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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).

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 ?

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Your first double inequality is an option or choice, not a must. –  DonAntonio Nov 9 '12 at 23:37

1 Answer 1

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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 $\arg(z)\in(-\pi,\pi]$.

For a geometric interpretation of the equation, consider this: your left hand side computes the angle between two points on the unit circle, as seen from the center of that circle. The right hand side computes the angle between those same two points, but this time as seen from a third point on the circle. Have a look at the inscribed angle theorem and you will recognize this configuration.

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