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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 271-300 of 337 questions

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271 Medium
271 of 337
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.
272 Medium
272 of 337
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.
273 Medium
273 of 337
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|.
274 Medium
274 of 337
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.
275 Medium
275 of 337
Application +4 | -1
If α\alpha and β\beta are the roots of the equation x2−2x+2=0, {x^2} - 2x + 2 = 0,\, then the least value of nn for which (αβ)n=1{\left( {\frac{\alpha }{\beta }} \right)^n} = 1 is
View answer and explanation
Correct option: C
4
Option C is correct because it accurately determines the least value of n for which the given condition is satisfied, using complex number properties.
276 Medium
276 of 337
Application +4 | -1
If a1,a2,…,ana_1, a_2, \ldots, a_n are real numbers and cos⁡α+isin⁡α\cos \alpha+i \sin \alpha is a root of

xn+a1xn−1+a2xn−2+…+an−1x+an=0x^n+a_1 x^{n-1}+a_2 x^{n-2}+\ldots+a_{n-1} x+a_n=0 (1)

then value of ∣a1cos⁡α+a2cos⁡2α+…+ancos⁡(nα)∣\left|a_1 \cos \alpha+a_2 \cos 2 \alpha+\ldots+a_n \cos (n \alpha)\right| is
View answer and explanation
Correct option: A
1
Option A (1) is correct because substituting the root into the polynomial and equating the real part to zero leads to the given expression being equal to -1, and taking the absolute value gives 1.
277 Medium
277 of 337
Application +4 | -1
If ∣z∣=1|z|=1 and arg⁡(z)=θ\arg (z)=\theta, then arg⁡(1+z1+zˉ)=\arg \left(\frac{1+z}{1+\bar{z}}\right)=
View answer and explanation
Correct option: C
θ\theta
Option C (θ\theta) is correct because using the property zˉ=1z\bar{z} = \frac{1}{z} when ∣z∣=1|z|=1, we can simplify the expression to arg⁡(z)=θ\arg(z) = \theta.
278 Medium
278 of 337
Analyzing +4 | -1
1−c2=nc−1\sqrt{1-c^2}=n c-1 and z=eiθz=e^{i \theta}, then c2n(1+nz)(1+nz)\frac{c}{2 n}(1+n z)\left(1+\frac{n}{z}\right) is equal to
View answer and explanation
Correct option: C
1+c cosθ1+c \ cos \theta
Option C (1+ccos⁡θ1+c \cos \theta) is the correct answer because after simplification and using the given equation, the expression reduces to 1+ccos⁡θ1+c \cos \theta.
279 Medium
279 of 337
Analyzing +4 | -1
Let z,wz, w be two complex numbers such that zˉ+iwˉ=0\bar{z}+i \bar{w}=0 and arg⁡(zw)=π\arg (z w)=\pi, then arg⁡z\arg z equals
View answer and explanation
Correct option: A
3π4\frac{3 \pi}{4}
Option A (3π4\frac{{3\pi}}{{4}}) is correct because the given conditions lead to arg⁡(z)=3π4\arg(z) = \frac{{3\pi}}{{4}} after substituting and simplifying the arguments.
280 Medium
280 of 337
Application +4 | -1
For any complex number zz, the minimum value of ∣z∣+∣z−1∣\left| z \right| + \left| {z - 1} \right| is
View answer and explanation
Correct option: A
1
Option B ((a2+b2)(∣z1∣2+∣z2∣2)(a^2 + b^2)(|z_1|^2 + |z_2|^2)) is correct because expanding and simplifying the given expression using properties of complex numbers leads to this result.
281 Medium
281 of 337
Understanding +4 | -1
STATEMENT-I: Let m,nm, n be two natural numbers such that m=a2+b2m=a^2+b^2 and n=c2+d2n=c^2+d^2 for some a,b,c,d∈Ia, b, c, d \in I, then there exist e,f∈Ie, f \in I such that mn=e2+f2m n=e^2+f^2

