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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Medium

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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 301-330 of 337 questions

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301 Medium
301 of 337
Application +4 | -1
If zz lies on the circle ∣z∣=1|z|=1, then 2/z2 / z lies on a
View answer and explanation
Correct option: A
Circle
Option A (Circle) is correct because if ∣z∣=1|z|=1, then 2/z2/z also lies on a circle, specifically a circle with radius 2.
302 Medium
302 of 337
Analyzing +4 | -1
Two conjugate complex numbers z1z_1 and z1‾\overline{z_1} represent two adjacent vertices of a polygon of n\mathrm{n} sides, whose centre is origin. If Im⁡(z1)Re⁡(zˉ1)=2−1\frac{\operatorname{Im}\left(z_1\right)}{\operatorname{Re}\left(\bar{z}_1\right)}=\sqrt{2}-1 then ratio of the areas of the circles inscribed and circumscribed the polygon is
View answer and explanation
Correct option: C
2+122\frac{\sqrt{2}+1}{2 \sqrt{2}}
Option C (2+122\frac{\sqrt{2}+1}{2 \sqrt{2}}) is correct because it represents the square of the cosine of π8\frac{\pi}{8}, which is the ratio of the inscribed to circumscribed circle areas.
303 Medium
303 of 337
Application +4 | -1
If zz is a complex number in the Argand plane, then the equation ∣z−2∣2+∣z+2∣2=8{\left| {z - 2} \right|^2} + {\left| {z + 2} \right|^2} = 8 represents
View answer and explanation
Correct option: D
A circle
The correct answer is D (A circle) because the simplified equation ∣z−2∣2+∣z+2∣2=8{\left| {z - 2} \right|^2} + {\left| {z + 2} \right|^2} = 8 results in the equation of a circle.
304 Medium
304 of 337
Application +4 | -1
If z=32+i2(i=−1)z=\frac{\sqrt{3}}{2}+\frac{i}{2}(i=\sqrt{-1}), then (1+iz+z5+iz8)9\left(1+i z+z^{5}+i z^{8}\right)^{9} is equal to:
View answer and explanation
Correct option: D
-1
Option D (-1) is correct because substituting z=32+i2z=\frac{\sqrt{3}}{2}+\frac{i}{2} into the expression and simplifying using properties of complex numbers and cube roots of unity leads to -1.
305 Medium
305 of 337
Application +4 | -1
If 3+i=(a+ib)(c+id)\sqrt{3}+i=(a+i b)(c+i d), then tan⁡−1ba+tan⁡−1dc\tan ^{-1} \frac{b}{a}+\tan ^{-1} \frac{d}{c} is equal to
View answer and explanation
Correct option: C
π6+nπ\frac{\pi}{6}+n \pi, for some integer nn
Option C (π6+nπ\frac{\pi}{6}+n \pi, for some integer nn) is correct because the argument of a product of complex numbers is the sum of their arguments.
306 Medium
306 of 337
Application +4 | -1
For complex numbers z1=x1+iyz_1=x_1+i y, and z2z_2 =x2+iy2=x_2+i y_2, we write z1∩z2z_1 \cap z_2 if x1≤x2x_1 \leq x_2 and y1≤y2y_1 \leq y_2. Let zz be a complex number such that 1∩z1 \cap z, then
View answer and explanation
Correct option: C
1−z1+z∩0\frac{1-z}{1+z} \cap 0
Option C (1−z1+z∩0\frac{{1-z}}{{1+z}} \cap 0) is correct because, given 1∩z1 \cap z, the real and imaginary parts of 1−z1+z\frac{{1-z}}{{1+z}} are both non-positive.
307 Medium
307 of 337
Application +4 | -1
If the 5th 5^{\text {th }} roots of unity are 1,z1,z2,z3,z41, z_1, z_2, z_3, z_4 then z12+z22+z32+z42z_1{ }^2+z_2{ }^2+z_3{ }^2+z_4{ }^2 is equal to
View answer and explanation
Correct option: B
-1
Option B (-1) is correct because the sum of the squares of the non-real fifth roots of unity is -1.
