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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Easy

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Showing 151-180 of 612 questions

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151 Easy
151 of 612
Analyzing +4 | -1
The points represented by the complex numbers 1+i,  −2+3i,53i1 + i,\; - 2 + 3i,\frac{5}{3}i on the argand diagram are
View answer and explanation
Correct option: C
Collinear
Option C (Collinear) is correct because the determinant of the matrix formed by the coordinates of the points is zero, indicating they lie on the same line.
152 Easy
152 of 612
Application +4 | -1
The locus of zz satisfying Im⁡(z2)=4\operatorname{Im}\left(z^2\right)=4 is
View answer and explanation
Correct option: B
a rectangular hyperbola
Option B (a rectangular hyperbola) is correct because the condition Im(z2z^2) = 4 translates to 2xy=42xy = 4, which is the equation of a rectangular hyperbola.
153 Easy
153 of 612
Application +4 | -1
If z=cos⁡θ+isin⁡θz=\cos \theta+i \sin \theta is a root of xn+p1xn−1+p2xn−2+…+pn=0,x^n+p_1 x^{n-1}+p_2 x^{n-2}+\ldots+p_n=0,

where p1,p2,…,pn∈Rp_1, p_2, \ldots, p_n \in \mathbf{R},then the value of

7(p1sin⁡θ+p2sin⁡2θ+…+pnsin⁡(nθ))7\left(p_1 \sin \theta+p_2 \sin 2 \theta+\ldots+p_n \sin (n \theta)\right) is
View answer and explanation
Correct option: D
0
Option D (0) is correct because since the coefficients are real, the complex conjugate is also a root, leading to the given expression equaling 0.
154 Easy
154 of 612
Application +4 | -1
If z=x+iyz=x+i y and 0≤sin⁡−1(z−42i)≤π20 \leq \sin ^{-1}\left(\frac{z-4}{2 i}\right) \leq \frac{\pi}{2} then
View answer and explanation
Correct option: A
x=4,0≤y≤2x=4,0 \leq y \leq 2
Option A (x=4,0≤y≤2x=4, 0 \leq y \leq 2) is correct because it satisfies the given condition 0≤sin⁡−1(z−42i)≤π20 \leq \sin^{-1}(\frac{{z-4}}{{2i}}) \leq \frac{{\pi}}{{2}}.
155 Easy
155 of 612
Application +4 | -1
Dividing f(z)f(z) by z−iz-i, we obtain the remainder 1−i1-i and dividing it by z+iz+i, we get the remainder 1+i1+i. Then, the remainder upon the division of f(z)f(z) by z2+1z^2+1, is
View answer and explanation
Correct option: C
1−z1-z
Option C (1−z1-z) is correct because the remainder is found by using the polynomial remainder theorem and solving a system of equations to determine the coefficients of the remainder polynomial.
156 Easy
156 of 612
Analyzing +4 | -1
If x,y,zx,y,z are real and distinct, then x2+4  y2+9  z2−6  yz−3  zx−2  xy{x^2} + 4\;{y^2} + 9\;{z^2} - 6\;yz - 3\;zx - 2\;xy is always
View answer and explanation
Correct option: A
Non-negative
Option A (Non-negative) is correct because the expression can be rewritten as a sum of squares, which is always non-negative.
157 Easy
157 of 612
Application +4 | -1
If x+iy=a+ibc+idx+i y=\sqrt{\frac{a+i b}{c+i d}}, then (x2+y2)2(c2+d2)\left(x^2+y^2\right)^2\left(c^2+d^2\right) equals
View answer and explanation
Correct option: A
a2+b2a^2+b^2
