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  4. If (z-i/z+i)(z ≠-i) is a purely imaginary number, then z barz is equal to

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Q. If $\frac{z-i}{z+i}(z \neq-i)$ is a purely imaginary number, then $z \bar{z}$ is equal to

Complex Numbers and Quadratic Equations

Solution:

$\frac{z-i}{z+i}=\lambda i$
$\frac{z-i}{z+i}+\frac{\bar{z}+i}{\bar{z}-i}=0$
$\frac{(z-i)(\bar{z}-i)+(\bar{z}+i)(z+i)}{(z+i)(\bar{z}-i)}=0$
$z \bar{z}-i \bar{z}-i z-1+z \bar{z}+i z+i \bar{z}-1=0$
$2 z \bar{z}-2=0$
$z \bar{z}=1$