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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Mixed difficulty

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Showing 151-180 of 979 questions

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151 Easy
151 of 979
Analyzing +4 | -1
The value of cos⁡2π7+cos⁡4π7+cos⁡6π7\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7} is equal to
View answer and explanation
Correct option: B
−12\frac{-1}{2}
Option B (-1/2) is correct because the value is obtained by multiplying and dividing by 2sin⁡(2π7)2\sin(\frac{2\pi}{7}) and using trigonometric identities.
152 Medium
152 of 979
Application +4 | -1
If 1−iα1+iα=A+iB\frac{1-i \alpha}{1+i \alpha}=A+i B, then A2+B2A^2+B^2 equals
View answer and explanation
Correct option: A
1
Option A (1) is correct because the magnitude of a complex number divided by its conjugate is always 1.
153 Medium
153 of 979
Application +4 | -1
The number of values of θ∈(0,π]\theta \in(0, \pi], such that ( cos⁡θ+\cos \theta+ isin⁡θ)(cos⁡3θ+isin⁡3θ)(cos⁡5θ+isin⁡5θ)(cos⁡7θi \sin \theta)(\cos 3 \theta+i \sin 3 \theta)(\cos 5 \theta+i \sin 5 \theta)(\cos 7 \theta +isin⁡7θ)(cos⁡9θ+isin⁡9θ)=−1+i \sin 7 \theta)(\cos 9 \theta+i \sin 9 \theta)=-1 is
View answer and explanation
Correct option: B
13
Option B (13) is correct because solving the equation cos(25θ) = -1 within the given interval (0, π] yields 13 possible values for θ.
154 Easy
154 of 979
Analyzing +4 | -1
Let a=ei2π3  a = {e^{i\frac{{2\pi }}{3}\;}} Then, the equation whose roots are a+a−2a + {a^{ - 2}} and a2+a−4{a^2} + {a^{ - 4}} is
View answer and explanation
Correct option: D
x2+2x+4=0{x^2} + 2x + 4 = 0
Option D (x2+2x+4=0{x^2} + 2x + 4 = 0) is the correct equation, found by determining the roots a+a−2a + a^{-2} and a2+a−4a^2 + a^{-4} and then constructing the quadratic equation.
155 Medium
155 of 979
Application +4 | -1
If ∣z+4∣≤3\left| {z + 4} \right| \le 3, then the greatest and the least value of ∣z+1∣\left| {z + 1} \right| are respectively
View answer and explanation
Correct option: B
6,06,0
Option B (6,06,0) is correct because these are the greatest and least values of ∣z+1∣|z+1| based on the given inequality and geometric interpretation.
156 Easy
156 of 979
Application +4 | -1
If z=cos⁡θ+isin⁡θz=\cos \theta+i \sin \theta, then
View answer and explanation
Correct option: A
zn+1zn=2cos⁡nθz^n+\frac{1}{z^n}=2 \cos n \theta
Option A (zn+1zn=2cos⁡nθz^n+\frac{1}{z^n}=2 \cos n \theta) is correct because it follows directly from De Moivre's theorem and Euler's formula.
157 Medium
157 of 979
Application +4 | -1
ei(2kcot⁡−1m)(mi+1mi−1)k=……e^{i\left(2 k \operatorname{cot}^{-1} m\right)}\left(\frac{m i+1}{m i-1}\right)^k=\ldots \ldots where k∈Z,m>0k \in Z, m>0
View answer and explanation
Correct option: B
1
Option B (1) is correct because the expression simplifies to 1 using Euler's formula and trigonometric identities.
158 Medium
158 of 979
Application +4 | -1
If the complex numbers z1=a+i,  z2=1++ib,z3=0{z_1} = a + i,\;{z_2} = 1 + + ib,{z_3} = 0 form the vertices of equilateral triangle (a,  ba,\;b are real numbers between 0 and 1), then
View answer and explanation
Correct option: B
a=2−3,  b=2−3a = 2 - \sqrt 3 ,\;b = 2 - \sqrt 3
Option B (a=2−3,  b=2−3a = 2 - \sqrt 3 ,\;b = 2 - \sqrt 3) is correct because by using the properties of equilateral triangles and complex numbers, we can derive the values of a and b.
159 Easy
159 of 979
Application +4 | -1
Let α\alpha be a root of the equation 1+x2+x4=01+x^2+x^4=0. Then the value of α1011+α2022−α3033\alpha^{1011}+\alpha^{2022}-\alpha^{3033} is equal to
View answer and explanation
Correct option: A
11
Option A (1) is correct because after simplifying the expression α1011+α2022−α3033\alpha^{1011}+\alpha^{2022}-\alpha^{3033} using the properties of roots of unity, the result is 1.
