HOW TO FIND THE INVERSE OF THE GIVEN COMPLEX NUMBER

Question 1 :

If z₁ = 2 − i and z₂ = −4 + 3i , find the inverse of z₁ z₂ and z₁/z₂

Solution : 

(i)

z₁ z₂  =  (2 - i)(-4 + 3i)

  =  -8 + 6i + 4i - 3i²

  =  -8 + 10i - 3(-1)

  =  -8 + 10i + 3

  =  -5 + 10i

z₁ z₂  =  -5(1 - 2i)

Inverse of z₁ z₂  =  1/z₁ z₂

   =  (-1/5)(1/(1 - 2i))

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

   =  (-1/5)(1 + 2i)/(1 - 4(-1))

   =  (-1/5)(1 + 2i)/(1 + 4)

   =  (-1/25)(1 + 2i)

  =  (-1 - 2i)/25

(ii)  Inverse of z₁/z₂  =  z₂/z₁

   =  [(-4 + 3i)/(2 - i)][(2 + i)/(2 + i)]

=  (-4 + 3i)(2 + i) / (2 - i)(2 +i)

=  (-8 - 4i + 6i + 3i²)/(4 -(-1))

=  (-8 + 2i - 3)/(4 + 1)

=  (-11 + 2i)/5

=  (1/5) (-11 + 2i)

Question 2 :

The complex numbers u,v , and w are related by (1/u)  =  (1/v) + (1/w)

if v  =  3 - 4i and w  =  4 + 3i, find u in rectangular form.

Solution :

Given that 

(1/u)  =  (1/v) + (1/w)

1/v  =  [1/(3 - 4i)][(3 + 4i)/(3 + 4i)]

  =  (3 + 4i)/(9 - 16(-1))

  =  (3 + 4i)/(9 + 16)

  =  (3 + 4i)/25

1/w  =  [1/(4 + 3i)][(4 - 3i)/(4 - 3i)]

  =  (4 - 3i)/(16 - 9(-1))

  =  (4 - 3i)/(16 + 9)

  =  (4 - 3i)/25

1/u  =  [(3 + 4i)/25] + [(4 - 3i)/25]

(1/u)  =  (7 + i)/25

By finding the inverse of 1/u, we get u

  =  [25/(7 + i)][(7 - i)/(7 - i)]

  =  25(7 - i)/(49 + 1)

  =  25(7 - i)/50

  =  (7 - i)/2

Hence the value of u in rectangular form is (1/2)(7 - i).

Question 4 :

Prove the following properties:

(i) z is real if and only if z = z bar

Solution :

Let us consider the complex number 4 + 3i

Here 4 is real part and 3 is the imaginary part.

z bar  =  4 - 3i

There is no changes in real parts of the above complex numbers.

Hence z is real if and only if z = z bar.

(ii) Re(z) = (z + z bar)/2

Solution :

z = 4 + 3i

z bar  =  4 - 3i

z + z bar  =  (4 + 3i)   + (4 - 3i)

  =  (4 + 4) + (3i - 3i)

  =  8

(z + z bar) / 2  =  8/2  =  4  ----(1)

Re(z)  =  4  ----(2)

(1)  =  (2)

Hence proved.

(iii)  im(z)  =  (z - z bar) / 2

z = 4 + 3i

z bar  =  4 - 3i

z - z bar  =  (4 + 3i)  - (4 - 3i)

  =  (4 - 4) + (3i + 3i)

=  6i

(z + z bar) / 2  =  6i/2  =  3  ----(1)

im(z)  =  3  ----(2)

(1)  =  (2)

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