For complex numbers $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$, write $z_1 \preceq z_2$ if $x_1 \leq x_2$ and $y_1 \leq y_2$. Then for all complex numbers $z$ with $1 \preceq z, ( 1 \neq z)$.
- $\begin{vmatrix} \dfrac{1+z}{1-z} \end{vmatrix} \preceq 1$
- $\dfrac{1+z}{1-z} \preceq 0$
- $\dfrac{1-z}{1+z} \preceq 0$
- none of these.