Let $z_1(\ne -1)$ and $z_2(\ne -1)$ be two complex numbers such that $\left|z_2\right|=1$ and $\frac{z_1-1}{z_1+1}={\left(\frac{z_2-1}{z_2+1}\right)}^2$, then the value of $|z_1-1|+|z_1+1|$ is equal to 

$$\begin{gathered} \frac{Z_1\_1}{Z_1+1}=\left(\frac{z_2^2+1-2z_2}{z_2^2+1+2z_2}\right)=\left(\frac{\frac{z_2^2+1}{2z_2}-1}{\frac{z_2^2+1}{2z_2}+1}\right)\Rightarrow z_1=\frac{1}{2}(z_2+\frac{1}{z_2}) \\ Let, z_2=\cos \theta +i\sin \theta \Rightarrow z_1=\cos \theta \\ \Rightarrow |z_1-1|+|z_1+1|=|\cos \theta -1|+|\cos \theta +1|=\left|2{\sin }^2\frac{\theta }{2}\right|+\left|2{\cos }^2\frac{\theta }{2}\right|=2 \end{gathered}$$