Q. If $|z| = 1$ and $w = \frac{z - 1}{z + 1}$ (where $z \neq - 1$), then $Re(w)$ is
A) 0
B) $\frac{1}{|z + 1|^{2}}$
C) $\left| \frac{1}{z + 1} \right|.\frac{1}{|z + 1|^{2}}$
D) $\frac{\sqrt{2}}{|z + 1|^{2}}$
Explanation
(a)
Since, $|z| = 1$ and $w = \frac{z - 1}{z + 1}\ \ \ \ \ \Rightarrow \ \ \ \ \ \ z = \frac{1 + w}{1 - w}$
$\Rightarrow \ \ \ \ \ \ |z| = \frac{|1 + w|}{|1 - w|}\ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ |1 - w| = |1 + w|\ \ \ \ \ \ \lbrack\because|z| = 1\rbrack$
$\Rightarrow \ \ \ \ \ \ 1 + |w|^{2} - 2\ Re\ (w) = 1 + |w|^{2} + 2\ Re\ (w)$
$\Rightarrow \ \ \ \ \ Re\ (w) = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.