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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Mixed difficulty

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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 751-780 of 979 questions

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751 Medium
751 of 979
Application +4 | -1
∑k=16(sin⁡(2kπ7)+icos⁡(2kπ7))=\sum\limits_{{k=1}}^6\left(\sin \left(\frac{2 k \pi}{7}\right)+i \cos \left(\frac{2 k \pi}{7}\right)\right)=
View answer and explanation
Correct option: C
−i-i
Option C (-i) is correct because by converting to exponential form and using geometric series formula, the sum simplifies to -i.
752 Medium
752 of 979
Analyzing +4 | -1
If z{z} satisfies the equation (z−2z+2)(zˉ−2zˉ+2)=1\left(\frac{z-2}{z+2}\right)\left(\frac{\bar{z}-2}{\bar{z}+2}\right)=1, then minimum value of ∣z∣|z| is equal to:
View answer and explanation
Correct option: A
0
The correct answer is A (0) because the given equation simplifies to |z-2| = |z+2|, which represents the perpendicular bisector of the line joining 2 and -2, i.e., the y-axis. The minimum value of |z| is 0, occurring at the origin.
753 Medium
753 of 979
Application +4 | -1
The multiplicative inverse of (6+5i)2(6+5 i)^2 is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the multiplicative inverse of (6+5i)2(6+5i)^2 is 11(61)2−60(61)2i\frac{11}{(61)^2} - \frac{60}{(61)^2}i, which is not among the given options.
754 Medium
754 of 979
Application +4 | -1
If ∣z+4∣≤3|\mathrm{z}+4| \leq 3 then maximum and minimum value of ∣z+1∣|z+1| is .
View answer and explanation
Correct option: C
6,0
Option C (6,0) is correct because the maximum and minimum values of ∣z+1∣|z+1| are found by using the triangle inequality on ∣(z+4)−3∣|(z+4)-3|.
755 Medium
755 of 979
Application +4 | -1
In a geometrical progression, first term and common ratio are both 12(3+i)\frac{1}{2}(\sqrt{3}+i). Then the absolute value of the nnth term of the progression is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the absolute value of the nth term of the geometric progression is 1, as the magnitude of the common ratio is 1.
756 Medium
756 of 979
Application +4 | -1
The value of x4+9x3+35x2−x+4{x^4} + 9{x^3} + 35{x^2} - x + 4 for x=−5+2−4x = - 5 + 2\sqrt { - 4} is
View answer and explanation
Correct option: B
−160- 160
Option B (−160- 160) is correct because substituting x=−5+2−4x = - 5 + 2\sqrt { - 4} into the expression x4+9x3+35x2−x+4{x^4} + 9{x^3} + 35{x^2} - x + 4 and simplifying results in -160.
757 Easy
757 of 979
Application +4 | -1
A value of θ\theta for which Z=2+3isin⁡θ1−2isin⁡θ\mathrm{Z}=\frac{2+3 i \sin \theta}{1-2 i \sin \theta} is purely imaginary is
View answer and explanation
Correct option: A
sin⁡−113\sin ^{-1} \frac{1}{\sqrt{3}}
Option A (sin⁡−113\sin^{-1} \frac{1}{\sqrt{3}}) is correct because it makes the real part of the complex number Z equal to zero, resulting in a purely imaginary number.
758 Easy
758 of 979
Application +4 | -1
If x+iy=3+i1+3ix+i y=\sqrt{\frac{3+i}{1+3 i}} then (x2+y2)2\left(x^2+y^2\right)^2 equals
View answer and explanation
Correct option: D
1
Option D (1) is correct because by simplifying the expression inside the square root and then calculating (x2+y2)2(x^2+y^2)^2, it can be shown that the result is 1.
759 Easy
759 of 979
Application +4 | -1
If zz is a complex number and i=−1i=\sqrt{-1}, then the minimum possible value of ∣z∣2+∣z−3∣2+∣z−6i∣2|z|^2+|z-3|^2+|z-6 i|^2 is
View answer and explanation
Correct option: B
30
Option B (30) is correct because the minimum value is found by completing the square and recognizing the minimum occurs when z = 1 + 2i.
760 Medium
760 of 979
Application +4 | -1
If ∣u∣<1,∣v∣<1|u| < 1,|v| < 1 and z=u−v1−uvz=\frac{u-v}{1-u v}, a lower bound of ∣z∣|z| is
View answer and explanation
Correct option: C
∥u∣−∣v∥1+∣u∣∣v∣\frac{\|u|-| v\|}{1+|u||v|}
Option C (∥u∣−∣v∥1+∣u∣∣v∣\frac{{\|u|-|v\|}}{{1+|u||v|}}) is correct because by using properties of complex numbers and the reverse triangle inequality, we can establish the lower bound for |z|.
