The equation $$\arg\left(\frac{z-1}{z+1}\right) = \frac{\pi}{4}$$ represents a circle with:

Let us denote the required complex number by $$z$$ and write it in the usual Cartesian form

$$z = x + iy, \qquad \text{where } x, y \in \mathbb R.$$

The given condition is

$$\arg\!\left(\dfrac{z-1}{\,z+1\,}\right)=\dfrac{\pi}{4}.$$

For any complex number $$W = A + iB,$$ we recall the basic fact

$$\arg(W)=\theta \;\Longrightarrow\; \dfrac{B}{A}=\tan\theta.$$

Since $$\tan\!\left(\dfrac{\pi}{4}\right)=1,$$ the above gives the simple relation

$$B = A.$$

Thus, for the complex quotient

$$W=\dfrac{z-1}{z+1},$$

we merely have to equate its imaginary and real parts. We therefore proceed to calculate these two parts explicitly.

First write the numerator and the denominator in terms of $$x$$ and $$y$$:

$$z-1 = (x-1)+iy, \qquad z+1 = (x+1)+iy.$$

To find $$W$$ conveniently, multiply numerator and denominator by the complex conjugate of the denominator:

$$\dfrac{z-1}{z+1}= \dfrac{(x-1)+iy}{(x+1)+iy}\; \times\; \dfrac{(x+1)-iy}{(x+1)-iy}.$$

The denominator thus becomes the real number

$$\bigl(x+1\bigr)^2 + y^2.$$

Let us expand the numerator carefully, multiplying the two binomials term by term:

$$ \bigl[(x-1)+iy\bigr]\bigl[(x+1)-iy\bigr] = (x-1)(x+1) + (x-1)(-iy) + iy(x+1) + iy(-iy). $$

Compute each piece:

$$(x-1)(x+1) = x^2-1,$$
$$(x-1)(-iy) = -i y x + i y,$$
$$iy(x+1) = i y x + i y,$$
$$iy(-iy) = -i^2 y^2 = +y^2.$$

Adding the real parts and the imaginary parts separately we obtain

$$ x^2 - 1 + y^2 \;+\; 2 i y. $$

Hence

$$W=\dfrac{(x^2+y^2-1) + 2iy}{(x+1)^2 + y^2}.$$

From this expression we read off

$$\operatorname{Re}(W)=\dfrac{x^2 + y^2 -1}{(x+1)^2 + y^2}, \qquad \operatorname{Im}(W)=\dfrac{2y}{(x+1)^2 + y^2}.$$

Because the denominator is strictly positive, the equality of the imaginary and real parts entails

$$2y = x^2 + y^2 -1.$$

Let us bring every term to one side:

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

To recognise a circle we complete the square in the $$y$$-variable. Observe that

$$y^2 - 2y = (y-1)^2 - 1.$$

Substituting this into the equation gives

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

Simplifying further,

$$x^2 + (y-1)^2 - 2 = 0,$$

or equivalently,

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

This is the standard form of a circle. Comparing with

$$ (x-h)^2 + (y-k)^2 = R^2,$$

we immediately read

Centre $$= (h,k) = (0,1), \qquad \text{Radius} \; R = \sqrt{2}.$$

Consequently the curve represented by the given argument condition is the circle centred at $$(0,1)$$ with radius $$\sqrt{2}$$.

Hence, the correct answer is Option D.