Option A (a2+b2a^2+b^2) is correct because it follows from taking the modulus of both sides of the equation and simplifying.
158 Easy
158 of 612
Application +4 | -1
(1+i)8+(1−i)8={\left( {1 + i} \right)^8} + {\left( {1 - i} \right)^8} =
View answer and explanation
Correct option: B
25{2^5}
Option B (25{{{2^5}}}) is correct because it is the result of simplifying the expression using De Moivre's theorem and trigonometric identities.
159 Easy
159 of 612
Application +4 | -1
Let an=i(n+1)2{a_n} = {i^{{{\left( {n + 1} \right)}^2}}}, where i=−1i = \sqrt { - 1} and n=1,  2,  3,…n = 1,\;2,\;3, \ldots Then the value of a1+a3+a5+...+a25{a_1} + {a_3} + {a_5} + ... + {a_{25}} is
View answer and explanation
Correct option: A
13
The correct answer is A (13) because the series a1+a3+a5+...+a25{a_1} + {a_3} + {a_5} + ... + {a_{25}} simplifies to a sum of 13 ones.
160 Easy
160 of 612
Understanding +4 | -1
If zz is a complex number, then (zˉ−1)(zˉ)\left( {{{\bar z}^{ - 1}}} \right)\left( {\bar z} \right) is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because the inverse of the conjugate multiplied by the conjugate equals 1.
161 Easy
161 of 612
Application +4 | -1
The complex number z=i−1cos⁡π3+isin⁡π3z=\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}} is equal to:
View answer and explanation
Correct option: A
2(cos⁡5π12+isin⁡5π12)\sqrt{2}\left(\cos \frac{5 \pi}{12}+i \sin \frac{5 \pi}{12}\right)
Option A (2(cos⁡5π12+isin⁡5π12)\sqrt{2}(\cos \frac{5 \pi}{12}+i \sin \frac{5 \pi}{12})) is correct because it involves converting the complex number to polar form and simplifying using Euler's formula.
162 Easy
162 of 612
Understanding +4 | -1
The equation ∣z+1−i∣=∣z+i−1∣|z+1-i|=|z+i-1| represents
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Correct option: A
a straight line
Option A (a straight line) is correct because the equation represents the locus of points equidistant from two fixed points, which is a straight line (perpendicular bisector).
163 Easy
163 of 612
Application +4 | -1
Let ω\omega be a complex cube root of unity. If the equation ∣z−ω∣2+∣z−ω2∣2=λ{\left| {z - \omega } \right|^2} + {\left| {z - {\omega ^2}} \right|^2} = \lambda represents a circle with points representing ω\omega and ω2{\omega ^2} as the end points of a diameter, then λ=\lambda =
View answer and explanation
Correct option: B
3
The correct answer is B (3) because the diameter squared is the distance between the cube roots of unity.
164 Easy
164 of 612
Application +4 | -1
If ω\omega is fifth root of unity, then log⁡2∣1+ω+ω2+ω3−ω−1∣\log _2\left|1+\omega+\omega^2+\omega^3-\omega^{-1}\right| is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because simplifying the expression using the properties of the fifth root of unity results in a value of 1.
165 Easy
165 of 612
Analyzing +4 | -1
Let z=cos⁡θ+isin⁡θz=\cos \theta+i \sin \theta. 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 by using the formula for the sum of a geometric series and trigonometric identities, the given sum simplifies to 14sin⁡2∘\frac{1}{4 \sin 2^{\circ}}.