761 Easy
761 of 979
Application +4 | -1
The locus of the complex number zz in an Argand plane satisfying the equation arg⁡(z+i)−arg⁡(z−i)=(π/2)\arg (z+i)-\arg (z-i)=(\pi / 2) is
View answer and explanation
Correct option: A
The boundary of a circle
Option A (The boundary of a circle) is correct because the locus satisfying the given condition represents a circle.
762 Easy
762 of 979
Application +4 | -1
Let z=cos⁡θ+isin⁡θz=\cos \theta+i \sin \theta, where i=−1i=\sqrt{-1}. Then the value of ∑m=115Im⁡(z2m−1)\sum\limits_{m=1}^{15} \operatorname{Im}\left(z^{2 m-1}\right) at θ=2∘\theta=2^{\circ} is
View answer and explanation
Correct option: D
14sin⁡2∘\frac{1}{4 \sin 2^{\circ}}
Option D (14sin⁡2∘\frac{{1}}{{4 \sin 2^{\circ}}}) is correct because the sum of the imaginary parts simplifies to this expression using trigonometric identities.
763 Easy
763 of 979
Analyzing +4 | -1
If P,  Q,  R,  SP,\;Q,\;R,\;S are represented by the complex numbers 4+i,  1+6i,  −4+3i,  −1−2i4 + i,\;1 + 6i,\; - 4 + 3i,\; - 1 - 2i respectively, then PQRSPQRS is a
View answer and explanation
Correct option: B
A square
The correct answer is a square because all sides are equal and the diagonals are equal, indicating that PQRS is a square.
764 Medium
764 of 979
Application +4 | -1
If ∣z+4i∣≤3|z+4 i| \leq 3 then maximum and minimum value of ∣z+3∣|z+3| is -
View answer and explanation
Correct option: A
8,2
Option A (8,2) is correct because the maximum and minimum values of ∣z+3∣|z+3| are found by using the triangle inequality on ∣(z+4i)+(3−4i)∣|(z+4i)+(3-4i)|.
765 Medium
765 of 979
Application +4 | -1
If z1,z2z_1, z_2 and z3z_3 represent the vertices of a triangle whose circumcenter is at the origin, then the complex number representing the orthocenter of the triangle is
View answer and explanation
Correct option: B
z1+z2+z3z_1+z_2+z_3
Option B (z1+z2+z3z_1+z_2+z_3) is correct because the orthocenter of a triangle with circumcenter at the origin is represented by the sum of the complex numbers representing the vertices.
766 Easy
766 of 979
Application +4 | -1
If S={z∈C:arg⁡(z−2z+2)=π3}S = \left\{ {z \in C:\arg \left( {\frac{{z - 2}}{{z + 2}}} \right) = \frac{\pi }{3}} \right\}, then SS is
View answer and explanation
Correct option: C
A circle
The correct answer is C (A circle) because the given condition represents a circle in the complex plane.
767 Medium
767 of 979
Application +4 | -1
The locus of the point zz which satisfy the condition arg (z−1z+1)=π3\left( {\frac{{z - 1}}{{z + 1}}} \right) = \frac{\pi }{3}, is
View answer and explanation
Correct option: B
A circle
Option B (A circle) is correct because the locus of the point zz that satisfies the given condition is a circle.
768 Easy
768 of 979
Application +4 | -1
The smallest positive integer nn for which (1+i1−i)n=−1\left(\frac{1+i}{1-i}\right)^n=-1 is
View answer and explanation
Correct option: B
2
Option B (2) is correct because when n=2, ((1+i)/(1-i))^n simplifies to -1, satisfying the given condition.
769 Easy
769 of 979
Application +4 | -1
If zk=cos⁡(kπ10)+isin⁡(kπ10)z_k=\cos \left(\frac{k \pi}{10}\right)+i \sin \left(\frac{k \pi}{10}\right), then z1z2z3z4z_1 z_2 z_3 z_4 is equal to
View answer and explanation
Correct option: A
-1
Option A (-1) is correct because by De Moivre's theorem, the product simplifies to cos⁡(π)+isin⁡(π)\cos(\pi) + i \sin(\pi), which equals -1.
770 Easy
770 of 979
Application +4 | -1
The value of ∑k=110(sin⁡2kπ11+icos⁡2kπ11)(\sum\limits_{k=1}^{10}\left(\sin \frac{2 k \pi}{11}+i \cos \frac{2 k \pi}{11}\right)( where i=−1)i=\sqrt{-1}) is
View answer and explanation
Correct option: D
−i-i
Option D (-i) is correct because the sum simplifies to -i using the properties of complex numbers and roots of unity.
