TITLE: Solve $\operatorname{Arg} (z-2) - \operatorname{Arg} (z+2) = \frac{\pi}{6}$
QUESTION [6 upvotes]: I'm trying to solve $$\operatorname{Arg}(z-2) - \operatorname{Arg}(z+2) = \frac{\pi}{6}$$ for $z \in \mathbb{C}$.
I know that
$$\operatorname{Arg} z_1 - \operatorname{Arg} z_2 = \operatorname{Arg} \frac{z_1}{z_2},$$
but that's only valid when $\operatorname{Arg} z_1 - \operatorname{Arg} z_2 \in (-\pi,\pi]$, so I'm not sure how to even begin solving this.
I'm not familiar with modular arithmetic so if it is possible to solve this without using it then that would be great! (not that I know whether it is required to solve this in the first place)
Thank you in advance.

REPLY [4 votes]: I suggest you draw a figure.
When $z$ lies in the lower half plane ${\rm Im}(z)<0$ then
$$-\pi<\arg(z-2)-\arg(z+2)<0\ .$$
It follows that there are no points in the lower half plane fulfilling your condition.
Consider now a point $z$ in the upper half plane $H:\ {\rm Im}(z)>0$. Then
$$0<\arg(z-2)-\arg(z+2)<\pi\ .$$ The condition
$$\arg(z-2)-\arg(z+2)={\pi\over 6}$$
means that the two segments connecting $z$ with the points $2$ and $-2$ enclose  an angle  of ${\pi\over 6}$. The set of $z$ fulfilling this condition is,  according to the theorem about peripheral angles (resp., its inverse), an arc of a circle $\gamma$. The midpoint $M$ of $\gamma$ lies on the imaginary axis such that $\angle(2,M,-2)={\pi\over3}$. It follows that $M=2\sqrt{3}i$, and the radius of $\gamma$ is obviously $4$. The equation of this circle $\gamma$ is
$$|z-2\sqrt{3}i|^2=16\ ,$$
and the set $S$  you are interested in is $\gamma\cap H$. One could provide a parametric representation of $S$ as follows:
$$S=\left\{z=2\sqrt{3}i+4e^{it}\>\biggm|\>-{\pi\over3}<t<{4\pi\over3}\right\}\ .$$