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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

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

Question 1. (5i)(\frac{-3i}{5})

Solution:

Let the given number be a,

a= (5i)*(\frac{-3i}{5})

a= \frac{-15i^2}{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. (\frac{1}{5}+\frac{2}{5}i  )-(4+\frac{5i}{2}  )

Solution:

Let the given number be a,

a = (\frac{1}{5}+\frac{2i}{5})-(4+\frac{5i}{2})

a = (\frac{1}{5}-4)+(\frac{2i}{5}-\frac{5i}{2})



a = (\frac{-19}{5})+(\frac{2}{5}-\frac{5}{2})i

a = (\frac{-19}{5}  )+(\frac{-21i}{10}  )

a =\frac{-38-21i}{10}

Question 7. [(\frac{1}{3}+\frac{7i}{3}  )+(4+\frac{i}{3}  ]-(\frac{-4}{3}  +i)

Solution:

Let the given number be a,

a = (\frac{1}{3}  +\frac{7i}{3}  )+(4+\frac{i}{3}  )-(\frac{-4}{3}  +i)

a = (\frac{1}{3}  +4+\frac{4}{3}  )+(\frac{7i}{3}+\frac{i}{3}  -i)

a = (\frac{5}{3}  +4)+(\frac{8i}{3}  -i)

a = \frac{17}{3}+ \frac{5i}{3}

a = \frac{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. (\frac{1}{3}  +3i)3

Solution:

Let the given number be a,

a = (\frac{1}{3}  +3i)3                              

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

a = ((\frac{1}{27}  )+(3i)3 +3(\frac{1}{3}  )*(3i)(\frac{1}{3}  +3i))

a = (\frac{1}{27}   +(-27i)+ 3i*(\frac{1}{3}  +3i))

a = (\frac{1}{27}  +(-27i)+i+9i2)

a = ((\frac{1}{27}  )-9+(-27)i+i)

a = ((\frac{-242}{27}  )-26i)

Question 10. (-2-(\frac{i}{3}  ))3

Solution:

Let the given number be a,



a = (-2-\frac{i}{3}  )3       

a = -((2+\frac{i}{3}  )3)                      

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

a = -((8)+(\frac{i}{3}  )3 +3(2)*(\frac{i}{3}  )(2+\frac{i}{3}  ))

a = -(8+(\frac{-i}{27}  )+ 2i*(2+\frac{i}{3}  ))

a = -(8-\frac{i}{27}  +4i+\frac{2i^2}{3}

a = -(8-\frac{2}{3}  +(\frac{-i}{27}  )+4i)

a = -(\frac{22}{3}   +(\frac{107i}{27}  ))

a = \frac{-22}{3}-\frac{107i}{27}

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

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