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166 Easy
166 of 612
Application +4 | -1
The least positive integer nn for which (1+i1−i)n{\left( {\frac{{1 + i}}{{1 - i}}} \right)^n} is real, is
View answer and explanation
Correct option: A
2
The correct answer is A (2) because it is the smallest positive integer nn for which (1+i1−i)n{\left( {\frac{{1 + i}}{{1 - i}}} \right)^n} is a real number.
167 Easy
167 of 612
Application +4 | -1
The vector z=−4+5  iz = - 4 + 5\;i is turned counterclockwise through an angle of 180∘180^\circ and stretched 1121\frac{1}{2} times. The complex number corresponding to newly obtained vector is
View answer and explanation
Correct option: A
6−152i6 - \frac{{15}}{2}i
Option A (6−152i6 - \frac{{15}}{2}i) is correct because it is the result of rotating the vector counterclockwise by 180 degrees and stretching it by a factor of 1.5.
168 Easy
168 of 612
Application +4 | -1
If ω=cos⁡πn+isin⁡πn\omega=\cos \frac{\pi}{n}+i \sin \frac{\pi}{n}, then value of 1+ω+ω2+…+ωn−11+\omega+\omega^2+\ldots+\omega^{n-1} is
View answer and explanation
Correct option: A
1+icot⁡(π2π)1+i \cot \left(\frac{\pi}{2 \pi}\right)
Option A (1+icot⁡(π2π)1+i \cot (\frac{{\pi}}{{2 \pi}})) is correct because it is derived from summing the geometric series and simplifying using trigonometric identities.
169 Easy
169 of 612
Analyzing +4 | -1
Suppose z∈Cz \in \mathbf{C}, and z∉Rz \notin \mathbf{R}. If w=1−z+z21+z+z2w=\frac{1-z+z^2}{1+z+z^2} is a real number, then ∣z∣|z| equals
View answer and explanation
Correct option: A
1
Option A (1) is correct because the condition that ww is real implies that z+1zz + \frac{{1}}{{z}} is real, which leads to ∣z∣=1|z| = 1 since zz is not real.
170 Easy
170 of 612
Analyzing +4 | -1
If z1=1+2i,    z2=2+3i,    z3=3+4i,{z_1} = 1 + 2i,\;\;{z_2} = 2 + 3i,\;\;{z_3} = 3 + 4i, then z1,  z2,  z3{z_1},\;{z_2},\;{z_3} represents the vertices of a/an
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the given complex numbers represent collinear points, not vertices of any triangle.
171 Easy
171 of 612
Analyzing +4 | -1
If 8iz3+12z2−18z+27i=08 i z^3+12 z^2-18 z+27 i=0, then
View answer and explanation
Correct option: A
∣z∣=32|z|=\frac{3}{2}
Option A (∣z∣=32|z|=\frac{{3}}{{2}}) is correct because by factoring the given equation, we find that ∣z∣=32|z| = \frac{{3}}{{2}} is a solution.
172 Easy
172 of 612
Application +4 | -1
If z1=2+3iz_1=2+3 i and z2=3+2iz_2=3+2 i, then z1+z2z_1+z_2 equals to a+ai.a+ ai. Value of 'aa' is equal to
View answer and explanation
Correct option: A
5
Option A (5) is correct because adding the two complex numbers z1 and z2 results in 5 + 5i, so the value of 'a' is 5.
173 Easy
173 of 612
Application +4 | -1
If zz lies on the circle ∣z−1∣=1,\left| {z - 1} \right| = 1, z−2z\frac{{z - 2}}{z} is
View answer and explanation
Correct option: B
Purely imaginary
Option B (Purely imaginary) is correct because the expression z−2z\frac{z-2}{z} simplifies to itan⁡(θ2)i \tan(\frac{\theta}{2}), which is purely imaginary.
174 Easy
174 of 612
Application +4 | -1
If (a1+ib1)(a2+ib2)⋯(an+ibn)=A+iB\left(a_1+i b_1\right)\left(a_2+i b_2\right) \cdots\left(a_n+i b_n\right)=A+i B, then (a12+b12)(a22+b22)⋯(an2+bn2)\left(a_1^2+b_1^2\right)\left(a_2^2+b_2^2\right) \cdots\left(a_n^2+b_n^2\right) is equal to
View answer and explanation
Correct option: B
A2+B2A^2+B^2
Option B (A2+B2A^2+B^2) is correct because it represents the product of the magnitudes of the complex numbers.
175 Easy
175 of 612
Application +4 | -1
If −π<arg⁡(z)<−π2- \pi < \arg \left( z \right) < - \frac{\pi }{2} then arg⁡(zˉ)−arg⁡(−zˉ)\arg \left( {\bar z} \right) - \arg \left( { - \bar z} \right) is
View answer and explanation
Correct option: A
π\pi
Option A (π\pi) is correct because given the range of arg(z), the expression simplifies to π\pi using properties of arguments of complex numbers.
176 Easy
176 of 612
Application +4 | -1
If zz is purely imaginary and Im⁡(z)<0\operatorname{Im}(z) < 0, then arg⁡(izˉ)+arg⁡(z)\arg (i \bar{z})+\arg (z) is equal to
View answer and explanation
Correct option: C
π/2\pi / 2
Option C (π/2\pi / 2) is correct because the arguments of the complex numbers add up to this value, given the conditions on z.
177 Easy
177 of 612
Understanding +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: B
1
Option B (1) is correct because by the triangle inequality, ∣z∣+∣z−1∣≥∣z−(z−1)∣=∣1∣=1|z| + |z-1| \ge |z - (z-1)| = |1| = 1.
178 Easy
178 of 612
Application +4 | -1
If ω(≠1)\omega(\neq 1) is a cube root of unity and (1+ω2)11=a+bω+cω2\left(1+\omega^2\right)^{11}=a+b \omega+c \omega^2, then (a,b,c)(a, b, c) equals
View answer and explanation
Correct option: A
(1,1,0)(1,1,0)
Option A (1,1,0) is correct because simplifying (1+ω^2)11 using the property 1+ω+ω^2=0 leads to the expression 1+ω, which corresponds to a=1, b=1, and c=0.
179 Easy
179 of 612
Application +4 | -1
One of the values of (1+i2)2/3{\left( {\frac{{1 + i}}{{\sqrt 2 }}} \right)^{2/3}} is
View answer and explanation
Correct option: C
ii
Option C (ii) is correct because it is one of the values obtained by converting the complex number to polar form and applying De Moivre's theorem.
180 Easy
180 of 612
Analyzing +4 | -1
The solution set of the inequation ∣x−2∣x−2<0,\frac{{\left| {x - 2} \right|}}{{x - 2}} < 0, is
View answer and explanation
Correct option: B
(−∞,2)\left( { - \infty ,2} \right)
Option C (18sin⁡(α+β+γ)18\sin \left( {\alpha + \beta + \gamma } \right)) is correct because it is derived from the complex number manipulation and equating imaginary parts.
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