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Question
If \[\frac{\left( 1 + i \right)^2}{2 - i} = x + iy\] find x + y.
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Solution
\[\frac{\left( 1 + i \right)^2}{2 - i} = \frac{1^2 + i^2 + 2i}{2 - i}\]
\[ = \frac{1 - 1 + 2i}{2 - i} [ \because i^2 = - 1]\]
\[ = \frac{2i}{2 - 1} \times \frac{2 + i}{2 + i} \]
\[ = \frac{2i(2 + i)}{2^2 - i^2}\]
\[ = \frac{4i + 2 i^2}{4 + 1} [ \because i^2 = - 1]\]
\[ = \frac{4i - 2}{5}\]
\[ = \frac{- 2}{5} + \frac{4}{5}i . . . . (1)\]
It is given that,
\[\frac{\left( 1 + i \right)^2}{2 - i} = x + iy\]
\[ \Rightarrow - \frac{2}{5} + \frac{4}{5}i = x + iy [\text { From }(1)]\]
\[ \Rightarrow x = - \frac{2}{5} \text { and } y = \frac{4}{5}\]
\[\therefore x + y = \frac{- 2}{5} + \frac{4}{5}\]
\[ = \frac{2}{5}\]
Thus, x + y = \[\frac{2}{5}\].
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