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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Mixed difficulty

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511 Easy
511 of 979
Application +4 | -1
The equation arg (z−1z+1)=π4\left( {\frac{{z - 1}}{{z + 1}}} \right) = \frac{\pi }{4} represents a circle with
View answer and explanation
Correct option: B
centre at (0, 1) and radius 2\sqrt 2
Option B is correct because the given equation represents a circle in the complex plane with center (0, 1) and radius 2\sqrt{2}, derived by converting to Cartesian coordinates.
512 Easy
512 of 979
Application +4 | -1
If C2+S2=1,{C^2} + {S^2} = 1, then 1+C+i  S1+C−i  S\frac{{1 + C + i\;S}}{{1 + C - i\;S}} is equal to
View answer and explanation
Correct option: A
C+i  SC + i\;S
Option A (C+i  SC + i\;S) is correct because it is the simplified form of the given expression using trigonometric identities.
513 Medium
513 of 979
Analyzing +4 | -1
If g  (x)\mathrm{g}\;\left( x \right) and h(x)h\left( x \right) are two polynomials such that the polynomial P(x)=g(x3)+x  h(x3)P\left( x \right) = \mathrm{g}\left( {{x^3}} \right) + x\;h\left( {{x^3}} \right) is divisible by x2+x+1,{x^2} + x + 1, then which one of the following is not true?
View answer and explanation
Correct option: B
g(1)=h(1)≠0\mathrm{g}\left( 1 \right) = h\left( 1 \right) \ne 0
Option B (g(1)=h(1)≠0\mathrm{g}\left( 1 \right) = h\left( 1 \right) \ne 0) is not true because the correct deduction is that g(1)=h(1)=0\mathrm{g}\left( 1 \right) = h\left( 1 \right) = 0.
514 Medium
514 of 979
Application +4 | -1
Let z=(32+i2)5+(32−i2)5z = {\left( {\frac{{\sqrt 3 }}{2} + \frac{i}{2}} \right)^5} + {\left( {\frac{{\sqrt 3 }}{2} - \frac{i}{2}} \right)^5}. If R (z)R\,(z) and I (z)I\,(z) respectively denote the real and imaginary parts of zz, then
View answer and explanation
Correct option: B
I(z)=0I\left( z \right) = 0
Option B (I(z)=0I\left( z \right) = 0) is correct because the imaginary part of the complex number zz is zero, as zz simplifies to a real number.
515 Medium
515 of 979
Application +4 | -1
If arg⁡(z)<0\arg (z) < 0, then arg⁡(−z)−arg⁡(z)\arg (-z)-\arg (z) is equal to
View answer and explanation
Correct option: A
π\pi
Option A (π\pi) is correct because when arg⁡(z)<0\arg(z) < 0, arg⁡(−z)\arg(-z) is arg⁡(z)+π\arg(z) + \pi.
516 Easy
516 of 979
Understanding +4 | -1
The argument of (1−i3)(1+i3)\left( {1 - i\sqrt 3 } \right)\left( {1 + i\sqrt 3 } \right) is
View answer and explanation
Correct option: D
2400{240^0}
Option D (2400{240^0}) is correct because the argument of the complex number 1−i31+i3\frac{{1 - i\sqrt 3 }}{{1 + i\sqrt 3 }} is calculated to be 2400{240^0} after rationalizing and finding the angle in the correct quadrant.
517 Medium
517 of 979
Application +4 | -1
Let f(x)=x4+ax3+bx2+cx+df(x)=x^4+a x^3+b x^2+c x+d be a polynomial with real coefficients and real zeroes. If ∣f(i)∣=1|f(i)|=1, (where i=−1i=\sqrt{-1} ) then a+b+c+da+b+c+d is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because if |f(i)| = 1, all real zeroes must be zero, leading to f(x) = x4, and thus a+b+c+d = 0.
518 Medium
518 of 979
Application +4 | -1
The closest distance of the origin from a curve given as azˉ+aˉz+∣a∣2=0a\bar z + \bar az + {\left| a \right|^2} = 0 is
View answer and explanation
Correct option: D
∣a∣2\frac{{\left| a \right|}}{2}
Option D (∣a∣2\frac{{\left| a \right|}}{2}) is correct because the closest distance of the origin from the given curve is found using the formula for the distance from a point to a line.
