Question

Show that $\Big(\frac{\sqrt{7}+i\sqrt{3}}{\sqrt{7}-i\sqrt{3}}+\frac{\sqrt{7}-i\sqrt{3}}{\sqrt{7}+i\sqrt{3}}\Big)$ is real.

Answer 1

Solution:

Complex numbers are solved by rationalising the denominator,but here there is no need,just take simple LCM and solve,as shown

$\Big(\frac{\sqrt{7}+i\sqrt{3}}{\sqrt{7}-i\sqrt{3}}+\frac{\sqrt{7}-i\sqrt{3}}{\sqrt{7}+i\sqrt{3}}\Big) \\ \\ = \Big(\frac{(\sqrt{7}+i\sqrt{3})( \sqrt{7} + i \sqrt{3}) + ( \sqrt{7} - i \sqrt{3})( \sqrt{7} - i \sqrt{3})}{(\sqrt{7}-i\sqrt{3})( \sqrt{7} + i \sqrt{3}) }\Big) \\ \\ = \frac{ {( \sqrt{7} + i \sqrt{3} )^{2} + ( \sqrt{7} - i \sqrt{3} )}^{2} }{ {( \sqrt{7}) }^{2} - ( {i \sqrt{3}) }^{2} } \\ \\ = \frac{7 - 3 + i \sqrt{21} + 7 - 3 - i \sqrt{21} }{7 + 3} \\ \\ = \frac{8}{10} \\ \\ \Big(\frac{\sqrt{7}+i\sqrt{3}}{\sqrt{7}-i\sqrt{3}}+\frac{\sqrt{7}-i\sqrt{3}}{\sqrt{7}+i\sqrt{3}}\Big)= \frac{4}{5}$

Since $i^{2} = -1$
keeping this in mind while solving identities.

Hence there is no i term left in the expression,thus given expression is real.

Hope it helps you.