# Is the following claim true in $\mathbb{C}^{2}$ about parallel lines?

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.

• "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. Feb 16 at 20:44
• "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. Feb 16 at 20:55
• I fixed the write up for lines in $\mathbb{R}^2$ Feb 17 at 2:43
• 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...? Feb 17 at 14:43

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.