - {z_2}}}{{{z_1} + {z_2}}}} \right| = 1 and i  z1=k  z2i\;{z_1} = k\;{z_2} where k∈R,k \in R, then the angle between z1−z2{z_1} - {z_2} and z1+z2{z_1} + {z_2} is
View answer and explanation
Correct option: C
−2tan⁡−1k- 2{\tan ^{ - 1}}k
Option C (−2tan⁡−1k- 2{\tan ^{ - 1}}k) is correct because by using componendo and dividendo and the given condition, the angle between the two complex numbers can be found to be −2tan⁡−1k- 2{\tan ^{ - 1}}k.
965 Medium
965 of 979
Application +4 | -1
If zz is a complex number of unit modulus and argument θ,\theta , then arg⁡(1+z1+z‾)\arg \left( {\frac{{1 + z}}{{1 + \overline z }}} \right) is equal to
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Correct option: C
θ\theta
Option C (θ\theta) is correct because using the properties of complex numbers and their arguments, arg⁡(1+z1+z‾)\arg \left( {\frac{{1 + z}}{{1 + \overline z }}} \right) simplifies to θ\theta.
966 Easy
966 of 979
Understanding +4 | -1
The values of (16)1/4(16)^{1 / 4} are
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Correct option: A
±2,±2i\pm 2, \pm 2 i
Option A is correct because the fourth roots of 16 are ±2\pm 2 and ±2i\pm 2i.
967 Easy
967 of 979
Application +4 | -1
If xx is an integer satisfying x2−6x+5≤0{x^2} - 6x + 5 \le 0 and x2−2x>0,{x^2} - 2x > 0, then the number of positive values of xx, is
View answer and explanation
Correct option: A
3
Option A (3) is correct because the integer values of x that satisfy both inequalities are 3, 4, and 5, resulting in 3 positive values.
968 Easy
968 of 979
Application +4 | -1
The value of (−1+i31−i)30\left(\frac{-1+i \sqrt{3}}{1-i}\right)^{30} is :
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Correct option: C
−215i-2^{15} i
Option C (-215 i) is correct because it is obtained by converting the complex numbers to polar form, simplifying the expression, and applying De Moivre's theorem.
969 Easy
969 of 979
Analyzing +4 | -1
Let z=x+iyz=x+i y be a complex number, where xx and yy are integers and i=−1i=\sqrt{-1}. Then, the area of the rectangle whose vertices are the roots of the equation zzˉ3+zˉz3=350z \bar{z}^3+\bar{z} z^3=350, is
View answer and explanation
Correct option: A
48
Option A (48) is correct because solving the equation leads to integer values for x and y, allowing calculation of the rectangle's area.
970 Easy
970 of 979
Analyzing +4 | -1
Let z=x+iyz = x + iy be a complex number where xx and yy are integers. Then the area of the rectangle whose vertices are the roots of the equation zz−3+zˉz3=350z{z^{ - 3}} + \bar z{z^3} = 350 is
View answer and explanation
Correct option: A
48
Option D (48) is correct because the area of the rectangle formed by the roots is calculated as 8 * 6 = 48.
971 Medium
971 of 979
Application +4 | -1
Let w1w_1 be the point obtained by the rotation of z1=5+4iz_1=5+4 i about the origin through a right angle in the anticlockwise direction, and w2w_2 be the point obtained by the rotation of z2=3+5iz_2=3+5 i about the origin through a right angle in the clockwise direction. Then the principal argument w1−w2w_1-w_2 is equal to
View answer and explanation
Correct option: A
π−tan⁡−189\pi-\tan ^{-1} \frac{8}{9}
Option A (π−tan⁡−189\pi - \tan^{-1} \frac{8}{9}) is correct because it represents the principal argument of the complex number w1−w2w_1 - w_2, calculated using rotations in the complex plane.
972 Easy
972 of 979
Application +4 | -1
If α\alpha and β\beta are the roots of the equation x2−x+1=0x^2-x+1=0, α2009+β2009\alpha^{2009}+\beta^{2009} is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because the expression simplifies to 1 using the properties of the roots of the given quadratic equation.
