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
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).
Can someone please help me understand the geometrical meaning of this ?
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 ?