Class 11 NCERT Solutions- Chapter 5 Complex Numbers And Quadratic Equations – Exercise 5.1 | Set 1

For Q.1 to Q.10 express each complex number in form of a+ib

Question 1. (5i)

Solution:

Let the given number be a,

a= (5i)*

a= 

a= (-3)*i²

a= (-3)*(-1)

a= 3+0i

Question 2. i⁹+i¹⁹

Solution:

Let the given number be a,

a = i⁹ * (1+i¹⁰)

a = ((i⁴)²*i )(1 + (i⁴)² (i²))

a = (1*i)(1+i²)

a = (i)*(0)

a = 0+0i

Question 3. i^-39

Solution:

Let the given number be a and let z = i³⁹ ,

z = (i)*(i²)¹⁹

z = (i)*(-1)¹⁹

z = -i

a = i^-39 

a = 1/i³⁹

a = 1/z

a = 1/-i

a = (i⁴)/-i

a = -i³ = -(i²*i)

a = -1*-i

a = 0+i

Question 4. 3(7+7i) + i(7+7i)

Solution:

Let the given number be a,

a = 3*(7+7i)+i*(7+7i)

a = 21+21i+7i+7i²

a = 21+7i²+28i

a = 21-7+28i

a = 14+28i

Question 5. (1-i)-(-1+i6)

Solution:

Let the given number be a,

a = (1-i)-(-1+6i)

a = 1-i+1-6i

a = 2-7i         

Question 6. ()-(4+)

Solution:

Let the given number be a,

a = 

a = 

a = 

a = ()+()

a =

Question 7. [()+(4+]-(+i)

Solution:

Let the given number be a,

a = (+)+(4+)-(+i)

a = (+4+)+(-i)

a = (+4)+(-i)

a = 

a = 

Question 8. (1-i)⁴

Solution:

Let the given number be a,

a = ((1-i)²)²                               

As we know , (a-b)²= (a²+b²-2ab)

a = (1+i²-2i)²

a = (1-1-2i)²

a = (-2i)²

a = 4i²

a = -4+0i

Question 9. (+3i)³

Solution:

Let the given number be a,

a = (+3i)³

As we know, (a+b)³= (a³+b³+3ab(a+b))     

a = (()+(3i)³ +3()*(3i)(+3i))

a = ( +(-27i)+ 3i*(+3i))

a = (+(-27i)+i+9i²)

a = (()-9+(-27)i+i)

a = (()-26i)

Question 10. (-2-())³

Solution:

Let the given number be a,

a = (-2-)³

a = -((2+)³)                      

As we know, (a+b)³= (a³+b³+3ab(a+b))  

a = -((8)+()³ +3(2)*()(2+))

a = -(8+()+ 2i*(2+))

a = -(8-+4i+) 

a = -(8-+()+4i)

a = -( +())

a = 

Chapter 5 Complex Numbers And Quadratic Equations – Exercise 5.1 | Set 2