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# 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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