Question
If $$\left| z \right| = 1\,\,{\text{and }}\omega = \frac{{z - 1}}{{z + 1}}$$ $$\left( {{\text{where }}z \ne - 1} \right),{\text{then Re}}\left( \omega \right)\,\,{\text{is}}$$
A.
$$0$$
B.
$$ - \frac{1}{{{{\left| {z + 1} \right|}^2}}}$$
C.
$$\left| {\frac{z}{{z + 1}}} \right|.\frac{1}{{{{\left| {z + 1} \right|}^2}}}$$
D.
$$\frac{{\sqrt 2 }}{{{{\left| {z + 1} \right|}^2}}}$$
Answer :
$$0$$
Solution :
Given that $$\left| z \right| = 1\,\,{\text{and }}\omega = \frac{{z - 1}}{{z + 1}}\left( {z \ne - 1} \right)$$
Now we know that $$z\overline z = {\left| z \right|^2}$$
$$\eqalign{
& \Rightarrow \,\,z\overline z = 1\,\,\,\,\,\,\,\left( {{\text{for }}\left| z \right| = 1} \right) \cr
& \therefore \,\,\omega = \left( {\frac{{z - 1}}{{z + 1}}} \right) \times \frac{{\left( {\overline z + 1} \right)}}{{\left( {\overline z + 1} \right)}} \cr
& \,\,\,\,\,\,\,\,\,\, = \frac{{z\overline z + z - \overline z - 1}}{{z\overline z + z + \overline z + 1}} = \frac{{2iy}}{{2 + 2x}} \cr
& \left[ {\because \,\,z\overline z = 1\,\,{\text{and taking }}z = x + i\,y\,\,\,{\text{so that}}\,\,z + \overline z = 2x\,\,{\text{and }}z - \overline z = 2i\,y} \right] \cr
& \Rightarrow \,\,{\text{Re}}\left( \omega \right) = 0 \cr} $$