Question

Prove that $[\frac{(3+2i)}{(2-5i)}+\frac{(3-2i)}{(2+5i)}]$is rational.

Answer 1

Determine the proving of the rational expression.

$$\begin{gathered} \frac{(3+2i)}{(2-5i)}+\frac{(3-2i)}{(2+5i)}=\frac{(3+2i)(2+5i)+(3-2i)(2-5i)}{(2-5i)(2+5i)}\mathbf{\because }\mathbf{}{\mathit{a}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}{\mathit{b}}^{\mathbf{2}}\mathbf{}=(a+b)(a-b) \\ \Rightarrow =\frac{6+15i+4i+10i^2+6-15i-4i+10i^2}{(2^2-{\left(5i\right)}^2)}\mathbf{\because }\mathbf{}{\mathit{i}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{1} \\ \mathbf{\Rightarrow }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\frac{6+15i+4i+10(-1)+6-15i-4i+10(-1)}{4-25(-1)} \\ \mathbf{\Rightarrow }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\frac{6+19i-10+6-19i-10}{4+25} \\ \mathbf{\Rightarrow }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\frac{12-20}{29} \\ \mathbf{\Rightarrow }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\frac{-8}{29}\mathbf{} \end{gathered}$$

Hence, the rational expression of the given value is $"\mathbf{-}\mathbf{}\frac{\mathbf{8}}{\mathbf{29}}"$.