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(D) Given $z = x + iy$. Then $\frac{z+i}{z-i} = \frac{x + iy + i}{x + iy - i} = \frac{x + i(y+1)}{x + i(y-1)}$.Multiplying the numerator and denominat

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(D) Given $z = x + iy$. Then $\frac{z+i}{z-i} = \frac{x + iy + i}{x + iy - i} = \frac{x + i(y+1)}{x + i(y-1)}$.Multiplying the numerator and denominator by the conjugate of the denominator, $x - i(y-1)$:$\frac{z+i}{z-i} = \frac{[x + i(y+1)][x - i(y-1)]}{x^2 + (y-1)^2} = \frac{x^2 - ix(y-1) + ix(y+1) - i^2(y+1)(y-1)}{x^2 + (y-1)^2}$.Since $i^2 = -1$, this simplifies to:$\frac{x^2 + y^2 - 1 + i[x(y+1) - x(y-1)]}{x^2 + (y-1)^2} = \frac{x^2 + y^2 - 1 + 2xi}{x^2 + (y-1)^2}$.For the expression to be purely imaginary, the real part must be zero:$	ext{Re}\left(\frac{z+i}{z-i}ight) = \frac{x^2 + y^2 - 1}{x^2 + (y-1)^2} = 0$.This implies $x^2 + y^2 - 1 = 0$, or $x^2 + y^2 = 1$.This is the equation of a circle with center $(0, 0)$ and radius $1$ (excluding the point $(0, 1)$ where the denominator is zero).

Let $z = x + iy$, where $x$ and $y$ are real. The points $(x, y)$ in the $X-Y$ plane for which $\frac{z+i}{z-i}$ is purely imaginary, lie on

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