160 Easy
160 of 979
Analyzing +4 | -1
If z3+(3+2i)z+(−1+ia)=0z^3+(3+2 i) z+(-1+i a)=0, where i=−1i=\sqrt{-1}, has one real root, the value of aa lies in the interval (a∈R)(a \in R)
View answer and explanation
Correct option: B
(−1,0)(-1,0)
Option B ((-1,0)) is correct because the real root condition leads to a cubic equation in 'a', and analyzing the function reveals that 'a' lies in the interval (-1, 0).
161 Medium
161 of 979
Application +4 | -1
Let complex number z=(a+cos⁡θ)+(2a−sin⁡θ)iz=(a+\cos \theta)+(2 a-\sin \theta) i. If ∣z∣≤2|z| \leq 2 for any θ∈R\theta \in R, then the range of aa is
View answer and explanation
Correct option: B
[−15,15]\left[-\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right]
Option B ([−15,15]\left[-\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right]) is correct because it represents the range of 'a' for which the magnitude of the complex number is less than or equal to 2 for all values of θ\theta.
162 Medium
162 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) are the vertices of a triangle such that ∣z1∣=∣z2∣=∣z3∣>0\left|z_1\right|=\left|z_2\right|=\left|z_3\right|>0, then orthocentre of △ABC\triangle A B C is
View answer and explanation
Correct option: A
z1+z2+z3z_1+z_2+z_3
Option A (z1+z2+z3z_1+z_2+z_3) is correct because the orthocenter is related to the centroid and circumcenter by the formula z_H = 3z_G - 2z_S, and since the circumcenter is 0, z_H = 3 * (1/3)(z_1 + z_2 + z_3).
163 Medium
163 of 979
Application +4 | -1
If 32+cos⁡θ+isin⁡θ=a+ib\frac{3}{2+\cos \theta+i \sin \theta}=a+i b, then (a−2)2+b2(a-2)^2+b^2 is equal to
View answer and explanation
Correct option: B
1
The value of (a−2)2+b2(a-2)^2 + b^2 is 1 because it is obtained by manipulating the given expression and using trigonometric identities.
164 Medium
164 of 979
Application +4 | -1
In the argand plane, the complex number z=4−3iz = 4 - 3i is turned in the clockwise sense through 1800{180^0} and stretched three times. The complex number represented by the new number is
View answer and explanation
Correct option: D
−12+9i- 12 + 9i
Option D (−12+9i- 12 + 9i) is correct because the complex number is rotated 180 degrees clockwise and then stretched by a factor of 3.
165 Medium
165 of 979
Application +4 | -1
If x+iy=u+ivu−ivx+i y=\frac{u+i v}{u-i v}, then x2+y2x^2+y^2 is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because multiplying x+iyx+iy by its conjugate x−iyx-iy results in 1.

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166 Easy
166 of 979
Application +4 | -1
The value of i\sqrt i is
View answer and explanation
Correct option: D
±12(1+i)\pm \frac{1}{{\sqrt 2 }}\left( {1 + i} \right)
Option D (±12(1+i)\pm \frac{1}{{\sqrt 2 }}\left( {1 + i} \right)) is correct because it is the simplified form of the square root of i.
167 Easy
167 of 979
Application +4 | -1
If x=a+b,  y+aα+bβ   and   z=aβ+bα,   where   α   and  βx = a + b,\;y + a\alpha + b\beta \;\,\mathrm{and\;}\,\mathrm{z} = a\beta + b\alpha ,\;\,\mathrm{where}\,\;\alpha \,\;\mathrm{and}\;\beta are complex cube roots of unity, then xyzxyz is equal to
View answer and explanation
Correct option: B
a3+b3{a^3} + {b^3}
The correct answer is B (a3+b3{a^3} + {b^3}) because it is obtained by multiplying the given expressions and using the properties of complex cube roots of unity.
168 Easy
168 of 979
Application +4 | -1
The value of (1+i31−i3)6+(1−i31+i3)6{\left( {\frac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right)^6} + {\left( {\frac{{1 - i\sqrt 3 }}{{1 + i\sqrt 3 }}} \right)^6} is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the expression simplifies to ω6+1ω6{\omega ^6} + \frac{1}{{{\omega ^6}}}, which equals 2 using properties of complex cube roots of unity.
169 Easy
169 of 979
Application +4 | -1
The locus of point zz satisfying Re(z2)=0,\mathrm{Re}\left( {{z^2}} \right) = 0, is
View answer and explanation
Correct option: A
A pair of straight lines
Option A (A pair of straight lines) is correct because the equation Re(z2) = 0 simplifies to y = ±x, representing two straight lines.
170 Easy
170 of 979
Application +4 | -1
Let z1=6+i{z}_1=6+{i} and z2=4−3i{z}_2=4-3 {i} (where i=−1{i}=\sqrt{-1} ). Let z{z} be a complex number such that arg⁡(z−z1z2−z)=π2\arg \left(\frac{z-z_1}{z_2-z}\right)=\frac{\pi}{2}, then zz satisfies
View answer and explanation
Correct option: B
∣z−(5−i)∣=5|z-(5-i)|=\sqrt{5}
Option B, ∣z−(5−i)∣=5|z-(5-i)|=\sqrt{5}, is correct because the locus of zz is a circle with center at the midpoint of z1z_1 and z2z_2 and radius equal to half the distance between them.
