If $$z \neq 1$$ and $$\frac{z^2}{z-1}$$ is real, then the point represented by the complex number $$z$$ lies

Let $$z = x + iy$$, where $$x, y \in \mathbb{R}$$ and $$y \neq 0$$ is allowed (but $$z \neq 1$$ means $$(x,y) \neq (1,0)$$).

The given expression is $$\dfrac{z^{2}}{z-1}$$. First find the numerator and denominator in Cartesian form:

$$z^{2} = (x + iy)^{2} = (x^{2} - y^{2}) + 2ixy$$
$$z - 1 = (x - 1) + iy$$

To check when the quotient is real, multiply numerator and denominator by the conjugate of the denominator:

$$\dfrac{z^{2}}{z-1} = \dfrac{\bigl[(x^{2}-y^{2}) + 2ixy\bigr]\bigl[(x-1) - iy\bigr]}{(x-1)^{2} + y^{2}}$$

The denominator $$(x-1)^{2}+y^{2}$$ is positive (because $$z \neq 1$$), so the quotient is real exactly when the numerator is real, i.e. its imaginary part is zero.

Compute the imaginary part of the numerator:
Numerator = $$[(x^{2}-y^{2}) + 2ixy]\,[\,(x-1) - iy\,]$$

Separate real and imaginary components:

Real part: $$ (x^{2}-y^{2})(x-1) + 2xy^{2} $$
Imaginary part: $$ i\Bigl[-(x^{2}-y^{2})y + 2xy(x-1)\Bigr] $$

For the fraction to be real, the imaginary part must vanish:

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

Factor out $$y$$:

$$y\Bigl[-(x^{2}-y^{2}) + 2x(x-1)\Bigr] = 0$$

Then $$z = x + i\cdot 0$$ lies on the real axis.

Simplify:

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

Complete the square in $$x$$:

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

This is the equation of a circle with centre at $$(1,0)$$ and radius $$1$$. Since the distance from the centre to the origin is also $$1$$, the circle passes through the origin.

Hence, the point corresponding to $$z$$ lies either on the real axis or on the circle $$(x - 1)^{2} + y^{2} = 1$$, which indeed passes through the origin.

Therefore, the correct choice is:
Option A which is: either on the real axis or on a circle passing through the origin.