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Question
If z1 is a complex number other than −1 such that \[\left| z_1 \right| = 1\] and \[z_2 = \frac{z_1 - 1}{z_1 + 1}\] then show that the real parts of z2 is zero.
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Solution
Let \[z = x + iy\]
Then,
\[z_2 = \frac{z_1 - 1}{z_1 + 1}\]
\[ = \frac{x + iy - 1}{x + iy + 1}\]
\[ = \frac{\left( x - 1 \right) + iy}{\left( x + 1 \right) + iy} \times \frac{\left( x + 1 \right) - iy}{\left( x + 1 \right) - iy}\]
\[ = \frac{x^2 + x - ixy - x - 1 + iy + ixy + iy - i^2 y^2}{\left( x + 1 \right)^2 - i^2 y^2}\]
\[ = \frac{x^2 + y^2 - 1 + 2iy}{x^2 + 1 + 2x + y^2} [ \because i^2 = - 1]\]
Now,
\[Re\left( z_2 \right) = \frac{x^2 + y^2 - 1}{x^2 + y^2 + 1 + 2x}\]
\[ = 0 [ \because \left| z_1 \right| = 1 \Rightarrow x^2 + y^2 = 1]\]
Thus, the real parts of z2 is zero.
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