Complex Numbers MCQ Quiz - Objective Question with Answer for Complex Numbers - Download Free PDF
Last updated on Jun 30, 2025
Latest Complex Numbers MCQ Objective Questions
Complex Numbers Question 1:
If α is the fifth root of unity, then
Answer (Detailed Solution Below)
|1 + α + α2 + α3| = 1
Complex Numbers Question 1 Detailed Solution
Concept:
- nth Root of Unity: The complex numbers satisfying αn = 1 are called the nth roots of unity.
- For n = 5, the fifth roots of unity are: 1, α, α2, α3, α4, where α = e2πi/5.
- The sum of all five roots of unity is always zero: 1 + α + α2 + α3 + α4 = 0
- We can use symmetry of these roots on the unit circle in the complex plane to evaluate modulus expressions involving them.
Calculation:
Given that α is a fifth root of unity,
⇒ α5 = 1
The 5 roots are: 1, α, α2, α3, α4
Sum of all roots: 1 + α + α2 + α3 + α4 = 0
⇒ 1 + α + α2 + α3 = −α4
Take modulus on both sides:
|1 + α + α2 + α3| = |−α4|
Since α4 lies on the unit circle, |α4| = 1
⇒ |1 + α + α2 + α3| = 1
∴ The correct answer is Option (2): |1 + α + α2 + α3| = 1
Complex Numbers Question 2:
If \(z\ne0\) is a complex number, then what is equal to?
Answer (Detailed Solution Below)
Complex Numbers Question 2 Detailed Solution
Concept:
1. The amplitude (or argument) of a complex number \( z = r(\cos\theta + i\sin\theta) \) is given by \( \text{amp}(z) = \theta \).
2. The conjugate of \( z \), denoted as \( \overline{z} \), has an amplitude \( \text{amp}(\overline{z}) = -\theta \), because conjugating a complex number reflects it across the real axis in the Argand plane.
Formula Used:
\( \text{amp}(z) + \text{amp}(\overline{z}) = \theta + (-\theta) = 0 \).
Calculation:
\( z = r(\cos\theta + i\sin\theta) \)
\( \overline{z} = r(\cos\theta - i\sin\theta) \)
\( \text{amp}(z) + \text{amp}(\overline{z}) = \theta + (-\theta) = 0 \)
Conclusion:
\( \therefore \text{amp}(z) + \text{amp}(\overline{z}) = 0 \).
Hence, the correct answer is Option 1.
Complex Numbers Question 3:
What is \( \left( \frac{\sqrt{3}+i}{\sqrt{3}-i} \right)^3 \) equal to?
Answer (Detailed Solution Below)
Complex Numbers Question 3 Detailed Solution
Concept:
The key concept involves simplifying complex numbers in the polar form. We use the argument (angle) of the complex number and De Moivre's theorem, which states:
(r(cosθ + i sinθ))n = rn(cos(nθ) + i sin(nθ))
Calculation:
⇒ \(\left( \frac{\sqrt{3} + i}{\sqrt{3} - i} \right)^3 \)
Multiply the numerator and the denominator by the conjugate of the denominator
⇒ \(\frac{\sqrt{3} + i}{\sqrt{3} - i} \times \frac{\sqrt{3} + i}{\sqrt{3} + i} \)
⇒ \(\frac{(\sqrt{3} + i)^2}{(\sqrt{3} - i)(\sqrt{3} + i)} \)
\(\frac{2 + 2\sqrt{3}i}{4} = \frac{1 + \sqrt{3}i}{2} \)
Convert to polar form. The modulus r is
\(r = \sqrt{\left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \)
The argument θ is
\(\theta = \tan^{-1}\left( \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} \right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \)
So the polar form of the complex number is
\(1 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) \)
To cube the complex number, we use De Moivre’s Theorem
In our case, r = 0 so,
\(\left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)^3 = \cos \pi + i \sin \pi = -1 + 0i = -1 \)
Thus, the result of cubing the complex number is \(\boxed{-1} \)
Hence, the correct answer is Option 1.
Complex Numbers Question 4:
Let z1 = 2 + 3i and z2 = 3 + 4i. The set
\(\rm S=\left\{z \in C:\left|z-z_{1}\right|^{2}-\left|z-z_{2}\right|^{2}=\left|z_{1}-z_{2}\right|^{2}\right\}\) represents a
Answer (Detailed Solution Below)
Complex Numbers Question 4 Detailed Solution
Concept:
Locus of Points Defined by Difference of Squares of Distances:
- In complex geometry, a point \( z \) such that \( |z - z_1|^2 - |z - z_2|^2 \) is constant represents a geometric locus.