308 Medium
308 of 337
Application +4 | -1
Let z1z_1 and z2z_2 be nnth roots of unity which subtend a right angle at the origin. Then nn must be of the form
View answer and explanation
Correct option: D
4k4 k
Option D (4k4k) is correct because the angle between the nth roots of unity must be a right angle, implying that n is a multiple of 4.
309 Medium
309 of 337
Application +4 | -1
For any two complex numbers z1{z_1} and z2{z_2} and any real numbers aa and bb, ∣az1−bz2∣2+∣bz1+az2∣2{\left| {a{z_1} - b{z_2}} \right|^2} + {\left| {b{z_1} + a{z_2}} \right|^2} is equal to
View answer and explanation
Correct option: B
(a2+b2)(∣z1∣2+∣z2∣2)\left( {{a^2} + {b^2}} \right)\left( {{{\left| {{z_1}} \right|}^2} + {{\left| {{z_2}} \right|}^2}} \right)
Option B ((a2+b2)(∣z1∣2+∣z2∣2)\left( {{a^2} + {b^2}} \right)\left( {{{\left| {{z_1}} \right|}^2} + {{\left| {{z_2}} \right|}^2}} \right)) is correct because expanding and simplifying the expression leads to this result.
310 Medium
310 of 337
Application +4 | -1
Let ωn=cos⁡(2πn)+isin⁡(2πn),  i2=−1,{\omega _n} = \cos \left( {\frac{{2\pi }}{n}} \right) + i\sin \left( {\frac{{2\pi }}{n}} \right),\;{i^2} = - 1, then (x+yω3+zω32)(x+yω32+zω3)\left( {x + y{\omega _3} + z\omega _3^2} \right)\left( {x + y\omega _3^2 + z{\omega _3}} \right) is equal to
View answer and explanation
Correct option: C
x2+y2+z2−yz−zx−xy{x^2} + {y^2} + {z^2} - yz - zx - xy
Option C (x2+y2+z2−yz−zx−xy{x^2} + {y^2} + {z^2} - yz - zx - xy) is correct because substituting ω3=ω\omega_3 = \omega and ω32=ω2\omega_3^2 = \omega^2 (where ω\omega is a cube root of unity) and expanding the product leads to the given expression.
311 Medium
311 of 337
Application +4 | -1
Locus of point zz so that z,iz, i, and izi z are collinear, is
View answer and explanation
Correct option: B
a circle
Option B (a circle) is correct because the condition for collinearity leads to an equation that represents a circle in the complex plane.
312 Medium
312 of 337
Application +4 | -1
Let α,  β\alpha ,\;\beta be the roots of x2−2xcos⁡ϕ+1=0{x^2} - 2x\cos \phi + 1 = 0, then the equation whose roots are αn,  βn,  {\alpha ^n},\;{\beta ^n},\; is
View answer and explanation
Correct option: B
x2−2xcos⁡nϕ+1=0{x^2} - 2x\cos n\phi + 1 = 0
Option B (x2−2xcos⁡nϕ+1=0{x^2} - 2x\cos n\phi + 1 = 0) is correct because the roots can be expressed using Euler's formula, and raising them to the power of n results in a new equation with coefficients involving cos(nφ).
313 Medium
313 of 337
Understanding +4 | -1
If f(x)f\left( x \right) is a polynomial of degree nn with rational coefficients and 1+2i,  2−31 + 2i,\;2 - \sqrt 3 and 5 are three roots of f(x)=0f\left( x \right) = 0, then the least value of nn is
View answer and explanation
Correct option: A
5
Option A (5) is correct because complex and irrational roots occur in conjugate pairs, so we must include 1−2i1-2i and 2+32+\sqrt{3} as roots, along with 5, 1+2i1+2i and 2−32-\sqrt{3}, giving a minimum degree of 5.
