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

Concept:

• n^th Root of Unity: The complex numbers satisfying αⁿ = 1 are called the n^th roots of unity.
• For n = 5, the fifth roots of unity are: 1, α, α², α³, α⁴, where α = e^2πi/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

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.

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.

Calculation:

Given,

The condition is \( \bigl|\frac{z - 2i}{z + i}\bigr| = 2,\quad z \neq -i\).

We need to find the locus of all complex numbers \(z\) satisfying this.

Step 1: Write in Cartesian form.

Let \(z = x + i\,y \) Then

\(z - 2i = x + i(y - 2),\quad z + i = x + i(y + 1).\)

Step 2: Translate the modulus equation.

\(\bigl|\tfrac{z - 2i}{z + i}\bigr| = 2 \)

\(⇒(\lvert z - 2i\rvert = 2\,\lvert z + i\rvert\)

Squaring both sides:

\(x^2 + (y-2)^2 = 4\bigl[x^2 + (y+1)^2\bigr].\)

Step 3: Expand and simplify.

Left: \(x^2 + y^2 - 4y + 4 \)
Right: \(4x^2 + 4y^2 + 8y + 4 \)

Bring all terms together and divide by 3:

\(x^2 + y^2 + 4y = 0.\)

Step 4: Complete the square.

\(x^2 + (y+2)^2 = 4.\)

This is a circle of radius \(2\) centered at \((0,-2) \)

Hence, the correct answer is Option 4. 

Concept:

Principal Argument and Cube Roots of Unity:

• Principal Argument (arg(z)): The principal argument of a complex number \( z \), denoted as \( \arg(z) \), is the angle formed by the complex number in the complex plane, typically in the range \( -\pi \leq \arg(z) \leq \pi \).
• Cube Root of Unity: The cube roots of unity are the solutions to the equation \( z^3 = 1 \), which are:
  • \( \omega = e^{2\pi i / 3} \), the primitive cube root of unity.
  • \( \omega^2 = e^{4\pi i / 3} \), the other cube root of unity.
• The cube roots of unity have the property that \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \).

Calculation:

Given,

• \( \alpha = \arg\left( \sum_{n=1}^{2025} (-\omega)^n \right) \)
• \( \omega \) is the cube root of unity, and the sum involves alternating powers of \( \omega \).

Let's express the sum:

\( \sum_{n=1}^{2025} (-\omega)^n = (-\omega) + (-\omega)^2 + (-\omega)^3 + \cdots \)

We notice that every pair of terms cancels out:

\( (-\omega)^1 + (-\omega)^2 = 0, (-\omega)^3 + (-\omega)^4 = 0, \dots \)

This simplifies the sum to a few remaining terms:

\( \beta = -\omega^{2023} - \omega^{2024} + \omega^{2025} \)

We now compute the exponents modulo 3, since \( \omega^3 = 1 \):

\( \omega^{2023} = \omega^2, \omega^{2024} = \omega, \omega^{2025} = 1 \)

The sum becomes:

\( \beta = -\omega^2 - \omega + 1 \)

The remaining terms give:

\( \beta = -1 - i\sqrt{3} \)

Next, we find \( \alpha \) using the argument of \( \beta \):

\( \alpha = \arg(\beta) = \arg(-1 - i\sqrt{3}) \)

The argument of \( \beta \) is \( -\frac{2\pi}{3} \).

Conclusion:

Hence, the value of \( 3\alpha / \pi \) is \( -2 \).

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 =  = 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) = \(\rm {-1\over13}-{18\over13}i\)