171 Medium
171 of 979
Analyzing +4 | -1
Let u=2z+iz−ki,  z=x+iy  and  k>0.u = \frac{{2z + i}}{{z - ki}},\,\,z = x + iy\mathrm{ }\,\,\mathrm{and }\,\,k > 0. if the curve represented by Re(u)+Im(u)=1{\mathop\mathrm{Re}\nolimits} (u) + {\mathop\mathrm{Im}\nolimits} (u) = 1 intersects the Y-axis at the point P and QP\mathrm{ }\,\mathrm{and}\,\mathrm{ Q} where PQ=5PQ=5 then the value of k is
View answer and explanation
Correct option: C
2
Option C (2) is correct because solving the equation derived from the given conditions yields k=2, considering k>0.
172 Medium
172 of 979
Application +4 | -1
Let zz be a complex number such that ∣z−2iz+i∣=2,z≠−i\left|\frac{z-2 i}{z+i}\right|=2, z \neq-i. Then zz lies on the circle of radius 2 and centre
View answer and explanation
Correct option: D
(0,−2)(0,-2)
Option D ((0,-2)) is correct because the equation represents a circle with center (0, -2) after simplification.
173 Medium
173 of 979
Application +4 | -1
Let z\mathrm{z} and w\mathrm{w} be non-zero complex numbers such that ∣z∣=∣w∣|\mathrm{z}|=|\mathrm{w}| and arg⁡(z)+arg⁡(w)=π\arg (\mathrm{z})+\arg (\mathrm{w})=\pi. Then z=\mathrm{z}=
View answer and explanation
Correct option: D
−w‾-\overline{\mathrm{w}}
Option D (−w‾-\overline{\mathrm{w}}) is correct because given the conditions, z is the negative of the conjugate of w.
174 Easy
174 of 979
Application +4 | -1
If (1+i)22−i=x+iy\frac{{{{\left( {1 + i} \right)}^2}}}{{2 - i}} = x + iy, then x+yx + y is equal to
View answer and explanation
Correct option: C
25\frac{2}{5}
Option C (25\frac{2}{5}) is correct because after simplifying the given expression and equating it to x + iy, the sum x + y equals 25\frac{2}{5}.
175 Medium
175 of 979
Application +4 | -1
If the imaginary part of 2z+1iz+1\frac{2 z+1}{i z+1} is −2,-2 , then the locus of the point representing zz in the complex plane is
View answer and explanation
Correct option: B
A straight line
Option B (A straight line) is correct because after substituting z = x + iy and simplifying, the imaginary part condition leads to a linear equation in x and y.
176 Easy
176 of 979
Analyzing +4 | -1
The triangle formed by the points 1, 1+i2\frac{{1 + i}}{{\sqrt 2 }} and ii as vertices in the Argand diagrams is
View answer and explanation
Correct option: C
Isosceles
Option C (Isosceles) is correct because two sides of the triangle formed by the given complex numbers have equal lengths.
177 Easy
177 of 979
Application +4 | -1
If ∣6i−3i143i−1203i∣=x+iy\left|\begin{array}{ccc}6 i & -3 i & 1 \\ 4 & 3 i & -1 \\ 20 & 3 & i\end{array}\right|=x+i y, then
View answer and explanation
Correct option: D
x=0,y=0x=0, y=0
Option D (x=0,y=0x=0, y=0) is correct because the determinant evaluates to 0, resulting in both the real and imaginary parts being zero.
178 Hard
178 of 979
Analyzing +4 | -1
If ∣z−4−3i∣≤1\left| {z - 4 - 3i} \right| \le 1 and mm and nn are the least and greatest value of ∣z∣\left| z \right| and λ\lambda is the least value of x4+x2+4x\frac{{{x^4} + {x^2} + 4}}{x} in the interval (0,  ∞)\left( {0,\;\infty } \right) then λ\lambda is equal to
View answer and explanation
Correct option: B
nn
Option B (nn) is correct because the least value of the given expression is 6, which is equal to the greatest value of |z|.
179 Medium
179 of 979
Analyzing +4 | -1
The number of integral solutions of x+1x2+2>14,\frac{{x + 1}}{{{x^2} + 2}} > \frac{1}{4}, is
View answer and explanation
Correct option: C
5
Option C (5) is correct because solving the inequality x+1x2+2>14\frac{{x + 1}}{{{x^2} + 2}} > \frac{1}{4} leads to the integral solutions 2, 3, 4, 5, and 6.
180 Easy
180 of 979
Understanding +4 | -1
If the complex numbers z1,  z2,  z3{z_1},\;{z_2},\;{z_3} are in AP, then they lie on
View answer and explanation
Correct option: C
A straight line
Option C (A straight line) is correct because if three complex numbers are in arithmetic progression (AP), they are collinear and lie on a straight line.
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