314 Medium
314 of 337
Analyzing +4 | -1
Let S={z∈C:zˉ=i(z2+Re⁡(zˉ))}S=\left\{z \in \mathbb{C}: \bar{z}=i\left(z^{2}+\operatorname{Re}(\bar{z})\right)\right\}. Then ∑z∈S∣z∣2\sum\limits_{{z \in S}}|z|^{2} is equal to
View answer and explanation
Correct option: B
4
The correct answer is B (4) because it is found by solving the given complex equation and summing the squared magnitudes of the solutions.
315 Medium
315 of 337
Application +4 | -1
The greatest and the least value of ∣z1+z2∣,\left| {{z_1} + {z_2}} \right|, if z1=24+7i{z_1} = 24 + 7i and ∣z2∣=6,\left| {{z_2}} \right| = 6, are respectively
View answer and explanation
Correct option: A
31,  1931,\;19
Option A (31, 19) is correct because the greatest value is the sum of the magnitudes and the least value is the difference of the magnitudes.
316 Medium
316 of 337
Application +4 | -1
The modulus of (3+2i)2(4−3i)\frac{(3+2 i)^2}{(4-3 i)} is
View answer and explanation
Correct option: A
135\frac{13}{5}
Option A (135\frac{13}{5}) is correct because the modulus of a quotient is the quotient of the moduli, and the modulus of a product is the product of the moduli.
317 Medium
317 of 337
Analyzing +4 | -1
If ∣z∣=1,z≠1|z|=1, z \neq 1, then value of arg⁡(11−z)\arg \left(\frac{1}{1-z}\right) cannot exceed
View answer and explanation
Correct option: A
π/2\pi / 2
Option A (π/2\pi / 2) is correct because the argument of the given expression is bounded by −π/2-\pi/2 and π/2\pi/2, and thus cannot exceed π/2\pi/2.
318 Medium
318 of 337
Application +4 | -1
If α\alpha and β\beta are the roots of x2−2x+2=0x^2-2 x+2=0, then αn+βn=\alpha^n+\beta^n=
View answer and explanation
Correct option: B
2n2+1cos⁡nπ42^{\frac{n}{2}+1} \cos \frac{n \pi}{4}
Option B, 2n2+1cos⁡nπ42^{\frac{n}{2}+1} \cos \frac{n \pi}{4}, is correct because it is the simplified form of the expression αn+βn\alpha^n + \beta^n where α\alpha and β\beta are the roots of the given quadratic equation.
319 Medium
319 of 337
Application +4 | -1
If x=ω−ω2−2x=\omega-\omega^2-2, then the value of x4+3x3+2x2−11x−6x^4+3 x^3+2 x^2-11 x-6 is
View answer and explanation
Correct option: A
1
Option A (1) is correct because by substituting x=ω−ω2−2x = \omega - \omega^2 - 2 and simplifying, the value of the expression is 1.
320 Medium
320 of 337
Application +4 | -1
If one vertex of a square whose diagonals intersect at the origin is 3(cos⁡θ+isin⁡θ),3\left( {\cos \mathrm{\theta } + i\sin \mathrm{\theta }} \right), then the two adjacent vertices are
View answer and explanation
Correct option: A
±3  (sin⁡θ−icos⁡θ)\pm 3\;\left( {\sin \mathrm{\theta } - i\cos \mathrm{\theta }} \right)
Option A (±3  (sin⁡θ−icos⁡θ)\pm 3\;\left( {\sin \mathrm{\theta } - i\cos \mathrm{\theta }} \right)) is correct because rotating the given vertex by ±π2\pm \frac{\pi}{2} gives the adjacent vertices.
321 Medium
321 of 337
Application +4 | -1
If arg⁡(z−2z+2)=π4\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{4} for any complex number zz, then ∣z−2i∣=|z-2 i|=
View answer and explanation
Correct option: A
222 \sqrt{2}
Option A (222 \sqrt{2}) is correct because by letting z=x+iyz=x+iy and using the given argument condition, we derive the equation of a circle and find ∣z−2i∣=22|z-2i| = 2\sqrt{2}.
