I am trying to prove / disprove the following in $\mathbb{C}^{2}$: The line formed between the points $(a_1,b_1)$ and $(a_2,b_2)$ is parallel to the line formed between the points $(a_1,b_1)$ and $(a_3,b_3)$ if and only if one of the following is true:

  1. $(a_2,b_2)=(a_3,b_3)$


  1. the slope formed by joining the point $(a_1,b_1)$ to $(a_2,b_2)$ and to $(a_3,b_3)$ is of the form $\omega^{j}$ where $\omega$ is the primitive $n$th root of unity and $j\in \{1,...,n\}$.

I am also assuming that not all of the three points $(a_1,b_1)$, $(a_2,b_2)$ and $(a_3,b_3)$ are on the same line in $\mathbb{R}^2$.

The reason I believe that this may be true is because if we choose $(a_1,b_1)$ to be a point $(0,\omega^{j})$ for $j\in\{1,...,n\}$ and the points $(a_2,b_2)$ and $(a_3,b_3)$ to be points $(\omega^{k},0)$ and $(\omega^{l},0)$ respectively for $k,l\in \{1,...,n\}$, then we do not have to have $k=l$ i.e. the points $(a_2,b_2)$ and $(a_3,b_3)$ do not have to be the same points since it is sufficient that

$\frac{\omega^{j}-0}{0-\omega^{k}}=\frac{\omega^{j}-0}{0-\omega^{l}}\Rightarrow -\omega^{j-k}=-\omega^{j-l}$ which implies that $j-k\equiv j-l \mod n \Rightarrow -k\equiv -l \mod n$.

Are there other ways to create such a construction or is this possibly the only construction of distinct points $(a_2,b_2)$ and $(a_3,b_3)$ in $\mathbb{C}^{2}$? I have not been able to find other such constructions, and hence I believe this to be only one (of course outside of the case where all the points $(a_i,b_i)$ for $i\in\{1,2,3\}$ are on the same line in $\mathbb{R}^2$).

I have started to write a proof for this argument as

$\frac{b_2-b_1}{a_2-a_1}=\frac{b_3-b_1}{a_3-a_1}$ where without the loss of generality one can choose $b_2=b_3=0$ so that we have $\frac{-b_1}{a_2-a_1}=\frac{-b_1}{a_3-a_1}\Rightarrow a_2-a_1=a_3-a_1\Rightarrow a_2=a_3$ and hence this idea seems to be going nowhere. I think the primitive roots of unity should come in somewhere, but I am not sure how to exactly argue this.

  • 1
    $\begingroup$ "I am also assuming that not all of the three points $(a_1,b_1), (a_2,b_2)$ and $(a_3,b_3)$ are on the same line." - If two parallel lines pass through the same point, then they must be the same line. The only way that $\overline{(a_1, b_1)(a_2,b_2)}$ can be parallel to $\overline{(a_1, b_1)(a_3,b_3)}$ is if the they are the same line. That is, all three points lie on one line. $\endgroup$ Feb 16 at 20:44
  • $\begingroup$ "or is this possibly the only construction of distinct points $(a_2,b_2)$ and $(a_3,b_3)$ in $\Bbb C^2$?" How could you possibly get this idea? $\Bbb C^2 \equiv \Bbb R^4$, That is four dimensional space. Your "lines" are actually 2-dimensional planes. There are uncountably many pairs of points on a plane, representing the parallel case. And for any plane, there are uncountably many points off the plane (the non-parallel case). Roots of unity are just a tiny few points in $\Bbb C$. Something that is true for them hardly need apply to all points. $\endgroup$ Feb 16 at 20:55
  • $\begingroup$ I fixed the write up for lines in $\mathbb{R}^2$ $\endgroup$ Feb 17 at 2:43
  • $\begingroup$ What is 'the slope formed by joining the point $(a1,b1)$ to $(a2,b2)$ and to $(a3,b3)$'...? I think I'm quite familiar with a 'slope' used in real analysis, but that involves two endpoints of a line, whose slope is considered. Can you, please, define algebraically a slope of two lines given with three points, one through Point1 and Point2, and the other one through Point1 and Point3...? $\endgroup$
    – CiaPan
    Feb 17 at 14:43

1 Answer 1


I'm sorry, but I cannot understand how you even came by this conception. You seem to think that all complex numbers must be roots of unity.

All it takes to construct a point in $\Bbb C^2$ is to pick four arbitrary real numbers. Any four numbers will do. They do not even need to be distinct. For example $0, 1, 0, \pi$ gives us the point $(0 + 1i, 0 + \pi i) = (i, \pi i)$.

Examples of parallel lines where the ratios do not have to be roots of unity are easy to construct. It is simplest to let the shared point $(a_1, b_1)$ be $(0,0)$.

Now choose some arbitrary other point for $(a_2, b_2)$. For example, $(i,e+i)$. Next choose some arbitrary complex number to serve as a ratio, say $w = 1 - i$. Then set $a_3 = wa_2, b_3 = wb_2: (a_3, b_3) = (1 + i, e+1 + (1-e)i)$, and there you go:

$$\dfrac{b_2 - b_1}{a_2 - a_1} = \dfrac{e+i - 0}{i - 0} = 1-ei\\\dfrac{b_3 - b_1}{a_3 - a_1} = \dfrac{e+1+(1-e)i - 0}{1+i - 0} = 1-ei$$

The lines are parallel (and since they both pass through $(a_1, b_1)$, that means they are in fact the same line). And the "slope" is not a root of unity. Indeed, it is not the root of any polynomial having rational coefficients.

Having $(a_1, b_1) = (0,0)$ is not the problem here. Just translate it by adding some non-zero value to all three points, say $(1,-e)$: $$\begin{align}(a_1,b_1) &= (1, -e)\\(a_2,b_2) &= (1+i, i)\\(a_3, b_3) &= (2+i, 1 + (1-e)i)\end{align}$$ The ratio calculations do not change.


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