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.

Answer 1

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.