If z - i - z - i z - i - i is a purely imaginary number- then Z - z is equal toA- 1B- 0C- 2D- None of these

Here z i/z+i=x+i(y 1/x + i(y + 1) . x i(y+1)/x i(y + 1) = (x^2 + y^2 1) + i ( 2x)/x^2 + (y + 1)^2 As z i/z + i is purely imaginary, we get x^2 + y^2 ...

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Byju's Answer

Standard XII

Mathematics

Purely Imaginary

If z - i / z ...

Question

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

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None of these

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Solution

The correct option is A 1
Here $\frac{z - i}{z + i} = \frac{x + i (y - 1}{x + i (y + 1)} . \frac{x - i (y + 1)}{x - i (y + 1)}$
$= \frac{(x^{2} + y^{2} - 1) + i (- 2 x)}{x^{2} + (y + 1)^{2}}$
As $\frac{z - i}{z + i}$ is purely imaginary, we get
$x^{2} + y^{2} - 1 = 0 \Rightarrow x^{2} + y^{2} = 1 \Rightarrow z ¯ z = 1.$

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