Complex Numbers MCQ Quiz - Objective Question with Answer for Complex Numbers - Download Free PDF
Last updated on Apr 22, 2025
Latest Complex Numbers MCQ Objective Questions
Complex Numbers Question 1:
Find the value of (1 - i)4, Where i = \(\sqrt {-1}\)
Answer (Detailed Solution Below)
Complex Numbers Question 1 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
Complex Numbers Question 2:
The value of i4n + 1, where \({\rm{i}} = \sqrt { - 1} \), is
Answer (Detailed Solution Below)
Complex Numbers Question 2 Detailed Solution
Concept:
Power of i:
- i = \(\sqrt{-1}\)
- i2 = -1
- i3 = -i × i2 = -i
- i4 = (i2)2 = (-1)2 = 1
- i4n = 1
Calculation:
Given that,
i4n + 1, where \({\rm{i}} = \sqrt { - 1} \)
= i4n × i
= 1 × i
= i
Complex Numbers Question 3:
Find the value of \((\rm \frac{1+i}{1-i})^{20}\), where \(\rm i = \sqrt{-1}\), is
Answer (Detailed Solution Below)
Complex Numbers Question 3 Detailed Solution
Concept:
a2 - b2 = (a - b)(a + b)
i2 = -1,i4 = 1
Calculation:
To find value of \((\rm \frac{1+i}{1-i})^{20}\)
\(\rm \frac{1+i}{1-i}\\=\frac{1+i}{1-i}\times \frac{1+i}{1+i}\)
\(\rm =\frac{(1+i)^2}{1^2-i^2}\) (∵ a2 - b2 = (a - b)(a + b))
\(\rm =\frac{1^2+i^2+2i}{1-(-1)}\\=\frac{1-1+2i}{2}\\=\frac{2i}{2}\\=i\)
\((\rm \frac{1+i}{1-i})^{20}\) = i20 = (i4)5 = 15 = 1
Complex Numbers Question 4:
If z1 = 9 + 5i and z2 = 3 + 5i , and \(\rm arg\left(\frac{z-z_1}{z-z_2}\right)\) = π/4, then the values of |z - 6 - 8i| is:
Answer (Detailed Solution Below)
Complex Numbers Question 4 Detailed Solution
Concept:
Let z = x + iy is a complex number.
Then, arg(z) = \(\tan^{-1}\frac{y}{x}\)
- arg(z1z2) = arg(z1) + arg(z2)
- \(\rm arg\left(\frac{z_1}{z_2}\right)\) = arg(z1) - arg(z2)
Calculation:
Let z = x + iy
∴ \(\rm arg\left(\frac{z-z_1}{z-z_2}\right)\) = π/4
⇒ arg(z - z1) - arg(z - z2) = π/4
⇒ \(\tan^{-1}\left(\frac{y-5}{x-9}\right)-\tan^{-1}\left(\frac{y-5}{x-3}\right)\) = π/4
⇒ \(\displaystyle\frac{\frac{y-5}{x-9}-\frac{y-5}{x-3}}{1+\frac{(y-5)^2}{(x-9)(x-3)}}\) = 1
⇒ 6(y - 5) = \((x-9)(x-3)+(y-5)^2\)
⇒ 6(y - 5) = (x - 9)(x - 3) + (y - 5)2
⇒ (x - 9)(x - 3) + (y - 5)2 = 6(y - 5)
⇒ x2 -12x + 27 + y2 - 10y + 25 = 6y - 30
⇒ x2 + y2 - 12x - 16y + 82 = 0
∴ |z - 6 - 8i|2 = (x - 6)2 + (y - 8)2
= x2 - 12x + 36 + y2 -16y + 64
= x2 + y2 - 12x - 16y + 100
= (x2 + y2 - 12x - 16y + 82) + 18
= 18
⇒ |z - 6 - 8i| = 3√2
∴ The value of |z - 6 - 8i| is 3√2.
The correct answer is Option 4.
Complex Numbers Question 5:
If z = x + iy, then minimum value of |z - 3| + |z - 4| is
Answer (Detailed Solution Below)
Complex Numbers Question 5 Detailed Solution
Explanation:
z = x + iy
Minimum value of |z - 3| + |z - 4| is
|3 - 4| = 1
Option (2) is true.
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.
If (1 + i) (x + iy) = 2 + 4i then "5x" is
Answer (Detailed Solution Below)
Complex Numbers Question 7 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
What is the value of (i2 + i4 + i6 +... + i2n), Where n is even number.
Answer (Detailed Solution Below)
Complex Numbers Question 8 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
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) = \(\rm {-1\over13}-{18\over13}i\)