Geometrical meaning of $\arg\left(\frac{z-1}{z+1}\right)=\frac{\pi }{3}$

The given condition characterizes a set of "admissible" $z\in\Bbb C$. Since division of complex numbers amounts to subtracting their arguments, a point $z$ is admissible when you obtain the vector $z-1$ by turning the vector $z+1$ by ${\pi\over3}\>\hat=\> 60^\circ$ counterclockwise, up to scaling by a factor $\lambda>0$. The following figure shows what this means in terms of elementary geometry:

I'm sure you know a theorem from high school that tells you under which circumstances the angle $\angle(-1,z,1)$ has a given size when $\pm1$ are fixed and $z$ is variable.

Note that $$ w=\frac{z-1}{z+1}\Longleftrightarrow z=\frac{1+w}{1-w}\tag{1} $$ So the equation in question says that $\arg(w)=\frac\pi3$; that is, $w$ lies on a line through the origin at an angle of $\frac\pi3$.

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Möbius transformations preserve generalized circles. That means that they send circles to circles if we include lines as the limits of circles. Thus, we know that the image of the line $\arg(w)=\frac\pi3$ is a circle or line.

Since the $w$-line $\arg(w)=\frac\pi3$ passes through $0$ and $\infty$, the $z$-circle $$ \arg\left(\frac{z-1}{z+1}\right)=\frac\pi3\tag{2} $$ passes through $z(0)=1$ and $z(\infty)=-1$.

The real axis (dotted line) is preserved by both transformations. Möbius transformations are conformal; that is, they preserve angles. Thus, the angle between the black and dotted lines at $w=0$ is equal to the angle between the red circle and dotted line at $z=1$.

Knowing two points and the tangent at one of the points allows us to determine the circle geometrically.

Suppose you have a circle $\mathscr{C}$ centered at $O$ and $A$, $B$ are two points on $\mathscr{C}$. $A$ and $B$ split $\mathscr{C}$ into two arcs. If you pick an arbitrary point $C$ from one of its arc, the angle $\measuredangle{ACB}$ will be independent of your choice of $C$ and equal to $\frac12\measuredangle{AOB}$.

Let's identify the complex plane $\mathbb{C}$ with the Eucliean plane $\mathbb{R}^2$ and take $A = 1+0i$, $B = -1+0i$ and $C = z$. Remember,

$\arg(z - 1)$ is the angle between the line $AC$ and the real axis.
$\arg(z + 1)$ is the angle between $BC$ and the real axis.

This means $ \arg(\frac{z-1}{z+1}) = \arg(z-1) - \arg(z+1)$ is nothing but the angle $\measuredangle{ACB}$.

Since the corresponding angle $\measuredangle{AOB} = 2 \measuredangle{ACB} = \frac{2\pi}{3}$ is one third of $360^\circ$, the locus for $\arg(\frac{z-1}{z+1}) = \frac{\pi}{3}$ is an arc on the circumcircle for some isosceles triangle $\triangle ABD$ which has $A$, $B$ as two of its vertices.