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.

  • $\begingroup$ I suggest you start by drawing a picture. $\endgroup$ – dfeuer Jun 30 '13 at 8:54
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    $\begingroup$ @dfeuer I thought the 'arg' function with the lowercase 'a' was generally used to denote the multi-valued argument function while the uppercase one was for the single-valued function in the range $(-\pi, \pi]$. That's why I used the uppercase one. $\endgroup$ – Alraxite Jun 30 '13 at 9:27
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    $\begingroup$ Alraxite, I tried to edit it again, but my edit was rejected. You can replace all instances of \arg with \operatorname{Arg} to get what you want. $\endgroup$ – dfeuer Jun 30 '13 at 18:46

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: $$S=\left\{z=2\sqrt{3}i+4e^{it}\>\biggm|\>-{\pi\over3}<t<{4\pi\over3}\right\}\ .$$

  • $\begingroup$ What do you mean by 'the segment $[-2, 2]$ as seen from $z$'? Do you mean the line joining the points $(x-2,\, y)$ and $(x+2,\, y)$ where $z=x+iy$? If so, then it's not obvious to me why this condition implies that $z$ lies on an arc of some circle. Or did you mean something else? $\endgroup$ – Alraxite Jul 7 '13 at 17:26
  • $\begingroup$ @Alraxite: See my edit. $\endgroup$ – Christian Blatter Jul 8 '13 at 13:00
  • $\begingroup$ I know that the condition $\operatorname{Arg}(z-2) - \operatorname{Arg}(z+2) = \frac{\pi}{6}$ means that the line segments joining $z-2$ and $z+2$ with the origin make an angle of $\frac{\pi}{6}$ with each other (in the upper half plane), but I don't see how this implies that the line segments joining $z$ with $(2,\,0)$ and $(-2,\,0)$ should make an angle of $\frac{\pi}{6}$. $\endgroup$ – Alraxite Jul 10 '13 at 12:02

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.

  • $\begingroup$ Then $z$ would be the mid-point of a line of length $4$ parallel to the $x$-axis and which subtends an angle of $\frac{\pi}{6}$ at the origin (and such that the left-end of the line makes a greater angle with the positive direction of the $x$-axis than the right-end). But it's not obvious to me the possible places this line could lie in the plane. $\endgroup$ – Alraxite Jun 30 '13 at 9:48

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?


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 $2\pi \pm \frac {\pi}6$.

  • $\begingroup$ Surely this only gives one of an infinite number of answers? $\endgroup$ – Peter Taylor Jun 30 '13 at 12:33
  • $\begingroup$ @PeterTaylor You can choose $V$ at any point on the circle radius 4 centre origin, subject only to getting the right sign etc for the angle in the end. $\endgroup$ – Mark Bennet Jun 30 '13 at 12:43

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