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