Solve this

Let z= a + ib

Now, $\frac{z-1}{z+1}=\frac{a+i b-1}{a+i b+1}$

$=\frac{(a-1)+i b}{(a+1)+i b}$

$\Rightarrow \frac{(a-1)+i b}{(a+1)+i b} \times \frac{(a+1)-i b}{(a+1)-i b}$

$=\frac{a^{2}+a-i a b-a-1+i b+i a b+i b-i^{2} b^{2}}{(a+1)^{2}+b^{2}}$

$=\frac{a^{2}+-1+i b+i b+b^{2}}{(a+1)^{2}+b^{2}}=\frac{a^{2}+b^{2}-1+2 i b}{(a+1)^{2}+b^{2}}$

Given that $\frac{z-1}{z+1}$ is purely imaginary $\Rightarrow$ real part $=0$

$\Rightarrow \frac{a^{2}+b^{2}-1}{(a+1)^{2}+b^{2}}=0$

$\Rightarrow a^{2}+b^{2}-1=0$

$\Rightarrow a^{2}+b^{2}=1$

$\Rightarrow|z|=1$

Hence proved.