Question

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

Answer 1

Answer:

LHS = $\frac{1 - \sqrt3i}{4}$ = RHS = $\frac{1 - \sqrt3i}{4}$

Step-by-step explanation:

Data provided in the question:

a = −1 + √3i and b = 2

to prove:

$\frac{1}{a+b}=\frac{1}{a} + \frac{1}{b}$

Considering the LHS

⇒ $\frac{1}{a+b}$

substituting the value of a and b, we get

⇒ $\frac{1}{−1 + \sqrt3i+2}$

or

⇒ $\frac{1}{1 + \sqrt3i}$

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

we get

$\frac{1}{1 + \sqrt3i}\times\frac{1 - \sqrt3i}{1 - \sqrt3i}$

or

$\frac{1 - \sqrt3i}{(1^2 - (\sqrt3i)^2}$

or

$\frac{1 - \sqrt3i}{1 + 3}$              (as (√i)² = -1 )

or

$\frac{1 - \sqrt3i}{4}$

Now,

considering the RHS

$\frac{1}{a} + \frac{1}{b}$

substituting the value of a and b, we get

⇒ $\frac{1}{-1 + \sqrt3i} + \frac{1}{2}$

or

⇒ $\frac{2\times1 + ( -1 + \sqrt3i)\times1}{(-1 + \sqrt3i)\times2}$

or

⇒ $\frac{2 + ( -1 + \sqrt3i)}{(-1 + \sqrt3i)\times2}$

or

⇒ $\frac{1 + \sqrt3i}{(-1 + \sqrt3i)\times2}$

now,

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

we get

$\frac{1 + \sqrt3i}{(−1 + \sqrt3i)\times2}\times\frac{-1 - \sqrt3i}{-1 - \sqrt3i}$

or

$\frac{1 + \sqrt3i}{(−1 + \sqrt3i)\times2}\times\frac{-1( 1 + \sqrt3i)}{-1 - \sqrt3i}$

or

$\frac{(1 + \sqrt3i}^2\times(-1){((-1)^2 - (\sqrt3i)^2)\times2}$

or

$\frac{(1^2 + (\sqrt3i)^2+2(1)(\sqrt3i)\times(-1)}{(1 + 3)\times2}$

or

$\frac{(1 - 3 + 2\sqrt3i)\times(-1)}{(4)\times2}$

or

$\frac{(-2 + 2\sqrt3i)\times(-1)}{(4)\times2}$

or

$\frac{-2( 1 - 2\sqrt3i)\times(-1)}{(4)\times2}$

or

$\frac{( 1 - 2\sqrt3i)}{(4)}$

Since, LHS = RHS

hence satisfied