771 Easy
771 of 979
Application +4 | -1
If z=x+iyz=x+i y is such that ∣z−4∣<∣z−2∣|z-4|<|z-2|, then
View answer and explanation
Correct option: D
x>3x>3 and yy is any real number
Option D (x>3x>3 and yy is any real number) is correct because simplifying the inequality ∣z−4∣<∣z−2∣|z-4|<|z-2| leads to the condition x>3x>3, with no restrictions on yy.
772 Easy
772 of 979
Application +4 | -1
Let (−2−13i)3=x+iy27(i=−1),{\left( { - 2 - \frac{1}{3}i} \right)^3} = \frac{{x + iy}}{{27}}\left( {i = \sqrt { - 1} } \right), where x and y are real numbers, then y – x equals
View answer and explanation
Correct option: A
91
Option A (91) is correct because after expanding the complex number and equating real and imaginary parts, y - x equals 91.
773 Easy
773 of 979
Analyzing +4 | -1
The triangle with vertices at the points z1,z2,  (1−i)z1+i  z2{z_1},{z_2},\;\left( {1 - i} \right){z_1} + i\;{z_2} is
View answer and explanation
Correct option: C
Right angled and isosceles
Option C (Right angled and isosceles) is correct because the sides PQ and RP are equal in length, and the Pythagorean theorem holds, indicating a right-angled isosceles triangle.
774 Medium
774 of 979
Application +4 | -1
If the area of the triangle on the complex plane formed by the points z,  z+izz,\;z + iz and iziz is 200, then the value of ∣3z∣\left| {3z} \right| must be equal to
View answer and explanation
Correct option: C
60
Option C (60) is correct because the area of the triangle is related to |z|^2, and given the area is 200, we can find |z| and then |3z|.
775 Easy
775 of 979
Analyzing +4 | -1
The polynomial x3m+x3n+1+x3k+2{x^{3m}} + {x^{3n + 1}} + {x^{3k + 2}} is exactly divisible by x2+x+1{x^2} + x + 1 if
View answer and explanation
Correct option: B
m,n,km,n,k are integers
Option B (m,n,km,n,k are integers) is correct because it ensures that the polynomial is divisible by x2+x+1{x^2} + x + 1 by making ω\omega and ω2\omega^2 roots of the polynomial.
776 Medium
776 of 979
Application +4 | -1
If the area of triangle on the argand place formed by the complex numbers −z,  iz,  z−iz- z,\;iz,\;z - iz is 600 sq. unit, then ∣z∣\left| z \right| is equal to
View answer and explanation
Correct option: B
20
Option B (7) is correct because the number of oscillations per minute is proportional to the square root of the horizontal component of the Earth's magnetic field, which depends on the total intensity and the angle of dip.
777 Easy
777 of 979
Analyzing +4 | -1
If n∈Z,n \in Z, then 2n(1−i)2n+(1+i)2n2n\frac{{{2^n}}}{{{{\left( {1 - i} \right)}^{2n}}}} + \frac{{{{\left( {1 + i} \right)}^{2n}}}}{{{2^n}}} is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the expression simplifies to 2in2i^n, which is not represented by any of the other options.
778 Easy
778 of 979
Application +4 | -1
If Im(2  z+1i  z+1)=−2,\mathrm{Im}\left( {\frac{{2\;z + 1}}{{i\;z + 1}}} \right) = - 2, then locus of zz is
View answer and explanation
Correct option: A
A circle
Option A (A circle) is correct because the given condition on the imaginary part of the complex expression leads to an equation that represents a circle in the complex plane.
779 Easy
779 of 979
Application +4 | -1
The conjugate of a complex number zz is 21−i\frac{2}{1-i}. Then Re⁡(z)\operatorname{Re}(z) equals
View answer and explanation
Correct option: C
1
Option C (1) is correct because the real part of z is equal to the real part of its conjugate, and the conjugate of z is given as 1+i, so Re(z) = 1.
780 Hard
780 of 979
Application +4 | -1
If A(z1),  B(z2)A\left( {{z_1}} \right),\;B\left( {{z_2}} \right) and C(z3)C\left( {{z_3}} \right) be the vertices of a triangle ABCABC in which ∠ABC=π4\angle ABC = \frac{\pi }{4} and ABBC=  2\frac{{AB}}{{BC}} = \;\sqrt 2, then the value of z2{z_2} is equal to
View answer and explanation
Correct option: C
z3+i(z1−z3){z_3} + i\left( {{z_1} - {z_3}} \right)
Option C (z3+i(z1−z3){z_3} + i\left( {{z_1} - {z_3}} \right)) is correct because considering the rotation about 'B' and using the given conditions leads to this result.
Revision notes and key takeaways

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