PRACTICE QUESTIONS BASED ON PROPERTIES OF COMPLEX NUMBERS

Question 1 :

If z₁  =  1 - 3i, z₂  =  -4i and z₃  =  5 show that 

(i)  (z₁ + z₂) + z₃  =  z₁ + (z₂ + z₃)

Solution :

L.H.S :

z₁ + z₂  =  1 - 3i - 4i  =  1 - 7i

(z₁ + z₂) + z₃  =  (1 - 7i) + 5

=  (1 + 5) - 7i

=  6 - 7i  ---(1)

R.H.S :

z₂ + z₃  =  - 4i + 5

z₁ + (z₂ + z₃)  =  (1 - 3i) + (-4i + 5)

=  (1 + 5) + i(-3 - 4)

=  6 - 7i  ---(2)

(1)  = (2)

(ii)  (z₁ z₂) z₃  =  z₁ (z₂ z₃)

Solution :

L.H.S :

z₁ z₂  =  (1 - 3i)(-4i)

  =  -4i + 12i²

  =  -4i + 12(-1)

  =  -4i - 12

  =  -4i + 12(-1)

z₁ z₂  =  -4i - 12

(z₁ z₂) z₃  =  (-4i - 12) (5)  = -20i - 60  ----(1)

R.H.S :

z₂ z₃  =  -4i (5)  =  -20i

 z₁ (z₂ z₃)  =  (1 - 3i)(-20i)

  =  -20i + 60i²

  =  -20i + 60(-1)

  = -20i - 60  -----(2)

(1)  =  (2)

Hence proved.

Question 2 :

If z₁ = 3, z₂ = -7i and z₃  =  5 + 4i

(i)  z₁ (z₂ + z₃)  =  z₁ z₂ +  z₁ z₃

Solution :

L.H.S :

z₂ + z₃  =  -7i + (5 + 4i)

  =  -7i + 5 + 4i

z₂ + z₃   =  -3i + 5

z₁ (z₂ + z₃)  =  3 (-3i + 5)

z₁ (z₂ + z₃)  =  15 - 9i  -----(1)

R.H.S :

z₁ z₂  =  3(-7i)  =  -21i

z₁ z₃  =  3(5 + 4i)  =  15 + 12i

z₁ z₂ + z₁ z₃  = -21i + 15 + 12i

=  -9i + 15  

=  15 - 9i ------(2)

(1)  =  (2)

(ii)  (z₁ + z₂)z₃  =  z₁ z₃ +  z₂ z₃

Solution :

L.H.S :

z₁ + z₂  =  3 - 7i

(z₁ + z₂)z₃  =  (3 - 7i) (5 + 4i)

=  15 + 12i - 35i - 28i²

=  15 - 23i - 28(-1)

=  15 - 23i + 28

(z₁ + z₂)z₃  =  43 - 23i  ----(1)

z₁ z₃  =  3(5 + 4i)  =  15 + 12i

z₂ z₃  = (-7i) (5 + 4i)

=  -35i - 28i²

z₂ z₃  =  -35i - 28(-1)

 =  -35i + 28

z₁ z₃ + z₂ z₃   =  15 + 12i - 35i + 28

=  43 - 23i  ---(2)

(1)  =  (2)

Hence proved.

Question 3 :

If z₁ = 2 + 5i, z₂ = -3 - 4i and z₃ = 1 + i , find the additive and multiplicative inverse of z₁, z₂ and z₃.

Solution :

z₁ = 2 + 5i

Additive inverse of (z₁) :

z₁ = 2 + 5i

Additive inverse  = -2-5i

Multiplicative inverse of (z₁) :

z₁ = 2 + 5i

1/z₁ = 1/(2 + 5i)

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

=  (2 - 5i)/(4 - 25(-1))

=  (2 - 5i)/29

Hence multiplicative inverse of 2 + 5i is (2 - 5i)/29.

z₂ = -3 - 4i 

Additive inverse of (z₁) :

z₂ = -3 - 4i

Additive inverse  = 3+4i

Multiplicative inverse of (z₁) :

z₂ = -3-4i

1/z₂ = -1/(3 + 4i)

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

=  -(3 - 4i)/(9 - 16(-1))

=  -(3 - 4i)/(9 + 16)

=  -(3 - 4i)/25

Hence multiplicative inverse of -3-4i is -(3 - 4i)/25.

z₃ = 1 + i

Additive inverse of (z₃) :

z₃ = 1 + i

Additive inverse  = -1 - i

Multiplicative inverse of (z₃) :

z₃ = -1 - i

1/z₂ = -1/(1 + i)

= (-1/(1 + i))((1 - i)/(1 - i))

=  (-1 + i)/(1 - (-1))

=  (-1 + i)/2

Hence multiplicative inverse of  1 + i is (-1 + i)/2

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