(A) Given,$\left|\frac{z-2i}{z+2i}\right|=1$ $\Rightarrow |z-2i| = |z+2i|$ Substituting $z=x+iy$: $|x+i(y-2)| = |x+i(y+2)|$ Squaring both sides: $x^2

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(A) Given,$\left|\frac{z-2i}{z+2i}\right|=1$ $\Rightarrow |z-2i| = |z+2i|$ Substituting $z=x+iy$: $|x+i(y-2)| = |x+i(y+2)|$ Squaring both sides: $x^2 + (y-2)^2 = x^2 + (y+2)^2$ $x^2 + y^2 - 4y + 4 = x^2 + y^2 + 4y + 4$ $-4y = 4y$ $8y = 0$ $y = 0$ This represents the $x$-axis.

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The locus of the point $z=x+iy$ satisfying $\left|\frac{z-2i}{z+2i}\right|=1$ is

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$x$-axis
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$y$-axis
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$y=2$
D
$x=2$

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The locus of the point $z=x+iy$ satisfying $\left|\frac{z-2i}{z+2i}\right|=1$ is

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