Show that

Given: $\frac{3+2 i}{2-3 i}+\frac{3-2 i}{2+3 i}$

Taking the L.C.M, we get

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

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

$\left[\because(a+b)(a-b)=\left(a^{2}-b^{2}\right)\right]$

$=\frac{6+9 i+4 i+6 i^{2}+6-9 i-4 i+6 i^{2}}{4-9 i^{2}}$

$=\frac{12+12 i^{2}}{4-9 i^{2}}$

Putting $i^{2}=-1$

$=\frac{12+12(-1)}{4-9(-1)}$

$=\frac{12-12}{4+9}$

$=0+0 \mathrm{i}$

Hence, the given equation is purely real as there is no imaginary part.

(ii) Given: $\frac{\sqrt{7}+i \sqrt{3}}{\sqrt{7}-i \sqrt{3}}+\frac{\sqrt{7}-i \sqrt{3}}{\sqrt{7}+i \sqrt{3}}$

Taking the L.C.M, we get

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

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

$\left[\because(a+b)(a-b)=\left(a^{2}-b^{2}\right)\right]$

Now, we know that,

$(a+b)^{2}+(a-b)^{2}=2\left(a^{2}+b^{2}\right)$

So, by applying the formula in eq. (i), we get

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

$=\frac{2\left[7+3 i^{2}\right]}{7-3(-1)}$

Putting $i^{2}=-1$

$=\frac{2[7+3(-1)]}{7+3}$

$=\frac{2[7-3]}{10}$

$=\frac{8}{10}+0 i$

$=\frac{4}{5}+0 i$

Hence, the given equation is purely real as there is no imaginary part.