Solve $\operatorname{Arg} (z-2) - \operatorname{Arg} (z+2) = \frac{\pi}{6}$ 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.
 A: 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:
$$S=\left\{z=2\sqrt{3}i+4e^{it}\>\biggm|\>-{\pi\over3}<t<{4\pi\over3}\right\}\ .$$
A: Think about the geometric significance of the difference between the arguments of two complex numbers. Then think about where in the plane $z-2$ and $z+2$ must lie to satisfy your equation.
A: Using this and this,
if $z=x+iy,$
Case $1:$ If $x>2,\text{Arg}(z-2)=\arctan \frac y{x-2}$ and $\text{Arg}(z+2)=\arctan \frac y{x+2}$
Case $2:$ If $x=2,\text{Arg}(z-2)=\text{sign}(y)\cdot\frac\pi2($  if $y\ne0)$ and $\text{Arg}(z+2)=\arctan \frac y{x+2}$
Case $3:$ If $ -2<x<2,$
$\text{Arg}(z-2)= \begin{cases} \arctan \frac y{x-2}+\pi &\mbox{if } y\ge0 \\
\arctan \frac y{x-2}-\pi & \mbox{if } y<0\end{cases}$ and $\text{Arg}(z+2)=\arctan \frac y{x+2}$
Case $4:$ If $x=-2,$
$\text{Arg}(z-2)= \begin{cases} \arctan \frac y{x-2}+\pi &\mbox{if } y\ge0 \\
\arctan \frac y{x-2}-\pi & \mbox{if } y<0\end{cases}$ and $\text{Arg}(z+2)=\text{sign}(y)\cdot\frac\pi2($  if $y\ne0)$
Case $5:$ If $x<-2,$ 
$\text{Arg}(z-2)= \begin{cases} \arctan \frac y{x-2}+\pi &\mbox{if } y\ge0 \\
\arctan \frac y{x-2}-\pi & \mbox{if } y<0\end{cases}$ and $\text{Arg}(z+2)= \begin{cases} \arctan \frac y{x+2}+\pi &\mbox{if } y\ge0 \\
\arctan \frac y{x+2}-\pi & \mbox{if } y<0\end{cases}$
Now can you deal the problem case by case?
A: Here is a way of proceeding which depends on special features of the particular problem, so is not really general.
Construct an equilateral triangle on the line segment between $z-2$ and $z+2$ choosing the one in which the third vertex $V$ is nearest to the origin. Then, given the angle subtended at the origin, $V$ is at the centre of a circle passing through the three points $0$, $z-2$ and $z+2$ (angle at centre is twice angle at circumference).
The radius of the circle is 4 (since it forms the side of an equilateral triangle with a segment of length 4). So one has the point $V$ on the circle radius 4 centre origin. The midpoint of the segment (so the point $z$) is either vertically above or vertically below this (case depends on which side of the $x$-axis we are), and all that is needed to calculate the co-ordinates using the simple geometry of the equilateral triangle.
Now take care to identify the sign of the difference in angles so that the differece comes out as $\frac{\pi}6$ rather than $-\frac{\pi}6$, and avoid cases where the angle becomes $2\pi \pm \frac {\pi}6$.
