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Question
If \[\left( 1 + i \right)z = \left( 1 - i \right) \bar{z}\],then show that \[z = - i \bar{z}\].
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Solution
\[\left( 1 + i \right)z = \left( 1 - i \right) \bar{z}\]
\[ \Rightarrow \frac{z}{\bar{z}} = \frac{1 - i}{1 + i}\]
\[ \Rightarrow \frac{z}{\bar{z}} = \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i}\]
\[ \Rightarrow \frac{z}{\bar{z}} = \frac{1 + i^2 - 2i}{1 - i^2}\]
\[ \Rightarrow \frac{z}{\bar{z}} = \frac{1 - 1 - 2i}{1 + 1} [ \because i^2 = - 1]\]
\[ \Rightarrow \frac{z}{\bar{z}} = \frac{- 2i}{2}\]
\[ \Rightarrow \frac{z}{\bar{z}} = - i\]
\[ \Rightarrow z = - i \bar{z}\]
Hence,
\[z = - i \bar{z}\].
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