322 Medium
322 of 337
Application +4 | -1
If (1+i1−i)3−(1−i1+i)3=x+iy\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3=x+i y, then (x,y)(x, y) is equal to
View answer and explanation
Correct option: C
(−2,0)(-2,0)
Option B ((0,−2)(0,-2)) is correct because simplifying the expression using 1+i1−i=i\frac{1+i}{1-i}=i and 1−i1+i=−i\frac{1-i}{1+i}=-i leads to i3−(−i)3=−2ii^3 - (-i)^3 = -2i, so x=0 and y=-2.
323 Medium
323 of 337
Application +4 | -1
If z=x+iyz = x + iy and ∣1−izz−i∣=1\left| {\frac{{1 - iz}}{{z - i}}} \right| = 1, the locus of zz is
View answer and explanation
Correct option: A
x−axisx - axis
Option C (−a-a) is correct because by using Vieta's formulas, we can express p2+q2+pq{p^2} + {q^2} + pq in terms of m and a, which simplifies to −a-a.
324 Medium
324 of 337
Understanding +4 | -1
If z1z_{1} and z2z_{2} are two non- zero complex numbers such that ∣z1+z2∣=∣z1∣+∣z2∣\left|z_{1}+z_{2}\right|=\left|z_{1}\right|+\left|z_{2}\right|, then arg⁡z1−arg⁡z2\arg z_{1}-\arg z_{2} is equal to
View answer and explanation
Correct option: C
0
Option C is correct because if |z1+z2| = |z1| + |z2|, then z1 and z2 are collinear and point in the same direction, meaning their arguments are equal, so arg(z1) - arg(z2) = 0.
325 Medium
325 of 337
Application +4 | -1
If arg⁡(z1z2)=0\arg \left( {{z_1}{z_2}} \right) = 0 and ∣z1∣=∣z2∣=1,\left| {{z_1}} \right| = \left| {{z_2}} \right| = 1, then
View answer and explanation
Correct option: C
z1=zˉ2{z_1} = {\bar z_2}
Option C (z1=zˉ2{z_1} = {\bar z_2}) is correct because if the argument of the product of two complex numbers is zero and their magnitudes are equal to 1, then the two numbers are complex conjugates.
326 Medium
326 of 337
Application +4 | -1
If Re  (z+42z−i)=12,\mathrm{Re\;}\left( {\frac{{z + 4}}{{2z - i}}} \right) = \frac{1}{2}, then zz is represented by a point lying on
View answer and explanation
Correct option: C
A straight line
The correct answer is A (A straight line) because the real part condition leads to a linear equation in x and y, representing a straight line.
327 Medium
327 of 337
Analyzing +4 | -1
If a,b,ca,b,c and u,υ,ωu,\upsilon ,\omega are complex numbers representing the vertices of two triangles such that c=(1−r)  a+rbc = \left( {1 - r} \right)\;a + rb and ω=(1−r)u+rυ,\omega = \left( {1 - r} \right)u + r\upsilon , where rr is a complex number, then the two triangles
View answer and explanation
Correct option: B
Are similar
Option B (Are similar) is correct because the given conditions imply that the two triangles have proportional sides and equal angles, thus they are similar.
328 Medium
328 of 337
Application +4 | -1
If the points z,−izz,-i z and 1 are collinear then zz lies on
View answer and explanation
Correct option: B
a circle
Option B (a circle) is correct because the collinearity condition leads to an equation of the form |z - c| = r, which represents a circle.
329 Medium
329 of 337
Understanding +4 | -1
If i=−1i=\sqrt{-1}, then exi+e−x2=\frac{e^{x i}+e^{-x}}{2}=
View answer and explanation
Correct option: B
1−x22!+x44!−…∞1-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\ldots \infty
Option A (1+x22!+x44!+…∞1+\frac{{x^2}}{{2 !}}+\frac{{x^4}}{{4 !}}+\ldots \infty) is incorrect. The correct answer should be 1−x22!+x44!−…∞1-\frac{{x^2}}{{2 !}}+\frac{{x^4}}{{4 !}}-\ldots \infty which represents cos⁡x\cos x. However, the provided option A is incorrect. The correct option should be B.
330 Medium
330 of 337
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.
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