519 Medium
519 of 979
Analyzing +4 | -1
If w=α+iβw = \alpha + i\beta, where β≠0\beta \ne 0 and z≠1z \ne 1, satisfies the condition that (w−wˉz1−z)\left( {\frac{{w - \bar wz}}{{1 - z}}} \right) is purely real, then the set of values of zz is
View answer and explanation
Correct option: B
∣z∣=1\left| z \right| = 1 and z≠1z \ne 1
Option B (∣z∣=1\left| z \right| = 1 and z≠1z \ne 1) is correct because the condition that (w−wˉz1−z)\left( {\frac{{w - \bar wz}}{{1 - z}}} \right) is purely real leads to ∣z∣=1|z| = 1, and since z≠1z \ne 1, the answer is ∣z∣=1\left| z \right| = 1 and z≠1z \ne 1.
520 Easy
520 of 979
Application +4 | -1
The complex number 1+2i1−i\frac{1+2 i}{1-i} lies in the
View answer and explanation
Correct option: B
II quadrant
Option B (II quadrant) is correct because the simplified complex number has a negative real part and a positive imaginary part.
521 Easy
521 of 979
Analyzing +4 | -1
If z1z_1 is a root of the equation z5+z3+z+3=0z^5+z^3+z+3=0, then
View answer and explanation
Correct option: B
∣z1∣≥1\quad\left|z_1\right| \geq 1
Option B (∣z1∣≥1\quad\left|z_1\right| \geq 1) is correct because if ∣z1∣<1|z_1| < 1, the inequality leads to a contradiction.
522 Easy
522 of 979
Application +4 | -1
If i=−1i = \sqrt { - 1} and nn is a positive integer, then in+in+1+in+2+in+3{i^n} + {i^{n + 1}} + {i^{n + 2}} + {i^{n + 3}} is equal to
View answer and explanation
Correct option: D
0
Option D (0) is correct because the sum of any four consecutive powers of ii is always 0.
523 Easy
523 of 979
Application +4 | -1
The points represents the complex numbers zz, for which ∣z−a∣2+∣z+a∣2=b2{\left| {z - a} \right|^2} + {\left| {z + a} \right|^2} = {b^2} lie on
View answer and explanation
Correct option: B
A circle
Option B (A circle) is correct because the given equation simplifies to the standard form of a circle's equation, x2+y2=constant{x^2} + {y^2} = constant.
524 Medium
524 of 979
Application +4 | -1
If −3+x2yi-3+x^2 y i and x2+y+4ix^2+y+4 i are conjugate of each other, then ∣x∣+∣y∣|x|+|y| is
View answer and explanation
Correct option: A
5
Option A (5) is correct because solving the equations derived from the conjugate property gives ∣x∣+∣y∣=5|x|+|y|=5.
525 Hard
525 of 979
Analyzing +4 | -1
Suppose f(x)f(x) is twice differentiable for all real Further suppose ∣f(x)∣≤1|f(x)| \leq 1 and ∣f′′(x)∣≤1\left|f^{\prime \prime}(x)\right| \leq 1 for all rea xx. Then ∣f′(x)∣\left|f^{\prime}(x)\right| is
where
f(x+1)=f(x)+f′(x)⌊+f′(ξ)⌊2f(x+1) =f(x)+\frac{f^{\prime}(x)}{\lfloor}+\frac{f^{\prime}(\xi)}{\lfloor 2}

 and f(x−1)=f(x)−f′(x)⌊+f′′(η)⌊2\text { and } \quad f(x-1) =f(x)-\frac{f^{\prime}(x)}{\lfloor}+\frac{f^{\prime \prime}(\eta)}{\lfloor 2}
for some ξ\xi and η\eta.
View answer and explanation
Correct option: C
less than or equal to 3/23 / 2
The correct answer is less than or equal to 3/23/2 because by using Taylor's expansion and the given conditions, we can bound the value of the first derivative.

