Question

If $$\frac{z-1}{z+1}$$ is purely imaginary number ($$z\ne -1$$), find the value of $$\left|z\right|$$.

Answer 1

Let $$z=x+iy$$

Then,  

$$\frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}$$

$$=\frac{(x-1)+iy}{(x+1)+iy}\times \frac{(x+1)-iy}{(x+1)-iy}$$

$$=\frac{x^2+x-ixy-x-1+iy+ixy+iy-i^2{y}^2}{{(x+1)}^2-i^2{y}^2}$$

$$=\frac{x^2+y^2-1+2iy}{x^2+1+2x+y^2}[\because i^2=-1]$$

If $$\frac{z-1}{z+1}$$ is purely imaginary number, then

$$\text{ Re }\left(\frac{z-1}{z+1}\right)=0$$

$$\Rightarrow x^2+y^2-1=0$$

$$\Rightarrow x^2+y^2=1$$

$$\Rightarrow {\left|z\right|}^2=1$$

$$\Rightarrow \left|z\right|=1$$

Thus, the value of $$\left|z\right|$$ is 1.