Question:
If $\frac{z-1}{z+1}$ is purely imaginary and $z=-1$, show that $|z|=1$
Solution:
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.