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526 Medium
526 of 979
Application +4 | -1
If x=(1+i2),    (where   i=−1)x = \left( {\frac{{1 + i}}{2}} \right),\;\;\left( {\mathrm{where}\,\;i = \sqrt { - 1} } \right), then the expression 2x4−2x2+x+32{x^4} - 2{x^2} + x + 3 equals
View answer and explanation
Correct option: A
3−(i2)3 - \left( {\frac{i}{2}} \right)
Option A (3−(i2)3 - \left( {\frac{i}{2}} \right)) is correct because it is obtained by substituting the value of x and simplifying the expression using the given condition.
527 Medium
527 of 979
Analyzing +4 | -1
Let SS be the set of all (α,β),π<α,β<2π(\alpha, \beta), \pi<\alpha, \beta<2 \pi, for which the complex number 1−isin⁡α1+2isin⁡α\frac{1-i \sin \alpha}{1+2 i \sin \alpha} is purely imaginary and 1+icos⁡β1−2icos⁡β\frac{1+i \cos \beta}{1-2 i \cos \beta} is purely real, Let Zαβ=sin⁡2α+iZ_{\alpha \beta}=\sin 2 \alpha+i cos⁡2β,(α,β)∈S\cos 2 \beta,(\alpha, \beta) \in S. Then ∑(α, β)∈S(iZαβ+1iZˉαβ)\sum\limits_{\left( {\alpha ,\,\beta } \right) \in S} {\left( {i{Z_{\alpha \beta }} + \frac{1}{{i{{\bar Z}_{\alpha \beta }}}}} \right)} is equal to
View answer and explanation
Correct option: C
1
Option C (1) is correct because after finding the values of alpha and beta that satisfy the given conditions, substituting them into the expression and simplifying yields 1.
528 Easy
528 of 979
Analyzing +4 | -1
Given z=q+ir1+p,  then  p+iq1+r=1+iz1−izz = \frac{{q + ir}}{{1 + p}},\;\mathrm{then}\;\frac{{p + iq}}{{1 + r}} = \frac{{1 + iz}}{{1 - iz}} if
View answer and explanation
Correct option: A
p2+q2+r2=1{p^2} + {q^2} + {r^2} = 1
Option A (p2+q2+r2=1{p^2} + {q^2} + {r^2} = 1) is correct because it is the condition derived from the given equation using componendo and dividendo.
529 Medium
529 of 979
Application +4 | -1
The least positive integer n\mathrm{n} for which (1+i31−i3)n=1\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right)^{n}=1, is
View answer and explanation
Correct option: D
3
Option D (3) is correct because the least positive integer n that satisfies the given equation is 3, as shown by simplifying the complex number and raising it to the power of 3.
530 Easy
530 of 979
Application +4 | -1
The solution of equation ∣z∣−z=1+2i\left| z \right| - z = 1 + 2i is
View answer and explanation
Correct option: B
32−2i\frac{3}{2} - 2i
Option B (3/2 - 2i) is correct because substituting z = x + iy into the equation and solving for x and y yields x = 3/2 and y = -2.
531 Easy
531 of 979
Application +4 | -1
If Z1≠0Z_{1} \neq 0 and Z2Z_{2} be two complex numbers such that Z2Z1\frac{Z_{2}}{Z_{1}} is a purely imaginary number, then ∣2Z1+3Z22Z1−3Z2∣\left|\frac{2 Z_{1}+3 Z_{2}}{2 Z_{1}-3 Z_{2}}\right| is equal to:
View answer and explanation
Correct option: D
1
Option D (1) is correct because when Z2Z1\frac{Z_{2}}{Z_{1}} is purely imaginary, the magnitude of the expression simplifies to 1.
532 Easy
532 of 979
Application +4 | -1
If (a+i)22a−1=p+iq\frac{(a+i)^2}{2 a-1}=p+i q, then p2+q2p^2+q^2 is equal to
View answer and explanation
Correct option: C
(a2+1)24a2−1\frac{\left(a^2+1\right)^2}{4 a^2-1}
Option C, (a2+1)24a2−1\frac{\left(a^2+1\right)^2}{4 a^2-1}, is the correct answer because it results from multiplying the complex conjugate of the given expression.
