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Question
If \[\frac{z - 1}{z + 1}\] is purely imaginary number (\[z \neq - 1\]), find the value of \[\left| z \right|\].
Solution
Let \[z = x + iy\]
Then,
\[\frac{z - 1}{z + 1} = \frac{x + iy - 1}{x + iy + 1}\]
\[ = \frac{\left( x - 1 \right) + iy}{\left( x + 1 \right) + iy} \times \frac{\left( x + 1 \right) - iy}{\left( x + 1 \right) - iy}\]
\[ = \frac{x^2 + x - ixy - x - 1 + iy + ixy + iy - i^2 y^2}{\left( x + 1 \right)^2 - i^2 y^2}\]
\[ = \frac{x^2 + y^2 - 1 + 2iy}{x^2 + 1 + 2x + y^2} [ \because i^2 = - 1]\]
If \[\frac{z - 1}{z + 1}\] is purely imaginary number, then
\[\text { Re }\left( \frac{z - 1}{z + 1} \right) = 0\]
\[ \Rightarrow x^2 + y^2 - 1 = 0\]
\[ \Rightarrow x^2 + y^2 = 1\]
\[ \Rightarrow \left| z \right|^2 = 1\]
\[ \Rightarrow \left| z \right| = 1\]
Thus, the value of \[\left| z \right|\] is 1.
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