Question

If $\left|z\right|=1$ and $w=\frac{z-1}{z+1},z\ne -1$, then $\mathrm{Re}\left(w\right)$ equals

• A. $0$
• B. $\frac{1}{{\left|z+1\right|}^2}$
• C. $\frac{1}{{\left|z+1\right|}^3}$
• D. $\frac{\sqrt{2}}{{\left|z+1\right|}^2}$

Answer 1

The correct option is A

$0$

Explanation for the correct option.

Step 1. Solve the given equation $w=\frac{z-1}{z+1}$ for $z$.

It is given that $w=\frac{z-1}{z+1}$, now cross multiply and solve for $z$.

$$\begin{gathered} w=\frac{z-1}{z+1} \\ \Rightarrow w(z+1)=z-1 \\ \Rightarrow wz+w=z-1 \\ \Rightarrow wz-z=-1-w \\ \Rightarrow -z(1-w)=-(1+w) \\ \Rightarrow z=\frac{-(1+w)}{-(1-w)} \\ \Rightarrow z=\frac{1+w}{1-w} \end{gathered}$$

Step 2. Find the real part of $w$.

Now, take the modulus of both sides:

$$\begin{gathered} \left|z\right|=\left|\frac{1+w}{1-w}\right| \\ \Rightarrow 1=\frac{\left|1+w\right|}{\left|1-w\right|}\left[\left|z\right|=1\right] \\ \Rightarrow \left|1-w\right|=\left|1+w\right| \end{gathered}$$

Now using ${\left|z_1\pm z_2\right|}^2={\left|z_1\right|}^2+{\left|z_2\right|}^2\pm 2z_1{z}_2\mathrm{Re}\left(\overline{z_1}{z}_2\right)$ square both sides and simplify.

$$\begin{gathered} {\left|1-w\right|}^2={\left|1+w\right|}^2 \\ \Rightarrow 1+{\left|w\right|}^2-2\left|w\right|\mathrm{Re}\left(w\right)=1+{\left|\mathrm{w}\right|}^2+2\left|\mathrm{w}\right|\mathrm{Re}\left(\mathrm{w}\right) \\ \Rightarrow 4\left|\mathrm{w}\right|\mathrm{Re}\left(\mathrm{w}\right)=0 \\ \Rightarrow \mathrm{Re}\left(\mathrm{w}\right)=0 \end{gathered}$$

So, the real part of $w$ is $0$.

Hence, the correct option is A.