Question

Show that (√7+i√3/√7-i√3 + √7-i√3/√7+i√3) is real​

Answer 1

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Consider,

$\rm \longmapsto\:\dfrac{ \sqrt{7} + i \sqrt{3} }{ \sqrt{7} - i \sqrt{3}} + \dfrac{ \sqrt{7} - i \sqrt{3} }{ \sqrt{7} + i \sqrt{3} }$

On taking LCM, we get

$\rm \: = \: \dfrac{ {( \sqrt{7} + i \sqrt{3})}^{2} + {( \sqrt{7} - i \sqrt{3})}^{2} }{( \sqrt{7} + i \sqrt{3})( \sqrt{7} - i \sqrt{3})}$

We know,

$\purple{\rm \longmapsto\:\boxed{\tt{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}}}$

and

$\purple{\rm \longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}}$

So, using these Identities, we get

$\rm \: = \: \dfrac{2\bigg[ {( \sqrt{7})}^{2} + {(i \sqrt{3}) }^{2} \bigg]}{ {( \sqrt{7}) }^{2} - {(i \sqrt{3})}^{2} }$

$\rm \: = \: \dfrac{2(7 + 3 {i}^{2})}{7 - {3i}^{2} }$

We know,

$\purple{\rm \longmapsto\:\boxed{\tt{ {i}^{2} = - 1}}}$

So, using this, we get

$\rm \: = \: \dfrac{2(7 - 3)}{7 + 3}$

$\rm \: = \: \dfrac{2 \times 4}{10}$

$\rm \: = \: \dfrac{4}{5}$

Hence,

$\red{\rm\implies \:\boxed{\sf{ \dfrac{ \sqrt{7} + i \sqrt{3} }{ \sqrt{7} - i \sqrt{3}} + \dfrac{ \sqrt{7} - i \sqrt{3} }{ \sqrt{7} + i \sqrt{3} } \: is \: purely \: real}}}$

Answer 2

Answer:

Step-by-step explanation: