Complex Numbers MCQ Quiz - Objective Question with Answer for Complex Numbers - Download Free PDF

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

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

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

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.

Explanation:

z = x + iy

Minimum value of |z - 3| + |z - 4| is

|3 - 4| = 1

Option (2) is true.

Concept:

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.

Concept:

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) + …. + (i^2n-2 + i²ⁿ)

= (-1 + 1) + (-1 + 1) + …. (-1 + 1)

= 0 + 0 + …. + 0

= 0

Concept:

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

Concept:

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

Concept:

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| = \(\rm \sqrt{0^2+2^2} = 2\)

1Concept:

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

Concept:

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

Concept 

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  

Concept:

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, i^{4n + 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

Concept: 

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) = \(\rm {-1\over13}-{18\over13}i\)