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 =
Last Updated :
05 Apr, 2021
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