Complex Numbers 2 Question 13
13. If $|z|=1$ and $w=\frac{z-1}{z+1}$ (where, $z \neq-1$ ), then $\operatorname{Re}(w)$ is
(a) 0
(b) $\frac{1}{|z+1|^{2}}$
(c) $\frac{1}{z+1} \cdot \frac{1}{|z+1|^{2}}$
(d) $\frac{\sqrt{2}}{|z+1|^{2}}$
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Answer:
Correct Answer: 13. (b)
Solution:
- Since, $|z|=1$ and $w=\frac{z-1}{z+1}$
$$ \begin{array}{lrrr} \Rightarrow & z-1=w z+w \Rightarrow z=\frac{1+w}{1-w} \Rightarrow & |z|=\frac{|1+w|}{|1-w|} \\ \Rightarrow & |1-w|=|1+w| & {[\because|z|=1]} \end{array} $$
On squaring both sides, we get
$$ \begin{aligned} & 1+|w|^{2}-2|w| \operatorname{Re}(w)=1+|w|^{2}+2|w| \operatorname{Re}(w) \\ & {\left[\text { using }\left|z _1 \pm z _2\right|^{2}=\left|z _1\right|^{2}+\left|z _2\right|^{2} \pm 2\left|z _1\right|\left|z _2\right| \operatorname{Re}\left(\bar{z} _1 z _2\right)\right]} \\ & \Rightarrow \quad 4|w| \operatorname{Re}|w|=0 \\ & \Rightarrow \quad \operatorname{Re}(w)=0 \end{aligned} $$
