If (z – 1)/(z + 1) is a purely imaginary number (z ≠ –1),

According to the question,

Let z = x + iy

Now,

$\frac{z-1}{z+1}=\frac{x+i y-1}{x+i y+1}$

$=\frac{[(x-1)+i y][(x+1)-i y]}{[(x+1)+i y][(x+1)-i y]}$

$=\frac{\left(x^{2}-1\right)+y^{2}+i[(x+1) y-(x-1) y]}{(x+1)^{2}+y^{2}}$

According to the question, we have,

$\frac{z-1}{z+1}$ is purely imaginary.

$\Rightarrow \frac{\left(x^{2}-1\right)+y^{2}}{(x+1)^{2}+y^{2}}=0$

⇒ x2 – 1 + y2 = 0

⇒ x2 + y2 = 1

⇒ √(x2 + y2) = 1

Hence, |z| = 1