973 Easy
973 of 979
Application +4 | -1
In which quadrant of the complex plane, the point 1+2i1−i\frac{{1 + 2i}}{{1 - i}} lies?
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Correct option: C
Second
Option C (Second) is correct because the simplified form of the complex number has a negative real part and a positive imaginary part.
974 Medium
974 of 979
Application +4 | -1
The real part of (1−cos⁡θ+2isin⁡θ)−1{\left( {1 - \cos \mathrm{\theta } + 2i\sin \mathrm{\theta }} \right)^{ - 1}} is
View answer and explanation
Correct option: D
15+3cos⁡θ\frac{1}{{5 + 3\cos \mathrm{\theta }}}
The correct answer is D (15+3cos⁡θ\frac{1}{{5 + 3\cos \mathrm{\theta }}}) because after rationalizing and simplifying, the real part of the given expression is 15+3cos⁡θ\frac{1}{{5 + 3\cos \mathrm{\theta }}}.
975 Medium
975 of 979
Understanding +4 | -1
For two complex numbers z1,  z2{z_1},\;{z_2} the relation ∣z1+z2∣=∣z1∣+∣z2∣\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right| holds, if
View answer and explanation
Correct option: A
arg⁡(z1)=arg⁡(z2)\arg \left( {{z_1}} \right) = \arg \left( {{z_2}} \right)
Option A is correct because the equality ∣z1+z2∣=∣z1∣+∣z2∣\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right| holds only when the arguments of the two complex numbers are equal.
976 Medium
976 of 979
Application +4 | -1
The solution set of ∣x2−10∣≤6,\left| {{x^2} - 10} \right| \le 6, is
View answer and explanation
Correct option: D
[−4,  −2]∪[2,  4]\left[ { - 4,\; - 2} \right] \cup \left[ {2,\;4} \right]
The correct answer is D ([−4,−2]∪[2,4][ { - 4, - 2} ] \cup [ {2,4} ]) because solving the absolute value inequality leads to this union of closed intervals.
977 Easy
977 of 979
Application +4 | -1
If x,y,a,b∈R,a≠0x, y, a, b \in \mathbf{R}, a \neq 0 and (a+ib)(x+iy)=(a2+b2)i(a+i b)(x+i y)=\left(a^2+b^2\right) i, then (x,y)(x, y) equals
View answer and explanation
Correct option: D
(b,a)(b, a)
Option D ((b,a)(b, a)) is correct because it is found by equating the real and imaginary parts after dividing both sides by (a+ib).
978 Medium
978 of 979
Analyzing +4 | -1
If z2+z+1=0z^{2}+z+1=0, where z\mathrm{z} is complex number, then the value of (z+1z)2+(z2+1z2)2+(z3+1z3)2+…….+(z6+1z6)2\left(z+\frac{1}{z}\right)^{2}+\left(z^{2}+\frac{1}{z^{2}}\right)^{2}+\left(z^{3}+\frac{1}{z^{3}}\right)^{2}+\ldots \ldots .+\left(z^{6}+\frac{1}{z^{6}}\right)^{2} is
View answer and explanation
Correct option: D
12
Option D (12) is correct because by finding the value of z and then substituting it into the expression, we can find the value of the given sum.
979 Easy
979 of 979
Application +4 | -1
Let z,wz,w be complex numbers such that zˉ+iw‾=0\bar z + \overline {iw} = 0 and   arg(zw)=π\mathrm{\;arg}\left( {zw} \right) = \pi. Then arg⁡(z)\arg \left( z \right) equals
View answer and explanation
Correct option: C
3π4\frac{{3\pi }}{4}
Option C (3π4\frac{{3\pi }}{4}) is correct because using the given conditions, arg⁡(z)\arg(z) can be found by relating arg⁡(zw)\arg(zw) to arg⁡(z)\arg(z) and using the property arg⁡(−i)=−π2\arg(-i) = -\frac{\pi}{2}.
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