The equation $$Im\left(\frac{iz - 2}{z - i}\right) + 1 = 0$$, $$z \in \mathbb{C}$$, $$z \neq i$$ represents a part of a circle having radius equal to:

Let us write the complex number $$z$$ in its Cartesian form

$$z = x + i\,y,\qquad x,\;y \in \mathbb R.$$

First compute the numerator

$$i z - 2 \;=\; i(x + i\,y) \;-\; 2 \;=\; i x \;-\; y \;-\; 2 \;=\; (-\,y - 2) \;+\; i\,x.$$

Next compute the denominator

$$z - i \;=\; x + i\,y - i \;=\; x \;+\; i\,(y - 1).$$

So the required fraction becomes

$$\frac{i z - 2}{z - i} \;=\; \frac{(-\,y - 2) + i\,x}{x + i\,(y - 1)}.$$

To extract its imaginary part, multiply numerator and denominator by the conjugate of the denominator:

$$\frac{(-\,y - 2) + i\,x}{x + i\,(y - 1)}\; \cdot\; \frac{x - i\,(y - 1)}{x - i\,(y - 1)} = \frac{\bigl[(-\,y - 2) + i\,x\bigr]\, \bigl[x - i\,(y - 1)\bigr]} {x^{2} + (y - 1)^{2}}.$$

Expand the numerator term-by-term:

$$\bigl[(-\,y - 2) + i\,x\bigr]\, \bigl[x - i\,(y - 1)\bigr]$$ $$=\;(-\,y - 2)\,x \;+\; (-\,y - 2)\,[-\,i\,(y - 1)] \;+\; i\,x\cdot x \;+\; i\,x\cdot[-\,i\,(y - 1)].$$

Calculate every product:

$$(-\,y - 2)\,x \;=\; -x\,y - 2x,$$

$$(-\,y - 2)\,[-\,i\,(y - 1)] \;=\; (+)\,i\,(y + 2)(y - 1),$$

$$i\,x\cdot x \;=\; i\,x^{2},$$

$$i\,x\cdot[-\,i\,(y - 1)] \;=\; +\,x\,(y - 1).$$

Gather the real and imaginary parts:

Real part:

$$-x\,y - 2x \;+\; x\,(y - 1) \;=\; (-x\,y + x\,y) - 2x - x \;=\; -3x.$$

Imaginary part:

$$i\Bigl[(y + 2)(y - 1) + x^{2}\Bigr].$$

Therefore

$$\frac{i z - 2}{z - i} \;=\; \frac{-\,3x \;+\; i\left[x^{2} + (y + 2)(y - 1)\right]} {x^{2} + (y - 1)^{2}}.$$

The denominator is purely real, hence

$$\operatorname{Im}\!\left( \frac{i z - 2}{z - i} \right) \;=\; \frac{x^{2} + (y + 2)(y - 1)} {x^{2} + (y - 1)^{2}}.$$

The given condition is

$$\operatorname{Im}\!\left( \frac{i z - 2}{z - i} \right) + 1 = 0 \;\;\Longrightarrow\;\; \operatorname{Im}\!\left( \frac{i z - 2}{z - i} \right) = -1.$$

Hence

$$\frac{x^{2} + (y + 2)(y - 1)} {x^{2} + (y - 1)^{2}} = -1.$$

Cross-multiply:

$$x^{2} + (y + 2)(y - 1) = -\bigl[x^{2} + (y - 1)^{2}\bigr].$$

Bring every term to the left side:

$$x^{2} + (y + 2)(y - 1) + x^{2} + (y - 1)^{2} = 0.$$

Combine like terms:

$$2x^{2} + \bigl[(y + 2)(y - 1) + (y - 1)^{2}\bigr] = 0.$$

Compute the brackets one by one:

$$(y + 2)(y - 1) = y^{2} + y - 2,$$

$$(y - 1)^{2} = y^{2} - 2y + 1.$$

Add them:

$$y^{2} + y - 2 + y^{2} - 2y + 1 = 2y^{2} - y - 1.$$

Hence

$$2x^{2} + 2y^{2} - y - 1 = 0.$$

Divide by 2 to simplify:

$$x^{2} + y^{2} - \frac{y}{2} - \frac{1}{2} = 0.$$

Move the constant to the right side:

$$x^{2} + y^{2} - \frac{y}{2} = \frac{1}{2}.$$

To obtain the standard circle form, complete the square in $$y$$:

$$y^{2} - \frac{y}{2} = y^{2} - \frac{1}{2}y = \left(y - \frac{1}{4}\right)^{2} - \frac{1}{16}.$$

Substitute this into the equation:

$$x^{2} + \Bigl[\bigl(y - \tfrac14\bigr)^{2} - \tfrac1{16}\Bigr] = \frac12.$$

Rearrange:

$$x^{2} + \bigl(y - \tfrac14\bigr)^{2} = \frac12 + \frac1{16} = \frac{9}{16}.$$

Thus the locus is a circle with centre

$$\bigl(0,\; \tfrac14\bigr)$$

and radius

$$\sqrt{\frac{9}{16}} = \frac34.$$

Consequently, the radius asked in the problem is $$\dfrac34$$.

Hence, the correct answer is Option C.