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• NCERT Solutions for Class 11 Maths

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

• Last Updated : 05 Apr, 2021

### 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

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