Question
If $\frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}=\text{x}+\text{iy},$ find the value of $\text{x}^2+\text{y}^2.$
✓
Answer
$\frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}=\text{x}+\text{iy} \ ...(1)$ $\Rightarrow\Big[\overline{\frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}}\Big]=\overline{\text{x}+\text{iy}}$ $\Rightarrow\frac{(\text{a}^2+1)^2}{2\text{a}+\text{i}}=\text{x}-\text{iy} \ ...(2)$ On multiplying (1) and (2), we get $\frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}\times\frac{(\text{a}^2+1)^2}{2\text{a}+\text{i}}=(\text{x}+\text{iy})(\text{x}-\text{iy})$ $\Rightarrow\frac{(\text{a}^2+1)^4}{(2\text{a})^2-\text{i}^2}=\text{x}^2-\text{i}^2\text{y}^2$ $\Rightarrow\frac{(\text{a}^2+1)^4}{(2\text{a})^2+1}=\text{x}^2+\text{y}^2$ Hence, $\text{x}^2+\text{y}^2=\frac{(\text{a}^2+1)^4}{4\text{a}^2+1}.$
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