- If the difference is constant, it can represent a straight line.
- The equation \( |z - z_1|^2 - |z - z_2|^2 = c \) simplifies to linear form in many cases.
- Here, it simplifies to a straight line equation.
Complex Number:
- Definition: A complex number is of the form \( z = x + iy \), where \( x \) is the real part and \( y \) is the imaginary part.
- SI Unit: Dimensionless
- Modulus: \( |z| = \sqrt{x^2 + y^2} \)
Distance in Complex Plane:
- Definition: Distance between two complex numbers \( z_1 \) and \( z_2 \) is \( |z_1 - z_2| \)
- Formula: \( |z_1 - z_2|^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 \)
Calculation:
Given,
\( z_1 = 2 + 3i,\quad z_2 = 3 + 4i \)
\( z \in \mathbb{C} \) such that \( |z - z_1|^2 - |z - z_2|^2 = |z_1 - z_2|^2 \)
Let \( z = x + iy \)
⇒ \( |z - z_1|^2 = (x - 2)^2 + (y - 3)^2 \)
⇒ \( |z - z_2|^2 = (x - 3)^2 + (y - 4)^2 \)
⇒ \( |z_1 - z_2|^2 = (-1)^2 + (-1)^2 = 2 \)
⇒ \( (x - 2)^2 + (y - 3)^2 - (x - 3)^2 - (y - 4)^2 = 2 \)
⇒ \( [x^2 - 4x + 4 + y^2 - 6y + 9] - [x^2 - 6x + 9 + y^2 - 8y + 16] = 2 \)
⇒ \( x^2 - 4x + 4 + y^2 - 6y + 9 - x^2 + 6x - 9 - y^2 + 8y - 16 = 2 \)
⇒ \( 2x + 2y - 12 = 2 \)
⇒ \( 2x + 2y = 14 \Rightarrow x + y = 7 \)
⇒ Equation of line: \( x + y = 7 \)
⇒ x-intercept = 7 (when y = 0), y-intercept = 7 (when x = 0)
∴ The locus represents a straight line with sum of intercepts = 14.
Complex Numbers Question 5:
Find the value of (1 - i)4, Where i = \(\sqrt {-1}\)
Answer (Detailed Solution Below)
Complex Numbers Question 5 Detailed Solution
Concept:
Power of i:
- i = \(\sqrt{-1}\)
- i2 = -1
- i3 = -i × i2 = -i
- i4 = (i2)2 = (-1)2 = 1
- i4n = 1
Calculation:
To Find: Value of (1 - i)4
(1 - i)4
= [(1 - i)2]2
= [12 + i2 - 2i]2 (∵ (a - b)2 = a2 + b2 - 2ab)
= [1 - 1 - 2i]2 (∵ i2 = -1)
= [-2i]2
= 4i2
= 4 × -1
= -4
Top Complex Numbers MCQ Objective Questions
Find the conjugate of (1 + i) 3
Answer (Detailed Solution Below)
Complex Numbers Question 6 Detailed Solution
Download Solution PDFConcept:
Let z = x + iy be a complex number.
- Modulus of z = \(\left| {\rm{z}} \right| = {\rm{}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = {\rm{}}\sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{Im\;}}{{\left( {\rm{z}} \right)}^2}}\)
- arg (z) = arg (x + iy) = \({\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)
- Conjugate of z = = x – iy
Calculation:
Let z = (1 + i) 3
Using (a + b) 3 = a3 + b3 + 3a2b + 3ab2
⇒ z = 13 + i3 + 3 × 12 × i + 3 × 1 × i2
= 1 – i + 3i – 3
= -2 + 2i
So, conjugate of (1 + i) 3 is -2 – 2i
NOTE:
The conjugate of a complex number is the other complex number having the same real part and opposite sign of the imaginary part.
What is the value of (i2 + i4 + i6 +... + i2n), Where n is even number.
