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

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

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

$=\frac{7-3+2i\sqrt{21}+7-3-2i\sqrt{21}}{7+3}$

=$\frac{8}{10}=\frac{4}{5}$

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.