Does $arg \left(\dfrac {z-1}{z+1}\right)=\dfrac{\pi}{3}$ represent a circle or arc of a circle? Does $arg \left(\dfrac {z-1}{z+1}\right)=\dfrac{\pi}{3}$ represent a circle or arc of a circle? 
By substituting $x+iy=z$ I do get equation of a circle. But I feel it should be one particular arc of the circle (the locus) depending on the direction of rotation of $(z-1)$ to $(z+1)$ by Coni's method.
Can someone clarify this issue? Thanks.
 A: $z=x+iy\implies\dfrac{z-1}{z+1}=\dfrac{(x-1+iy)(x+1-iy)}{(x+1)^2+y^2}=\dfrac{x^2+y^2-1+i(2y)}{(x+1)^2+y^2}$
Using this, arg$\dfrac{2y}{x^2+y^2-1}=\dfrac\pi3$ if $x^2+y^2-1>0$
In that case, $x^2+y^2-1=2\sqrt3y\iff x^2+(y-\sqrt3)^2=2^2$
WLOG any point on $x^2+(y-\sqrt3)^2=2^2$ can be represented as $2\cos t,\sqrt3+2\sin t$
Now $x^2+y^2-1=\cdots=6+4\sqrt3\sin t$ which will be $>0\iff\sin t>-\dfrac{\sqrt3}2$
So, its not the full circle, but an arc is represented by arg$\dfrac{z-1}{z+1}=\dfrac\pi3$
A: Observe that (taking principal value of argument)
$$\arctan\frac{2y}{x^2+y^2-1}=\frac\pi3\iff\frac{2y}{x^2+y^2-1}=\tan\frac\pi3=\sqrt3\iff$$
$$\iff x^2+y^2-\frac2{\sqrt3}y-1=0\iff x^2+\left(y-\frac1{\sqrt3}\right)^2=\frac43\;\;(**)$$
Which is a circle of radius $\;\frac2{\sqrt3}\;$ and center $\;\left(0,\,\frac1{\sqrt3}\right)\;$
Yet the first equality means both denominator and numerator in the left side have the same sign. and thus
$$\begin{cases}y>0\;\text{and}\;x^2+y^2>1\,,\;\;\text{or}\\{}\\y<0\;\text{and}\;x^2+y^2<1\end{cases}$$
and thus the options are, in both cases, an arc of the circle $\;(**)\;$...though the cases are not symmetric.
