0
$\begingroup$

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

or

  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.

$\endgroup$
4
  • 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

0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.