STATEMENT-II: If z1,z2∈Cz_1, z_2 \in C, then ∣z1+z2∣=∣z1∣+∣z2∣\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right| if and only if z1z2>0z_1 z_2>0
View answer and explanation
Correct option: B
I is true, II is false
Option B (I is true, II is false) is correct because statement I is based on the property of complex number magnitudes, while statement II is only true when z1 and z2 have the same argument.
282 Medium
282 of 337
Application +4 | -1
For all complex numbers z\mathrm{z} of the form 1+iα,α∈R1+\mathrm{i} \alpha, \alpha \in \mathrm{R}, if z2=x+iy\mathrm{z}^{2}=\mathrm{x}+\mathrm{iy}, then
View answer and explanation
Correct option: B
y2+4x−4=0y^{2}+4 \mathrm{x}-4=0
Option B (y2 + 4x - 4 = 0) is correct because substituting z = 1 + iα into z2 = x + iy and eliminating α results in the equation y2 + 4x - 4 = 0.
283 Medium
283 of 337
Application +4 | -1
If ∣z−25z−1∣=5\left|\frac{z-25}{z-1}\right|=5, then ∣z∣|z| is equal to
View answer and explanation
Correct option: A
5
Option A (5) is correct because simplifying the given equation leads to |z|^2 = 25, thus |z| = 5.
284 Medium
284 of 337
Application +4 | -1
If ω\omega is a complex cube root of unity, then (1−ω+ω2)(1−ω2+ω4)(1−ω4+ω8)\left(1-\omega+\omega^2\right)\left(1-\omega^2+\omega^4\right)\left(1-\omega^4+\omega^8\right) (1−ω8+ω16)\left(1-\omega^8+\omega^{16}\right) is equal to
View answer and explanation
Correct option: C
16
Option C (16) is correct because by simplifying the expression using the properties of complex cube roots of unity, the result is 16.
285 Medium
285 of 337
Application +4 | -1
If (1+x)5=a0+a1x+a2x2+a3x3+a4x4+a5x5(1+x)^5=a_0+a_1 x+a_2 x^2+a_3 x^3+a_4 x^4+a_5 x^5. Then the value of (a0−a2+a4)2+(a1−a3+a5)2\left(a_0-a_2+a_4\right)^2+\left(a_1-a_3+a_5\right)^2 is equal to
View answer and explanation
Correct option: B
32
Option B (32) is correct because substituting x=i into the binomial expansion and taking the magnitude squared results in 32.
286 Medium
286 of 337
Analyzing +4 | -1
The equation not representing a circle is given by
View answer and explanation
Correct option: D
∣z−1z+1∣=1\left|\frac{z-1}{z+1}\right|=1
Option D is correct because ∣z−1z+1∣=1\left|\frac{{z-1}}{{z+1}}\right|=1 represents a straight line, while the other options represent circles.
287 Medium
287 of 337
Application +4 | -1
The value of the expression (i+32)8+(i−32)8\left(\frac{i+\sqrt{3}}{2}\right)^8+\left(\frac{i-\sqrt{3}}{2}\right)^8 is
View answer and explanation
Correct option: A
−1-1
Option A (−1-1) is correct because it is the simplified form of the given expression using properties of complex cube roots of unity.
288 Medium
288 of 337
Application +4 | -1
The complex number satisfying arg (z+i)=π4(z+i)=\frac{\pi}{4} and arg⁡(2z+3−2i)=3π4\arg (2 z+3-2 i)=\frac{3 \pi}{4} simultaneously, is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the given conditions lead to contradictory requirements for x and y, indicating no solution exists for z.
289 Medium
289 of 337
Application +4 | -1
Let x1x_1 and y1y_1 be real numbers. If z1z_1 and z2z_2 are complex numbers such that ∣z1∣=∣z2∣=4\left|z_1\right|=\left|z_2\right|=4, then ∣x1z1−y1z2∣2+∣y1z1+x1z2∣2=\left|\mathrm{x}_1 \mathrm{z}_1-\mathrm{y}_1 \mathrm{z}_2\right|^2+\left|\mathrm{y}_1 \mathrm{z}_1+\mathrm{x}_1 \mathrm{z}_2\right|^2=
View answer and explanation
Correct option: A
32(x12+y12)32\left(\mathrm{x}_1^2+\mathrm{y}_1^2\right)
Option A (2) is correct because the minimum value is obtained when any two are zero and the third is a minimum magnitude integer i.e. 1.
290 Medium
290 of 337
Application +4 | -1
A complex number zz is such that arg⁡(z−2z+2)=π3\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}. The points representing this complex number will lie on
View answer and explanation
Correct option: C
A circle
Option C (A circle) is correct because the given condition leads to the equation of a circle when zz is expressed as x+iyx+iy.
291 Medium
291 of 337
Application +4 | -1
The principal argument of the complex number (1+i)5(1+3i)2−2i(−3+i)\frac{(1+i)^5(1+\sqrt{3} i)^2}{-2 i(-\sqrt{3}+i)} is
View answer and explanation
Correct option: C