Answer (Detailed Solution Below)
Complex Numbers Question 7 Detailed Solution
Download Solution PDFConcept:
i2 = -1
i3 = - i
i4 = 1
i4n = 1
Calculation:
We have to find the value of (i2 + i4 + i6 +... + i2n)
(i2 + i4 + i6 +... + i2n) = (i2 + i4) + (i6 + i8) + …. + (i2n-2 + i2n)
= (-1 + 1) + (-1 + 1) + …. (-1 + 1)
= 0 + 0 + …. + 0
= 0
If (1 + i) (x + iy) = 2 + 4i then "5x" is
Answer (Detailed Solution Below)
Complex Numbers Question 8 Detailed Solution
Download Solution PDFConcept:
Equality of complex numbers.
Two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 are equal if and only if x1 = x2 and y1 = y2
Or Re (z1) = Re (z2) and Im (z1) = Im (z2).
Calculation:
Given: (1 + i) (x + iy) = 2 + 4i
⇒ x + iy + ix + i2y = 2 + 4i
⇒ (x – y) + i(x + y) = 2 + 4i
Equating real and imaginary part,
x - y = 2 …. (1)
x + y = 4 …. (2)
Adding equation 1 and 2, we get
x = 3
Now,
5x = 5 × 3 = 15
The value of ω6 + ω7 + ω5 is
Answer (Detailed Solution Below)
Complex Numbers Question 9 Detailed Solution
Download Solution PDFConcept:
Cube Roots of unity are 1, ω and ω2
Here, ω = \(\frac{{ - {\rm{\;}}1{\rm{\;}} + {\rm{\;i}}\sqrt 3 }}{2}\) and ω2 = \(\frac{{ - {\rm{\;}}1{\rm{\;}} - {\rm{\;i}}\sqrt 3 }}{2}\)
Property of cube roots of unity
- ω3 = 1
- 1 + ω + ω2 = 0
- ω3n = 1
Calculation:
ω6 + ω7 + ω5
= ω5 (ω + ω2 + 1)
= ω5 × (1 + ω + ω2)
= ω5 × 0
= 0
What is the modulus of \(\rm \dfrac{4+2i}{1-2i}\) where \(\rm i=\sqrt{-1} ?\)
Answer (Detailed Solution Below)
Complex Numbers Question 10 Detailed Solution
Download Solution PDFConcept:
Let z = x + iy be a complex number, Where x is called real part of the complex number or Re (z) and y is called Imaginary part of the complex number or Im (z)Modulus of z = |z| = \(\rm \sqrt {x^2+y^2} = \sqrt {Re (z)^2+Im (z)^2}\)
Calculations:
Let \(\rm z= x + iy = \dfrac{4+2i}{1-2i}\)
\(\rm = \dfrac{4+2i}{1-2i}\times\dfrac{1+2i}{1+2i}\)
\(\rm= \dfrac{4+10i+4i^2}{1-4i^2}\)
As we know i2 = -1
\(\rm = \dfrac{4+10i-4}{1+4}\)
\(\rm x + iy =\dfrac{10i}{5} = 0 + 2i\)
As we know that if z = x + iy be any complex number, then its modulus is given by,|z| = \(\rm \sqrt{x^2+y^2}\)
∴ |z| = \(\rm \sqrt{0^2+2^2} = 2\)
Find the conjugate of (i - i2)3
Answer (Detailed Solution Below)
Complex Numbers Question 11 Detailed Solution
Download Solution PDF1Concept:
Let z = x + iy be a complex number.
- Modulus of z = \(\left| {\rm{z}} \right| = {\rm{}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = {\rm{}}\sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{Im\;}}{{\left( {\rm{z}} \right)}^2}}\)
- arg (z) = arg (x + iy) = \({\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)
- For calculating the conjugate, replace i with -i.
- Conjugate of z = x – iy
Calculation:
Let z = (i - i2)3
⇒ z = i3 (1 - i) 3 = - i (1 - i)3
For calculating the conjugate, replace i with -i.
