If $|z|=1$ and $w=\frac{z-1}{z+1}$ (where $z \neq-1$ ), then $\operatorname{Re}(\mathrm{w})$ is

If $|z|=1$ and $w=\frac{z-1}{z+1}$ (where $z \neq-1$ ), then $\operatorname{Re}(\mathrm{w})$ is
  1. 0
  2. $\quad-\frac{1}{|z+1|^2}$
  3. $\left|\frac{z}{z+1}\right| \cdot \frac{1}{|z+1|^2}$
  4. $\frac{\sqrt{2}}{|z+1|^2}$

Solution

$\begin{aligned} \mathrm{W} & =\frac{\mathrm{z}-1}{\mathrm{z}+1} \\ & =\frac{x+\mathrm{i} y-1}{x+\mathrm{i} y+1} \\ & =\frac{(x-1+\mathrm{i} y)}{(x+1+\mathrm{i} y)} \times \frac{(x+1-\mathrm{i} y)}{(x+1-\mathrm{i} y)}\end{aligned}$ $\begin{aligned} & =\frac{\left(x^2+y^2-1\right)+(-x y+1+x y+1) \mathrm{i}}{(x+1)^2+y^2} \\ & =\frac{x^2+y^2+1}{(x+1)^2+y^2}+\frac{2 y \mathrm{i}}{(x+1)^2+y^2} \end{aligned}$ Given that $|z|=1$ $\begin{aligned} & \Rightarrow x^2+y^2=1 \\ & \Rightarrow x^2+y^2-1=0 \end{aligned}$ $\therefore \quad \operatorname{Re}(\mathrm{w})=0$

Asked in: MHT CET 2024 (03 May Shift 1)

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