Question

If \[\frac{\left( 1 + i \right)^2}{2 - i} = x + iy\]  find x + y.

Answer 1

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

\[ = \frac{1 - 1 + 2i}{2 - i} [ \because i^2 = - 1]\]

\[ = \frac{2i}{2 - 1} \times \frac{2 + i}{2 + i} \]

\[ = \frac{2i(2 + i)}{2^2 - i^2}\]

\[ = \frac{4i + 2 i^2}{4 + 1} [ \because i^2 = - 1]\]

\[ = \frac{4i - 2}{5}\]

\[ = \frac{- 2}{5} + \frac{4}{5}i . . . . (1)\]

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 = \[\frac{2}{5}\].