Respuesta :
Answer:
LHS = [tex]\frac{1 - \sqrt3i}{4}[/tex] = RHS = [tex]\frac{1 - \sqrt3i}{4}[/tex]
Step-by-step explanation:
Data provided in the question:
a = −1 + √3i and b = 2
to prove:
[tex]\frac{1}{a+b}=\frac{1}{a} + \frac{1}{b}[/tex]
Considering the LHS
⇒ [tex]\frac{1}{a+b}[/tex]
substituting the value of a and b, we get
⇒ [tex]\frac{1}{−1 + \sqrt3i+2}[/tex]
or
⇒ [tex]\frac{1}{1 + \sqrt3i}[/tex]
on multiplying and dividing by conjugate ( 1 - √3i )
we get
[tex]\frac{1}{1 + \sqrt3i}\times\frac{1 - \sqrt3i}{1 - \sqrt3i}[/tex]
or
[tex]\frac{1 - \sqrt3i}{(1^2 - (\sqrt3i)^2}[/tex]
or
[tex]\frac{1 - \sqrt3i}{1 + 3}[/tex] (as (√i)² = -1 )
or
[tex]\frac{1 - \sqrt3i}{4}[/tex]
Now,
considering the RHS
[tex]\frac{1}{a} + \frac{1}{b}[/tex]
substituting the value of a and b, we get
⇒ [tex]\frac{1}{-1 + \sqrt3i} + \frac{1}{2}[/tex]
or
⇒ [tex]\frac{2\times1 + ( -1 + \sqrt3i)\times1}{(-1 + \sqrt3i)\times2}[/tex]
or
⇒ [tex]\frac{2 + ( -1 + \sqrt3i)}{(-1 + \sqrt3i)\times2}[/tex]
or
⇒ [tex]\frac{1 + \sqrt3i}{(-1 + \sqrt3i)\times2}[/tex]
now,
on multiplying and dividing by conjugate ( -1 - √3i )
we get
[tex]\frac{1 + \sqrt3i}{(−1 + \sqrt3i)\times2}\times\frac{-1 - \sqrt3i}{-1 - \sqrt3i}[/tex]
or
[tex]\frac{1 + \sqrt3i}{(−1 + \sqrt3i)\times2}\times\frac{-1( 1 + \sqrt3i)}{-1 - \sqrt3i}[/tex]
or
[tex]\frac{(1 + \sqrt3i}^2\times(-1){((-1)^2 - (\sqrt3i)^2)\times2}[/tex]
or
[tex]\frac{(1^2 + (\sqrt3i)^2+2(1)(\sqrt3i)\times(-1)}{(1 + 3)\times2}[/tex]
or
[tex]\frac{(1 - 3 + 2\sqrt3i)\times(-1)}{(4)\times2}[/tex]
or
[tex]\frac{(-2 + 2\sqrt3i)\times(-1)}{(4)\times2}[/tex]
or
[tex]\frac{-2( 1 - 2\sqrt3i)\times(-1)}{(4)\times2}[/tex]
or
[tex]\frac{( 1 - 2\sqrt3i)}{(4)}[/tex]
Since, LHS = RHS
hence satisfied
