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

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4-1.Complex numbers

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

$0$

$1$

$2$

None of these

Solution

(b) Here $\frac{{z – i}}{{z + i}} = \frac{{x + i(y – 1)}}{{x + i(y + 1)}}.\frac{{x – i(y + 1)}}{{x – i(y + 1)}}$
$ = \frac{{({x^2} + {y^2} – 1) + i( – 2x)}}{{{x^2} + {{(y + 1)}^2}}}$
As $\frac{{z – i}}{{z + i}}$ is purely imaginary, we get
${x^2} + {y^2} – 1 = 0$==> ${x^2} + {y^2} = 1$==>$z\overline z = 1$.

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