If Arg $(z + i) -$ Arg $(z - i)$ $= \frac{π}{2}$ , then $z$ lies on a circle.
If statement is True, enter $1$ , else enter $0$

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Solution

We have
Arg $(z + i) -$ Arg $(z - 1) = \frac{π}{2}$ $\Rightarrow$ Arg $(\frac{z + i}{z - i}) = \frac{π}{2}$
$∴ R e (\frac{z + i}{z - i}) = 0$ $\Rightarrow \frac{(\frac{z + i}{z - i}) + ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ (\frac{z + i}{z - i})}{2} = 0$
$\Rightarrow (\frac{z + i}{z - i}) + (\frac{¯ ¯ ¯ z - i}{¯ ¯ ¯ z + i}) = 0$
$\Rightarrow (z + i) (¯ ¯ ¯ z + i) + (z - i) (¯ ¯ ¯ z - i) = 0$
$\Rightarrow z ¯ ¯ ¯ z + i (¯ ¯ ¯ z + z) - 1 + z ¯ ¯ ¯ z - i (z + ¯ ¯ ¯ z) - 1 = 0$
$\Rightarrow 2 (z ¯ ¯ ¯ z) = 2$ $\Rightarrow z ¯ ¯ ¯ z = 1$ $| z |^{2} = 1$ $\Rightarrow | z | = 1$
The equation represents a circle centered at origin and radius $1$ units.

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