Question:

<p><strong>Step 1:</strong> Let z = x + iy where x, y ∈ ℝ.</p><p><strong>Step 2:</strong> Calculate iz - 2 = i(x + iy) - 2 = ix - y - 2 = (-y - 2) + ix</p><p><strong>Step 3:</strong> Calculate z - i = x + i(y - 1)</p><p><strong>Step 4:</strong> Find the complex fraction: $$\frac{iz-2}{z-i} = \frac{(-y-2)+ix}{x+i(y-1)}$$</p><p><strong>Step 5:</strong> Rationalize by multiplying by conjugate: $$\frac{[(-y-2)+ix][x-i(y-1)]}{[x+i(y-1)][x-i(y-1)]}$$</p><p><strong>Step 6:</strong> Numerator = $(-y-2)x + i[-y-2][-（y-1)] + ix·x + i^2x(y-1)$<br/>= $(-y-2)x - x(y-1) + i[x^2 + (y+2)(y-1)]$<br/>= $-xy - 2x - xy + x + i[x^2 + y^2 + y - 2]$<br/>= $(-2xy - x) + i(x^2 + y^2 + y - 2)$</p><p><strong>Step 7:</strong> Denominator = $x^2 + (y-1)^2 = x^2 + y^2 - 2y + 1$</p><p><strong>Step 8:</strong> The imaginary part is: $$\text{Im}\left(\frac{iz-2}{z-i}\right) = \frac{x^2 + y^2 + y - 2}{x^2 + y^2 - 2y + 1}$$</p><p><strong>Step 9:</strong> Given condition: $\frac{x^2 + y^2 + y - 2}{x^2 + y^2 - 2y + 1} + 1 = 0$</p><p><strong>Step 10:</strong> Therefore: $\frac{x^2 + y^2 + y - 2 + x^2 + y^2 - 2y + 1}{x^2 + y^2 - 2y + 1} = 0$</p><p><strong>Step 11:</strong> Numerator must be zero: $2x^2 + 2y^2 - y - 1 = 0$</p><p><strong>Step 12:</strong> Rearrange: $2x^2 + 2y^2 - y - 1 = 0$<br/>$x^2 + y^2 - \frac{1}{2}y - \frac{1}{2} = 0$</p><p><strong>Step 13:</strong> Complete the square: $x^2 + \left(y - \frac{1}{4}\right)^2 - \frac{1}{16} - \frac{1}{2} = 0$<br/>$x^2 + \left(y - \frac{1}{4}\right)^2 = \frac{1}{16} + \frac{8}{16} = \frac{9}{16}$</p><p><strong>Step 14:</strong> This is a circle with center $(0, \frac{1}{4})$ and radius $r = \sqrt{\frac{9}{16}} = \frac{3}{4}$</p><p><strong>∴ Answer:</strong> A</p>