NCERT Class 11 Mathematics Chapter 4 Complex Number and Quadratic Equation

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NCERT Class 11 Mathematics Chapter 4 Complex Number and Quadratic Equation

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Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 11 Mathematics Chapter 4 Complex Number and Quadratic Equation Textual Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 11 Mathematics Chapter 4 Complex Number and Quadratic Equation Solutions for All Subject, You can practice these here.

Complex Number and Quadratic Equation

Chapter – 4

Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.

Ans: 

= – 3i²

= -3(-1)     [i² = – 1]

= 3

Q2: i⁹ + i¹⁹.

Ans: i⁹ + i¹⁹ = i⁴ˣ²⁺¹ + i⁴ˣ⁴⁺³

= (i⁴)² .i + (i⁴)⁴ .i³

= 1 x i + 1x (- i)        [ i⁴ =1, i³ = -i]

= 0

Q3: i⁻³⁹.

Ans: 

Q4. 3(7 + i7) + i(7 + i7).

Ans: 3(7+i7)+i(7+i7)= 21 + 21i + 7i + 7i²

= 21 + 28i + 7x(- 1)                [∵i² = -1]

= 14 + 28i

Q5. (1 – i) -(-1+i6).

Ans : (1 – i)-(-1+i6)= 1 – i + 1 – 6i 

= 2 – 7i

Q6:

Ans: 

Q7

.

Ans: 

Q8. (1 – i)⁴.

Ans: (1 – i)⁴ = [(1 – i)²]²

= [I² + i² – 2i]²

= [1 – 1 – 2i]²

=(-2i)²

= (- 2i) × (- 2i)

= 4i² = – 4

[i² = – 1]

Q9:

Ans: 

Q10

: 

Ans: 

Find the multiplicative inverse of each of the complex numbers given in the Exercises 11 to 13.

Q11: 4 – 3i.

Ans: Let z= 4 – 3i 

Then, Z̅ = 4 + 3i and |z|² = 4² + (- 3)² = 16 + 9 = 25

Therefore, the multiplicative inverse of 4 – 3i is given by

Q12: √5 + 3i

Ans: Let z = √5 + 3i

Then, Z̅ = √5 – 3i and |z|² = (√5)² + 3² = 5 + 9 = 14

Therefore, the multiplicative inverse of √5 + 3i

is given by 

Q13: – i.

Ans: Let z = – i 

Then, Z̅ = i and |z|² = 1² = 1 

Therefore, the multiplicative inverse of – i is given by

Q14: Express the following expression in the form of a + ib.

Ans: 

Ans: 

= [ -1-i ]³

=(-1)³ [1+I]³

= – [1³ + i³ + 3.1.i(1 + i)]

= – [1 – i³ + 3i – 3i²]

= – [1 – i + 3i – 3]

= – [- 2 + 2i]

= 2 – 2i

Q2: For any two complex numbers Z₁and Z₂ prove that Re (z₁ z₂) = Re z₁Re z₂ – Im z₁ Im Z₂

Ans: Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂ 

∴ z₁ z₂ = (x₁ + iy₁)(x₂ + iy₂) 

= x₁ (x₂ + iy₂) + iy₁(x₂ + iy₂) 

= x₁ x₂ + ix₁ y₂ + iy₁x₂ + i² y₁y₂ 

= x₁x₂ + ix₁ y₂ + iy₁ x₂ – y₁y₂    [i²= – 1]

= (x₁x₂ – y₁y₂) + i(x₁y₂ + y₁x₂) 

⇒Re(z₁z₂) = x₁x₂ – y₁y₂

⇒Re(z₁z₂) = Re z₁Rez₂ – Im z₁ Im z₂

Hence, proved. 

Ans: 

Ans: 

Ans: 

Ans: 

7. Let z₁ = 2 – i, z₂= – 2 + i Find.

Ans:  z₁ z₂ = (2 – i)(- 2 + i) = – 4 + 2i + 2i – i² = – 4 + 4i – (- 1) = – 3 + 4i 

Z̅₁= 2 + i

On multiplying numerator and denominator by (2 – i), we obtain

Ans: 

8. Find the real numbers x and y if (x – iy)(3 + 5i) is the conjugate of – 6 – 24i.

Ans: Let z = (x – iy)(3 + 5i) 

z = 3x + 5xi – 3yi – 5yi² = 3x + 5xi – 3yi + 5y = (3x + 5y) + i(5x – 3y) 

∴ Z̅ = (3x + 5y) – i(5x – 3y) 

It is given that, Z̅ = – 6 – 24i 

∴ (3x + 5y) – i(5x – 3y) = – 6 – 24i 

Equating real and imaginary parts, we obtain 

3x + 5y = – 6                     ……(i)

5x – 3y = 24                     ……(ii) 

Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain 

9x + 15y = – 18 

Putting the value of x in equation (i), we obtain 

3(3) + 5y = – 6 

⇒ 5y = – 6 – 9 = – 15 

⇒ y = – 3 

Thus, the values of x and y are 3 and – 3 respectively.

Ans: 

Ans: (x + iy)³  = u + iv 

⇒x³ + (iy)³+ 3.x.iy(x + iy) = u + iv 

⇒x³ + i³ y³ + 3x² yi + 3xy² i² = u + iv 

⇒x³ – iy³ + 3x² yi – 3xy² = u + iv 

⇒ (x³ – 3xy²) + i(3x² y – y³) = u + iv 

On equating real and imaginary parts, we obtain

11. If a and ẞ are different complex numbers with |ẞ| = 1 , then find

Ans: For any complex number z, z z¹ = |z|²

It is given that |β| = 1 .

Then

|β|² = ββ’ = 1 

Now, (β-α)/(1 – α’β)| = |(β-α)/(ββ’ – α’β)| [because 1 = ββ’ ]

= Ιβ – α| / |β| β’ – α’

= |β – α| / 1 β’ – α’| [because |β| = 1 ]

= |β – α| / | (β-α)’ |

= 1 (because |z| = |z’|, for any complex number z)

Therefore, |(β-α)/(1 – α’β)| = 1

12. Find the number of non-zero integral solutions of the equation |1 – i|ˣ = 2ˣ.

Ans: |1 – i|ˣ = 2ˣ

⇒ (√1² + (- 1)²)ˣ = 2ˣ

⇒ (√2)ˣ = 2ˣ

⇒ 2ˣ/² = 2ˣ

⇒ x/2 = x

⇒ x = 2x

⇒2x – x = 0

⇒ x = 0

Thus, 0 is the only integral solution of the given equation. Therefore, the number of non-zero integral solutions of the given equation is 0.

13. If (a + ib) (c + id) (e + if) (g + i h) = A + iB then show that (a² + b²) (c²+ d²)(e² + f²) (g² + h²) = A² + B².

Ans: (a + ib)(c + id)(e + if)(g + i h) = A + iB

∴ |(a + ib)(c + id)(e + if)(g + i h)| = |A + iB|

 ⇒|(a + ib)|x|(c + id)|x|(e + if)|x|(g + i h)| = |A + iB|

[|z₁z₂| = |z₁||z₂|]

⇒ √a² + b² × √c² + d² × √e² + f² × √g² + h² = √A² + B²

On squaring both sides, we obtain 

(a² + b²)(c² + d²)(e² + f²)(g² + h²) = A² + B²

Hence, proved.

Ans: 

⇒iᵐ =1 

∴ m = 4k where k is some integer. 

Therefore, the least positive integer is 1. 

Thus, the least positive integral value of m is 4(= 4 x1 ).