Geometrical meaning of $\arg\left(\frac{z-1}{z+1}\right)=\frac{\pi }{3}$ If $\displaystyle \arg\left(\frac{z-1}{z+1}\right)=\frac{\pi }{3}$. Then locus of complex no. $z$ is:
My try: Let $z = x+iy\;, $ We get $\displaystyle \displaystyle\arg\left(\frac{x+iy-1}{x+iy+1}\right)=\frac{\pi }{3}$
$$\displaystyle \arg\left(\frac{(x^2+y^2-1)+2ixy}{(x+1)^2+y^2}\right) = \frac{\pi}{3}$$
$$\displaystyle x^2+y^2-\frac{2}{\sqrt{3}}y-1=0$$
So Locus of $z$ is a Circle.
My question is: by using geometry how can I interpret the given equation $$\displaystyle \arg\left(\frac{z-1}{z+1}\right)=\frac{\pi }{3}$$

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.
