Find the locus of $z$ such that $$\arg\left( \frac{ z^2 - 1}{ z^2 + 1} \right) = 0~, \qquad z \neq \pm i$$ I am able to solve this by substituting $z$ as $x+iy$ and proceeding algebraically.
My question is
How can I solve for this loci in terms of vectors (maybe with a bit plane geometry)?
For example, if I see an expression of the form |$z$|=$r$, I know that it's a circle.
$\arg\left( \frac{ z - z_A}{ z - z_B} \right)$ expresses the oriented angle between the vectors $\vec{AM}, \vec{BM},$ where the points $A,B,M$ represent the complex numbers $z_A, z_B, z$ respectively.
In particular, $\arg\left( \frac{ z - z_A}{ z - z_B} \right)=0$ says that the points $A,B$ lie on the same half-line with the bound $M.$ Moreover, $M\neq A$ because $\arg$ is not defined for $0,$ and $M\neq B$ for a trivial reason.
Return to the given equation.
The points $Q(z^2)$ such that $0=\arg\left( \frac{ z^2 - 1}{ z^2 + 1} \right) = \arg\left( \frac{ z^2 - 1}{ z^2 - (-1)} \right)$ are collinear with the points representing $1$ or $-1,$ where $Q$ is the limit point of a convenient half-line. This gives $z^2\in (-\infty,-1)\cup (1,\infty).$
From this one concludes easily for $z:$
The locus is the union of four half-lines which are parts of $x-$axis or $y-$axis, and do not contain the segments $[-1,1]$ nor $[-i,i].$