⇒ z̅ = -(- i) (1 - (- i))3
⇒ z̅ = i(1 + i)3
Using (a + b) 3 = a3 + b3 + 3a2b + 3ab2
⇒ z̅ = i(1 + i3 +3 ×12 × i + 3 × i2 × 1 )
⇒ z̅ = i(1 - i + 3i - 3)
⇒ z̅ = i(-2 + 2i)
⇒ z̅ = -2i + 2i2
⇒ z̅ = -2 - 2 i
So, the conjugate of (i - i2)3 is -2 - 2i
The value of ω3n + ω3n+1 + ω3n+2, where ω is cube roots of unity, is
Answer (Detailed Solution Below)
Complex Numbers Question 12 Detailed Solution
Download Solution PDFConcept:
Cube Roots of unity are 1, ω and ω2
Here, ω = \(\frac{{ - {\rm{\;}}1{\rm{\;}} + {\rm{\;\;i}}\sqrt 3 }}{2}\) and ω2 = \(\frac{{ - {\rm{\;}}1{\rm{\;}} - {\rm{\;\;i}}\sqrt 3 }}{2}\)
Property of cube roots of unity
- ω3 = 1
- 1 + ω + ω2 = 0
- ω = 1 / ω 2 and ω2 = 1 / ω
- ω3n = 1
Calculation:
We have to find the value of ω3n + ω3n+1 + ω3n+2
⇒ ω3n + ω3n+1 + ω3n+2
= ω3n (1 + ω + ω2) (∵ 1 + ω + ω2 = 0)
= 1 × 0 = 0
If 1, ω, ω2 are the cube roots of unity then the roots of the equation (x - 1)3 + 8 = 0 are
Answer (Detailed Solution Below)
Complex Numbers Question 13 Detailed Solution
Download Solution PDFConcept
Cube Roots of unity are 1, ω and ω2
Here, ω = \(\frac{{ - \;1\; + \;i\sqrt 3 }}{2}\) and ω2 = \(\frac{{ - \;1\; - \;i\sqrt 3 }}{2}\)
Property of cube roots of unity:
- ω3 = 1
- 1 + ω + ω2 = 0
- ω = 1 / ω 2 and ω2 = 1 / ω
- ω3n = 1
Calculation:
Given that,
(x - 1)3 + 8 = 0
⇒ (x - 1)3 = (-2)3
⇒ (x - 1) = -2(1)1/3
(x - 1) = -2(1, ω, ω2)
⇒ x = -1, 1 - 2ω, 1 - 2ω2
The smallest positive integer for which \(\rm \left(\frac{1-i}{1+i}\right)^{n}=-1\), where i = √-1, is:
Answer (Detailed Solution Below)
Complex Numbers Question 14 Detailed Solution
Download Solution PDFConcept:
Complex Numbers:
- A complex number is a number of the form a + ib, where a and b are real numbers and i is the complex unit defined by i = √-1.
- i2 = -1, i3 = -i, i4 = 1 etc.
- In general, i4n + 1 = i, i4n + 2 = -1, i4n + 3 = -i, i4n = 1.
-
For a complex number z = a + ib, conjugate of z is z̅ = a - ib.
Calculation:
Rationalizing the complex number \(\rm \frac{1-i}{1+i}\), by multiplying and dividing by the conjugate of the denominator, we get:
\(\rm \frac{1-i}{1+i}=\frac{(1-i)(1-i)}{(1+i)(1-i)}=\frac{1^2-2i+i^2}{1^2-i^2}=\frac{-2i}{1+1}\) = -i.
Now, \(\rm \left(\frac{1-i}{1+i}\right)^{n}=-1\).
⇒ \(\rm (-i)^{n}=-1\)
⇒ (-i)n for n = 2.
(-i)2 = (-1)2 × (i)2 = 1 × -1 = -1.
∴ n = 2
The conjugate of the complex number \(\rm 3i+4\over2-3i\) is:
Answer (Detailed Solution Below)
Complex Numbers Question 15 Detailed Solution
Download Solution PDFConcept:
Let z = x + iy be a complex number.
- Modulus of z = \(\left| {\rm{z}} \right| = {\rm{}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = {\rm{}}\sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{Im\;}}{{\left( {\rm{z}} \right)}^2}}\)
- arg (z) = arg (x + iy) = \({\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)
- Conjugate of z = z̅ = x – iy
Calculation:
Given complex number is z = \(\rm 3i+4\over2-3i\)
z = \(\rm {3i+4\over2-3i}\times{2+3i\over2+3i}\)
z = \(\rm 6i+8-9+12i\over2^2-(3i)^2\)
z = \(\rm 18i-1\over13\)
z = \(\rm {-1\over13}+{18\over13}i\)
Conjugate of z = (z̅) = \(\rm {-1\over13}-{18\over13}i\)