If z = x + iy is a complex number,then the eq | vedclass

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(A) Given $\left|\frac{z+i}{z-i}\right| = \sqrt{3}$.Squaring both sides,we get $\frac{|z+i|^2}{|z-i|^2} = 3$.Substituting $z = x + iy$,we have $|x + i

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circle with centre $(0, 2)$ and radius $\sqrt{3}$

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(A) Given $\left|\frac{z+i}{z-i}\right| = \sqrt{3}$.Squaring both sides,we get $\frac{|z+i|^2}{|z-i|^2} = 3$.Substituting $z = x + iy$,we have $|x + i(y+1)|^2 = 3|x + i(y-1)|^2$.This simplifies to $x^2 + (y+1)^2 = 3(x^2 + (y-1)^2)$.$x^2 + y^2 + 2y + 1 = 3(x^2 + y^2 - 2y + 1)$.$x^2 + y^2 + 2y + 1 = 3x^2 + 3y^2 - 6y + 3$.$2x^2 + 2y^2 - 8y + 2 = 0$.Dividing by $2$,we get $x^2 + y^2 - 4y + 1 = 0$.Completing the square for $y$: $x^2 + (y-2)^2 - 4 + 1 = 0$.$x^2 + (y-2)^2 = 3$.This is a circle with centre $(0, 2)$ and radius $\sqrt{3}$.

If $z = x + iy$ is a complex number,then the equation $\left|\frac{z+i}{z-i}\right| = \sqrt{3}$ represents the

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