Question

Show that a = −1 + √3i and b = 2 satisfy 1/a+b=1/a + 1/b

Answer 1

LHS =
`(1 - \sqrt3i)/(4)` = RHS =
`(1 - \sqrt3i)/(4)`

Explanation:

Data provided in the question:

a = −1 + √3i and b = 2

to prove:

`(1)/(a+b)=(1)/(a) + (1)/(b)`

Considering the LHS

⇒
`(1)/(a+b)`

substituting the value of a and b, we get

⇒
`(1)/(−1 + \sqrt3i+2)`

or

⇒
`(1)/(1 + \sqrt3i)`

on multiplying and dividing by conjugate ( 1 - √3i )

we get

`(1)/(1 + \sqrt3i)*(1 - \sqrt3i)/(1 - \sqrt3i)`

or

`(1 - \sqrt3i)/((1^2 - (\sqrt3i)^2)`

or

`(1 - \sqrt3i)/(1 + 3)` (as (√i)² = -1 )

or

`(1 - \sqrt3i)/(4)`

Now,

considering the RHS

`(1)/(a) + (1)/(b)`

substituting the value of a and b, we get

⇒
`(1)/(-1 + \sqrt3i) + (1)/(2)`

or

⇒
`(2*1 + ( -1 + \sqrt3i)*1)/((-1 + \sqrt3i)*2)`

or

⇒
`(2 + ( -1 + \sqrt3i))/((-1 + \sqrt3i)*2)`

or

⇒
`(1 + \sqrt3i)/((-1 + \sqrt3i)*2)`

now,

on multiplying and dividing by conjugate ( -1 - √3i )

we get

`(1 + \sqrt3i)/((−1 + \sqrt3i)*2)*(-1 - \sqrt3i)/(-1 - \sqrt3i)`

or

`(1 + \sqrt3i)/((−1 + \sqrt3i)*2)*(-1( 1 + \sqrt3i))/(-1 - \sqrt3i)`

or

`\frac{(1 + \sqrt3i}^2*(-1){((-1)^2 - (\sqrt3i)^2)*2}`

or

`((1^2 + (\sqrt3i)^2+2(1)(\sqrt3i)*(-1))/((1 + 3)*2)`

or

`((1 - 3 + 2\sqrt3i)*(-1))/((4)*2)`

or

`((-2 + 2\sqrt3i)*(-1))/((4)*2)`

or

`(-2( 1 - 2\sqrt3i)*(-1))/((4)*2)`

or

`(( 1 - 2\sqrt3i))/((4))`

Since, LHS = RHS

hence satisfied