If $\left(\frac{z-1}{z+1}\right)$ is purely an imaginary number and $z \neq-1$ then find the value of |z|.
Given: $\frac{z-1}{z+1}$ is purely imaginary number
Let $z=x+i y$
So, $\frac{z-1}{z+1}=\frac{x+i y-1}{x+i y+1}$
$=\frac{(x-1)+i y}{(x+1)+i y}$
Now, rationalizing the above by multiply and divide by the conjugate of [(x + 1) + iy]
$=\frac{(x-1)+i y}{(x+1)+i y} \times \frac{(x+1)-i y}{(x+1)-i y}$
$=\frac{[(x-1)+i y][(x+1)-i y]}{[(x+1)+i y][(x+1)-i y]}$
Using $(a-b)(a+b)=\left(a^{2}-b^{2}\right)$
$=\frac{(x-1)[(x+1)-i y]+i y[(x+1)-i y]}{(x+1)^{2}-(i y)^{2}}$
$=\frac{(x-1)(x+1)+(x-1)(-i y)+i y(x+1)+(i y)(-i y)}{x^{2}+1+2 x-i^{2} y^{2}}$
$=\frac{x^{2}-1-i x y+i y+i x y+i y-i^{2} y^{2}}{x^{2}+1+2 x-i^{2} y^{2}}$
Putting $i^{2}=-1$
$=\frac{x^{2}-1+2 i y-(-1) y^{2}}{x^{2}+1+2 x-(-1) y^{2}}$
$=\frac{x^{2}-1+2 i y+y^{2}}{x^{2}+1+2 x+y^{2}}$
$=\frac{x^{2}-1+y^{2}}{x^{2}+1+2 x+y^{2}}+i \frac{2 y}{x^{2}+1+2 x+y^{2}}$
Since, the number is purely imaginary it means real part is 0
$\therefore \frac{x^{2}-1+y^{2}}{x^{2}+1+2 x+y^{2}}=0$
$\Rightarrow x^{2}+y^{2}-1=0$
$\Rightarrow x^{2}+y^{2}=1$
$\Rightarrow \sqrt{x^{2}+y^{2}}=\sqrt{1}$
$\Rightarrow \sqrt{x^{2}+y^{2}}=1$
$\therefore|z|=1$