533 Medium
533 of 979
Application +4 | -1
If z=cos⁡20∘+isin⁡20∘z=\cos 20^{\circ}+i \sin 20^{\circ}, then ∣z+2z2+3z3+………+18z18∣−1\left|z+2 z^2+3 z^3+\ldots \ldots \ldots+18 z^{18}\right|^{-1} is:
View answer and explanation
Correct option: B
19sin⁡10∘\frac{1}{9} \sin 10^{\circ}
Option B (19sin⁡10∘\frac{{1}}{{9}} \sin 10^{\circ}) is correct because after simplifying the given expression using geometric progression and properties of complex numbers, the magnitude of the resulting complex number is found to be 19sin⁡10∘\frac{{1}}{{9}} \sin 10^{\circ}.
534 Easy
534 of 979
Application +4 | -1
If z1,z2z_1, z_2 are complex numbers such that Re⁡(z1)=∣z1−2∣\operatorname{Re}\left(z_1\right)=\left|z_1-2\right|, Re⁡(z2)=∣z2−2∣\operatorname{Re}\left(z_2\right)=\left|z_2-2\right| and arg⁡(z1−z2)=π/3\arg \left(z_1-z_2\right)=\pi / 3, then Im⁡(z1−z2)=\operatorname{Im}\left(z_1-z_2\right)=
View answer and explanation
Correct option: B
4/34 / \sqrt{3}
Option B (4/34 / \sqrt{3}) is correct. By using the given conditions and properties of complex numbers, we can find the imaginary part of the difference between the two complex numbers.
535 Easy
535 of 979
Application +4 | -1
If the area of the triangle on the argand plane formed by the complex numbers −z,iz,z−iz-z, i z, z-i z is 600 sq units, then ∣z∣|z| is equal to
View answer and explanation
Correct option: B
20
Option B (20) is correct because the area of the triangle formed by -z, iz, and z-iz is (3/2)|z|^2, which equals 600, leading to |z| = 20.
536 Medium
536 of 979
Application +4 | -1
The value of ∑k=110(sin⁡2πk11−icos⁡2πk11)\sum_{k=1}^{10}\left(\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11}\right) is
View answer and explanation
Correct option: C
ii
Option C (ii) is correct because the summation simplifies to ii using properties of complex numbers and roots of unity.
537 Medium
537 of 979
Application +4 | -1
If i=−1i=\sqrt{-1}, then 4+5(−12+i32)334+3(−12+i32)3654+5\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)^{334}+3\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)^{365} is equal to
View answer and explanation
Correct option: C
i3i \sqrt{3}
Option C (i3i \sqrt{3}) is correct because it is the simplified form of the given complex number expression using properties of cube roots of unity.
538 Hard
538 of 979
Application +4 | -1
If a=cos⁡4π3+isin⁡4π3a=\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}, then the value of (1+a2)3n\left(\frac{1+a}{2}\right)^{3 n} is
View answer and explanation
Correct option: B
(−1)n23n\frac{(-1)^n}{2^{3 n}}
Option B ((−1)n23n\frac{{(-1)^n}}{{2^{3n}}}) is correct because after simplification using trigonometric identities and De Moivre's theorem, the expression equals (−1)n23n\frac{{(-1)^n}}{{2^{3n}}}.
539 Medium
539 of 979
Analyzing +4 | -1
If a complex number zz statisfies the equation z+2∣z+1∣+i=0z+\sqrt{2}|z+1|+i=0, then ∣z∣|z| is equal to :
View answer and explanation
Correct option: C
5\sqrt{5}
Option C (5\sqrt{5}) is correct because by substituting z=x+iyz = x + iy and equating real and imaginary parts, we can solve for xx and yy and then find ∣z∣|z|.
540 Easy
540 of 979
Application +4 | -1
If x+iy=3cos⁡θ+isin⁡θ+2x+i y=\frac{3}{\cos \theta+i \sin \theta+2}, then 4x−x2−y24 x-x^2-y^2 reduces to
View answer and explanation
Correct option: B
3
Option B (3) is correct because after simplifying the expression and substituting, the value reduces to 3.
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