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Class 11 NCERT Solutions- Chapter 5 Complex Numbers And Quadratic Equations – Exercise 5.1 | Set 1
  • Last Updated : 13 Jan, 2021

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

Question 1. (5i)(-3i/5)

Solution:

Let the given number be a,

a= (5i)*(-3i/5)

a= (-15i2)/5

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. (1/5+2i/5)-(4+5i/2)

Solution:

Let the given number be a,

a = (1/5+2i/5)-(4+5i/2)

a = (1/5-4)+(2i/5-5i/2)

a = (-19/5)+(2/5-5/2)i

a = -19/5+(-21/10)i

a = (-38-21i)/10 

Question 7. [(1/3+7i/3)+(4+i/3)]-(-4/3+i)

Solution:

Let the given number be a,

a = (1/3+7i/3)+(4+i/3)-(-4/3+i)

a = (1/3+4+4/3)+(7i/3+i/3-i)

a = (5/3+4)+(8i/3-i)

a = 17/3+ (5i/3)

a = (17+5i)/3

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. (1/3+3i)3

Solution:

Let the given number be a,

a = (1/3+3i)3                              

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

a = ((1/27)+(3i)3 +3(1/3)*(3i)(1/3+3i))

a = (1/27 +(-27i)+ 3i*(1/3+3i))

a = (1/27+(-27i)+i+9i2)

a = ((1/27)-9+(-27)i+i)

a = ((-242/27)-26i)

Question 10. (-2-(i/3))3

Solution:

Let the given number be a,

a = (-2-i/3)3       

a = -((2+i/3)3)                      

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

a = -((8)+(i/3)3 +3(2)*(i/3)(2+i/3))

a = -(8+(-i/27)+ 2i*(2+i/3))

a = -(8-i/27+4i+2i2/3) 

a = -(8-2/3+(-i/27)+4i)

a = -(22/3 +(107i/27))

a = -22/3-107i/27

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