−5π12-\frac{5 \pi}{12}
Option C (−5π12-\frac{5 \pi}{12}) is correct because calculating the argument of each complex number and using the properties of arguments leads to the principal argument being −5π12-\frac{5 \pi}{12}.
292 Medium
292 of 337
Application +4 | -1
If z1=8+4i,z2=6+4iz_1=8+4 i, z_2=6+4 i and arg⁡(z−z1z−z2)=π4\arg \left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi}{4}, then zz satisfies
View answer and explanation
Correct option: B
∣z−7−5i∣=2|z-7-5 i|=\sqrt{2}
Option B (∣z−7−5i∣=2|z-7-5 i|=\sqrt{2}) is correct because it represents the locus of the complex number zz satisfying the given argument condition.
293 Medium
293 of 337
Application +4 | -1
The value of ∑k=16(sin⁡2πk7−icos⁡2πk7)  \sum \limits_{k = 1}^6 \left( {\sin \frac{{2\mathrm{\pi }k}}{7} - i\cos \frac{{2\mathrm{\pi }k}}{7}} \right)\; is
View answer and explanation
Correct option: D
ii
Option D (ii) is correct because the sum is simplified using De Moivre's theorem and the properties of geometric progressions.
294 Medium
294 of 337
Analyzing +4 | -1
Let a≠ba \neq b be two non-zero real numbers. Then the number of elements in the set X={z∈C:Re⁡(az2+bz)=aX=\left\{z \in C: \operatorname{Re}\left(a z^{2}+b z\right)=a\right. and Re⁡(bz2+az)=b}\left.\operatorname{Re}\left(b z^{2}+a z\right)=b\right\} is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the given equations lead to an impossible condition, indicating no solution for z.
295 Medium
295 of 337
Application +4 | -1
Let z1,z2z_1, z_2 be two distinct non-zero complex numbers such that the origin, z1z_1 and z2z_2 form an equilateral triangle, then z12+z22−z1z2z_1^2+z_2^2-z_1 z_2 is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the given condition implies that z_1/z_2 is a complex cube root of unity, which leads to the equation z_12 + z_22 - z_1z_2 = 0.
296 Medium
296 of 337
Application +4 | -1
Suppose z1,z2,z3z_1, z_2, z_3 represent the vertices A,BA, B, and CC respectively of a △ABC\triangle A B C with centroid at GG. If the mid point of AGA G is the origin, then
View answer and explanation
Correct option: D
4z1+z2+z3=04 z_1+z_2+z_3=0
Option D (4z1+z2+z3=04z_1+z_2+z_3=0) is correct because the midpoint of AG being the origin leads to the given relationship between the complex numbers.
297 Medium
297 of 337
Application +4 | -1
If 1,ω1, \omega and ω2\omega^2 are the three cube roots of unity, then (1+ω)(1+ω2)(1+ω4)(1+ω8)…(1+\omega)\left(1+\omega^2\right)\left(1+\omega^4\right)\left(1+\omega^8\right) \ldots to 2n2 n factors is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because the product simplifies to 1 using the properties of complex cube roots of unity.
298 Medium
298 of 337
Analyzing +4 | -1
If z=x−iyz = x - iy and z1/3=p+iq{z^{1/3}} = p + iq, then (xp+yq)/(p2+q2)\left( {\frac{x}{p} + \frac{y}{q}} \right)/\left( {{p^2} + {q^2}} \right) is equal to
View answer and explanation
Correct option: D
−2- 2
Option D (−2- 2) is correct because after substituting z=x−iyz = x - iy and z1/3=p+iq{z^{1/3}} = p + iq and simplifying, the expression (xp+yq)/(p2+q2)\left( {\frac{x}{p} + \frac{y}{q}} \right)/\left( {{p^2} + {q^2}} \right) evaluates to -2.
299 Medium
299 of 337
Application +4 | -1
If z=−1z=-1, then the principal value of arg (z2/3)\left(z^{2 / 3}\right) is equal to
View answer and explanation
Correct option: B
2π3\frac{2 \pi}{3} or 2π2 \pi or 10π3\frac{10 \pi}{3}
Option B (2π3\frac{2 \pi}{3} or 2π2 \pi or 10π3\frac{10 \pi}{3}) is correct because it represents the possible values of the argument of z2/3z^{2/3} when z=−1z=-1.
300 Medium
300 of 337
Application +4 | -1
The centre of a square is at the origin and 1+i1 + i is one of its vertices. The extremities of its diagonals which does not pass through this vertex are
View answer and explanation
Correct option: A
1−i,  −1+i1 - i,\; - 1 + i
Option A (1−i,−1+i1 - i, - 1 + i) is correct because these are the vertices obtained by rotating the given vertex by 90 and 270 degrees around the origin.
Revision notes and key takeaways

Review COMPLEX NUMBERS in your textbook or class notes, then use this set to check your understanding.

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