NCERT Solutions for Chapter 4, Miscellaneous Exercise, Class 11, Maths

Miscellaneous Exercise
Complex Numbers and Quadratic Numbers
Question and Answers
Class 11 – Maths

Class	Class 11
Subject	Mathematics
Chapter Name	Complex Numbers and Quadratic Numbers
Chapter No.	Chapter 4
Exercise	Miscellaneous Exercise
Category	Class 11 Maths NCERT Solutions

Question 1 Evaluate 

Answer 

Question 2  For any two complex numbers z ₁ and z ₂, prove that

Re (z₁z₂) = Re z₁ Re z₂ – Im z₁ Im z₂

Answer

Let Let z₁=x₁+iy₁  and   z₂=x₂+iy₂
Product of these two complex numbers, z₁ z₂
z₁ z₂ = ( x₁+iy₁ ) (x₂+iy₂)
z₁ z₂ = x₁ (x₂+iy₂) + iy₁ (x₂+iy₂)
z₁ z₂ = x₁x₂+ix₁y₂+iy₁x₂+i²y₁y₂
z₁ z₂ = x₁x₂+ix₁y₂+iy₁x²−y₁y₂
z₁ z₂ = (x₁x₂−y₁y₂)+i(x₁y²+y₁x²)
Re(z₁z₂)=x₁x₂−y₁y₂
Re (z₁z₂) = Re z₁ Re z₂ – Im z₁ Im z₂

Question 3 Reduce to the standard form

Answer

Question 4

Answer

Question 5 If z ₁ = 2 – i, z ₂ = 1 + i, find

Answer

Question 6

Answer

Question 7 Let z₁ = 2 – i,  z₂ = –2 + i. Find

(i) 

z₁ = 2 – i,  z₂ = –2 + i
z₁ z₂ = (2-i) (-2 +1)
z₁ z₂= −4+2i+2i−i²
z₁ z₂ = −4+4i−(−1)
z₁ z₂= -3 + 4i

(ii) 

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

Answer

z = (x – iy) (3 + 5i)
z = 3x + 5 xi− 3 yi− 5yi²
z = 3x + 5xi − 3yi + 5y
z = (3x+5y) + i (5x−3y)
z̅ = − 6 − 24i
(3x+5y) + i (5x−3y) = − 6 − 24i
On equating real and imaginary parts, we have
3x + 5y = -6 …… (i)
5x – 3y = 24 …… (ii)
Performing (i) x 3 + (ii) x 5, we get
(9x + 15y) + (25x – 15y) = -18 + 120
34x = 102
x = 102/34 = 3
Putting the value of x in equation (i), we get
3(3) + 5y = -6
5y = -6 – 9 = -15
y = -3
Therefore, the values of x and y are 3 and –3 respectively.

Question 9 Find the modulus of

Answer

Question 10 If (x + iy)³ = u + iv, then show that 

Answer

(x + iy)³ = u +iv
x³ + (iy)³+ 3 ×  x × iy (x+iy) = u+iv
x³ + i³y³ + 3x²yi + 3xy²i² = u+iv
x³−iy³+3x²yi−3xy²=u+iv
(x³−3xy²)+i(3x²y−y³)=u+iv
On equating real and imaginary
u=x³−3xy² ,v=3x²y−y³

= x²−3y²+3x²−y²
=4x² -4y²
=4 (x²– y²)

Question 11 If α and β are different complex numbers with |β| = 1, then find

Answer

Let α = a + ib and β = x +iy

|β|= 1 

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

Answer
|1 – i|^x = 2^x

2^x/2 = 2^x
 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.

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

Answer

(a + ib)(c + id)(e + if)(g + ih) = A + iB
∴ |(a + ib)(c + id)(e + if)(g + ih)| = |A + iB| 
⇒ |(a + ib)|× |(c + id)|× |(e + if)| × |(g + ih)| = |A + B|             [|z₁ z₂| z₁||z₂|] 

On squaring both sides, we obtain 
(a² + b² )(c² + d² )(e² + f² )(g² + h² ) = A² + B² . 
Hence, proved. 

Question 14 If [(1 +i)/(1 – i)]^m = 1, then find the least positive integral value of m. 

Answer 

i^m =1 

i^m= i^4k

m=4k where k is some integer. Therefore, the least positive is one. Thus, the least positive integral value of  m  is  4=( 4× 1)