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Question
Write the conjugate of \[\frac{2 - i}{\left( 1 - 2i \right)^2}\] .
Solution
\[\frac{2 - i}{\left( 1 - 2i \right)^2} = \frac{2 - i}{1 + 4 i^2 - 4i}\]
\[ = \frac{2 - i}{1 - 4 - 4i}\]
\[ = \frac{2 - i}{- 3 - 4i}\]
\[ = \frac{- 2 + i}{3 + 4i}\]
\[ = \frac{i - 2}{3 + 4i} \times \frac{3 - 4i}{3 - 4i}\]
\[ = \frac{3i - 4 i^2 - 6 + 8i}{3^2 - 4^2 i^2}\]
\[ = \frac{11i + 4 - 6}{9 + 16}\]
\[ = \frac{- 2}{25} + \frac{11}{25}i\]
∴ Conjugate of \[\frac{2 - i}{\left( 1 - 2i \right)^2} = \left( \bar{{- \frac{2}{25} + \frac{11}{25}i}} \right) = - \frac{2}{25} - \frac{11}{25}i\]
Hence, Conjugate of \[\frac{2 - i}{\left( 1 - 2i \right)^2}\] is \[- \frac{2}{25} - \frac{11}{25}i\].
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