Modulus of Complex Number MCQ Quiz - Objective Question with Answer for Modulus of Complex Number - Download Free PDF

Explanation:

= 

= 

= 

∴ Option (b) is correct.

Explanation:

⇒ Z ₁  = ω and Z ₂  = ω ²

Now

⇒ 

Option (a) is correct.

Explanation:

Given

|Z| =  √2 

∴ Option (b) is correct.

Concept:

If z = a + ib , |z| = 

If z = a + ib ,  =  

|z₁z₂| = |z₁| × |z₂|

Calculation:

Given z1 = 1 - 2i , z2 = 1 + i and z3 = 3 + 4i

∴  =  = 

Similarly   = 2 ×    = 2 ×  = 2 ×   = (1 - i)

 = (1 - i)

⇒ \(\frac{1}{z_{2}} \) = 

 = z₃ ×  \(\frac{1}{z_{2}} \)  = (3 + 4i) × 

⇒  =  

 We need to find 

  =  × 

=   × 

=  × 

=  × 

= 

= 

= 

=  

=  

= 

Evaluating , we get .

Concept

The general form of a complex number is z = x + iy. The polar representation of z

is z = r(cos θ + i sin θ). Here, r is the modulus of z and θ is called the amplitude or

argument of the complex number. The formula to find the amplitude of a complex

number is:  and ∣z∣ = 

Calculation

Given:

|z| = 4 and arg z = 

Let z = |z| (cos θ + i sin θ) where θ = arg(z)

⇒ z = 

⇒ z = 

⇒ z = - 2√3 + 2i

∴ z = - 2√3 + 2i

Concept:

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| = 

Calculations:

Let 

As we know i2 = -1 

As we know that if z = x + iy be any complex number, then its modulus is given by,|z| = 

∴ |z| = 

Concept;

Modulus of a complex number z = x + iy is given by:

|z| = 

(a + b)(a - b) = a² - b²

Calculation: 

If z =  so, modulus of |z| = 

z = 

|z| =  

Concept:

Let z = x + iy be a complex number, Where x is called the real part of the complex number or Re (z) and y is called the Imaginary part of the complex number or Im (z)

Modulus of z = |z| = 

Calculations:

Let 

z = x + iy = 0 + 2i

As we know that if z = x + iy be any complex number, then its modulus is given by, |z| = 

∴ |z| = 

Concept:

Iota power:

• i² = -1, i³ = -i, i⁴ = 1, i⁴ⁿ = 1
• Number of the form 2n is always even, n ∈ N
• Numebr of the form 2n +1 or 2n - 1 is always odd, n ∈ N
• (-a)^{2n -1} = -(a)^{2n -1}

Calculation:

Let,

Z = i2n + 1(-i)2n - 1

⇒ Z = i2n+1 × (i)2n-1 × (-1)^2n-1

⇒ Z =  i2n +1+2n-1 ×(-1)^2n-1

⇒ Z = i⁴n × (-1)          [∵ (-1)2n-1 = -1]

⇒ Z = (-1)⁴ⁿ ×

⇒ Z = i4n × (-1)

⇒ Z = -(i)4n × (-1)

⇒  Z = 1 × (-1)         [∵ i⁴ⁿ = 1] 

⇒ Z = -1

⇒ |Z| = |-1| = 1

Hence, the modulus of the complex number i2n + 1(-i)^{2n - 1} is 1.

Concept;

Modulus of a complex number z = x + iy is given by:

|z| = 

(a + b)(a - b) = a² - b²

Calculation:

Given: z = 

Mutiply by (3 + 4i) in denominator and numerator.

z =  =  =                   (∵ i² = -1) 

z = 

|z| = 

|z| = √2 

If z =  so, modulus of |z| = 

z = 

|z| =  

|z| = √2  

Concept:

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| = 

Calculations:

Let z = x + iy = (1 + i)² = 1² + 2i + i²         (∵ (a + b)² = a² + b² + 2ab)

z = 1 + 2i - 1                                (∵ i² = -1)

∴ z = x + iy = 2i

So, x = 0  and y = 2

As we know that if z = x + iy be any complex number, then its modulus is given by, |z| = 

∴ |z| =  

CONCEPT:

• i2 = - 1
• If z = x + iy then  

CALCULATION:

Given: z = (1 - i)4 First let's simplify the expression (1 - i)⁴ 

⇒ (1 - i)² = 1 + i² - 2i

As we know that, i² = - 1

⇒ (1 + i)² = -2i

Since (1 - i)4 = (1 - i)2 × (1 - i)2 we get:

⇒ (1 + i)⁴ = (-2i)² = - 4

⇒ z = - 4 + 0i

As we know that, if z = x + iy then  

Here, x = - 4 and y = 0

⇒  

As we know that, |z| denotes the distance between origin and z in the argand plane. So, |z| cannot be negative

⇒ |z| = 4

Hence, correct option is 2.

Concept:

The modulus of a complex number z = x + iy

|z| = 

The conjugate  = x - iy

Calculation:

z = 1+ 2i

= 1 - 2i

S = 

S = 

S =  

S = 3

Concept:

If z = a + ib, then |z| = 

Solution:

The given equation can be solved as:

iz3 + z2 - z + i = 0

⇒ iz3 + i²z + z2 + i = 0

⇒ iz(z² + i) + (z2 + i) = 0

⇒ (z² + i)(zi + 1) = 0

⇒ z² + i = 0 or zi + 1 = 0

⇒ z2 = -i or z = i

|z| = 1 in both the cases.

Concept: 

Modulus of complex no. z =  a + ib is given by |z| =  . 

Property of complex number:

Calculation:

Let z = 

Taking modulus on both sides, we get

⇒ |z| = 

= 

= 

= 1