Question

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

Answer 1

Consider $\frac{\sqrt{7}+i\sqrt{3}}{\sqrt{7}-i\sqrt{3}}+\frac{\sqrt{7}-i\sqrt{3}}{\sqrt{7}+i\sqrt{3}}$

Taking LCM

$\frac{\sqrt{7}+i\sqrt{3}}{\sqrt{7}-i\sqrt{3}}+\frac{\sqrt{7}-i\sqrt{3}}{\sqrt{7}+i\sqrt{3}}=\frac{(\sqrt{7}+i\sqrt{3}{)}^2+(\sqrt{7}-i\sqrt{3}{)}^2}{(\sqrt{7}-i\sqrt{3})(\sqrt{7}+i\sqrt{3})}$

$=\frac{7-3+2i\sqrt{7}\sqrt{3}+7-3-2i\sqrt{7}\sqrt{3}}{(\sqrt{7}{)}^2-(i\sqrt{3}{)}^2}$

$=\frac{14-6}{7-(-3)}$

$=\frac{8}{7+3}$

$=\frac{8}{10}$

$=\frac{4}{5}$

$\frac{\sqrt{7}+i\sqrt{3}}{\sqrt{7}-i\sqrt{3}}+\frac{\sqrt{7}-i\sqrt{3}}{\sqrt{7}+i\sqrt{3}}=0.8$ which is a real number.

Hence $\frac{\sqrt{7}+i\sqrt{3}}{\sqrt{7}-i\sqrt{3}}+\frac{\sqrt{7}-i\sqrt{3}}{\sqrt{7}+i\sqrt{3}}$ is real.