# Geometric meaning of $Arg(\frac{z_{1}}{z_{2}})=2Arg(\frac{z_{3}-z_{1}}{z_{3}-z_{2}})+2\pi k$ for $z_{i}$ on the unit circle

I was given the following question:

Let $z_{1},z_{2},z_{3}\in\mathbb{C}$ s.t $|z_{1}|=|z_{2}|=|z_{3}|=1$, it is known that $Arg(\frac{z_{1}}{z_{2}})=2Arg(\frac{z_{3}-z_{1}}{z_{3}-z_{2}})+2\pi k$ where $k\in\mathbb{Z}$.

Determine the possible values of $k$ and give geometric meaning to the equation.

My thoughts:

Since for every $z\in\mathbb{C}$ we have $$-\pi<Arg(z)\leq\pi$$ and $$arg(z^{2})=2arg(z)$$ then $$2Arg(\frac{z_{3}-z_{1}}{z_{3}-z_{2}})\in(-2\pi,2\pi]$$ so we may have $k=0,1,-1$ since for any other $k$ and $\theta\in(-2\pi,2\pi]$ we have $\theta+2\pi k\not\in(-\pi,\pi]$.

What I can't seem to understand is the geometric meaning of this equation (in the tutorial we proved $Arg(\frac{z_{1}}{z_{2}})=2Arg(\frac{z_{3}-z_{1}}{z_{3}-z_{2}})+2\pi k$ where $k\in\mathbb{Z}$ in an algebraic way).

Edit: I am having the feeling that $arg(z^{2})=2arg(z)$ may not suffice to imply $2Arg(\frac{z_{3}-z_{1}}{z_{3}-z_{2}})\in(-2\pi,2\pi]$, is the justification correct ?
The argument of a fraction is, geometrically speaking, the (oriented) angle between two vectors. This explains geometrically why the number is only defined up to a multiple of $2\pi$. Thus restricting the range of $\arg(z)$ to the one you specified is a reasonable choice. From that range, it automatically follows that twice the argument covers twice the range, even without considering $\arg(z^2)$. So your equation $2\arg\left(\frac{z_3-z_1}{z_3-z_2}\right)\in(-2\pi,2\pi]$ is a direct consequence of the convention $\arg(z)\in(-\pi,\pi]$.