$\frac{Z - i}{Z - 1}$ is purely imaginary then the locus of $z$ is

x^{2} + y^{2} - x - y = 0

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x^{2} + y^{2} + x + y = 0

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x^{2} + y^{2} + 2 x - 3 y = 0

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x^{2} + y^{2} x + 3 y = 0

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Solution

The correct option is B $x^{2} + y^{2} - x - y = 0$
$L e t z = x + i y$
$t h e n \frac{z - i}{z - 1}$
$= \frac{x + i y - i}{x + i y - 1}$
$= \frac{x + (y - 1) i}{(x - 1) - i y}$
$= \frac{[x + (y - 1) i] [(x - 1) i y]}{[(x - 1) + i y] [(x - 1) - i y]}$
$= \frac{x (x - 1) - i x y + (x - 1) (y - 1) i + y (y - 1)}{{(x - 1)}^{2} + y^{2}}$
$= \frac{(x^{2} - x + y^{2} - y) + i [- x y + x y - x - y + 1]}{{(x - 1)}^{2} + y^{2}}$
For purely imaginary
$= \frac{x^{2} - x + y^{2} - y}{{(x - 1)}^{2} + y^{2}} = 0$
$x^{2} + y^{2} - x - y = 0$

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