Show that the complex number $z$ satisfying the condition $\arg\Big(\dfrac{z-1}{z+1}\Big)=\dfrac{\pi}{4}$ is a circle
I can prove it by expressing $z=x+iy$, $$ \arg\Big(\frac{x-1+iy}{x+1+iy}\Big)=\arg(x-1+iy)-\arg(x+1+iy)=\frac{\pi}{4}\\ 1=\frac{\frac{y}{x-1}-\frac{y}{x+1}}{1+\frac{y^2}{x^2-1}}=\frac{xy+y-xy+y}{x^2-1+y^2}\implies x^2-1+y^2=2y\\\implies x^2+y^2-2y-1=0 $$ But can I show that using geometry and the concept of Apollonius circle ?
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Let $M(z), A(-1)$ and $B(1)$.
$\arg(\frac{z-1}{z+1})=(\overrightarrow{MA},\overrightarrow{MB})=\frac{\pi}{4} [\pi]$ is a circle $\Gamma$ passing through $A$ and $B$ except those 2 points.
How to get this circle: let $\Delta$ the perpendicular bisector of $[AB]$, and choose $T$ a point such that $(\overrightarrow{AB},\overrightarrow{AT})=\frac{\pi}{4}$, define $\Delta’$ as the line perpendicular to $(AT)$ at $A$ and finally $\Omega=\Delta\bigcap\Delta’$. $\Gamma$ is the circle centered a $\Omega$ passing through $A$ or/and $B$.
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If $\arg(\frac{z-1}{z+1})=(\overrightarrow{MA},\overrightarrow{MB})=\frac{\pi}{4} [2\pi]$ $\Gamma$ is the arc located in the half plan delimited by $(AB)$ and not containing the point $T$. This method is still valid if we change $\frac{\pi}{4}$ by any value $\theta$
${\rm arg}\left(\frac{z-1}{z+1}\right)$ is a measure of the angle between $\overrightarrow{AM}$ and $\overrightarrow{BM}$, where $A$ has affix $-1$ and $B$ has affix $1$. Find one point $M_0$ verifying this equation which is not $A$ or $B$. The circumcircle of triangle $M_0AB$, excluding the arc between $A$ and $B$, is the solution, because if $\Omega$ is its center, for every $M$ on this circle, the angle between $\overrightarrow{AM}$ and $\overrightarrow{BM}$ is half the angle between $\overrightarrow{\Omega A}$ and $\overrightarrow{\Omega B}$, which is $\frac\pi2$.