Question

express (2+i) /(2-i) in the form a+ib​

Answer 1

To Express

$\dfrac{2 + i}{2 - i}$

in a simple complex number form

i.e.

$a + ib$

we have to rationalize it

As

$\dfrac{2 +i }{2 - i} \\ \\ =\dfrac{2 +i }{2 - i} \times \dfrac{2 +i }{2 + i} \\ \\ = \dfrac{(2 +i ) {}^{2} }{(2 - i)(2 + i)} \\ \\ = \frac{2 {}^{2} + {i}^{2} + 4i}{ {2}^{2} - {i}^{2} } \\ \\ = \frac{4 - 1 + 4i}{ 4 + 1} \: \: \: \: \: \: \: ( \because \: {i}^{2} = - 1) \\ \\ = \frac{3 + 4i}{5 } \\ \\ = \frac{3}{5} + \frac{4}{5} i$

Which is in the form

$a + ib$

Where

$a = \dfrac{3}{5}$

And

$b = \dfrac{4}{5}$

Answer 2

Rationalise the given form (denominator) to get the required form.

as shown below in the solution.

$\dfrac{2 +i }{2 - i} \\ \\ =\dfrac{2 +i }{2 - i} \times \dfrac{2 +i }{2 + i} \\ \\ = \dfrac{(2 +i ) {}^{2} }{(2 - i)(2 + i)} \\ \\ = \frac{2 {}^{2} + {i}^{2} + 4i}{ {2}^{2} - {i}^{2} } \\ \\ = \frac{4 - 1 + 4i}{ 4 + 1} \\ \\ = \frac{3 + 4i}{5 } \\ \\ = \frac{3}{5} + \frac{4}{5} i$

Now it is in the required form