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serious [3.7K]
3 years ago
15

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

Mathematics
1 answer:
Zarrin [17]3 years ago
3 0

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

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