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Question
The complex number z which satisfies the condition \[\left| \frac{i + z}{i - z} \right| = 1\] lies on
Options
circle x2 + y2 = 1
the x−axis
the y−axis
the line x + y = 1
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Solution
\[\left| \frac{i + z}{i - z} \right| = 1\]
\[ \Rightarrow \left| \frac{i + z}{i - z} \right|^2 = 1^2 \]
\[ \Rightarrow \left( \frac{i + z}{i - z} \right) \bar{\left( \frac{i + z}{i - z} \right)} = 1\]
\[ \Rightarrow \left( \frac{i + z}{i - z} \right)\left( \frac{- i + \bar{z}}{- i - \bar{z}} \right) = 1\]
\[ \Rightarrow \left( \frac{- i^2 - zi + \bar{z}i + z \bar{z}}{- i^2 + zi - \bar{z}i + z \bar{z}} \right) = 1\]
\[ \Rightarrow - i^2 - zi + \bar{z}i + z \bar{z} = - i^2 + zi - \bar{z}i + z \bar{z}\]
\[ \Rightarrow - zi + \bar{z}i = zi - \bar{z}i\]
\[ \Rightarrow \bar{z}i + \bar{z}i = zi + zi\]
\[ \Rightarrow 2 \bar{z}i = 2zi\]
\[ \Rightarrow \bar{z} = z\]
\[ \Rightarrow \text { z is purely real }\]
Hence, the correct option is (b).
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