# How to write the equation of a line in $\mathbb C^n$

I want to write the equation of a line in $\mathbb C^n$ passing through a point $(z_1,z_2,...,z_n)$. Actually I have a set of points and I suspect they all lie in the same line which passes through this point and I want a convenient way to check it.

Thank you

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You're missing a way to indicate direction. Many possible lines can pass through one point. Having that, it's easy to set up parametric equations. –  Ｊ. Ｍ. Oct 28 '10 at 5:55
As I said, I have a set of points and I want to check if they are collinear. How should I go about it? Also there are usually many ways to write equations of a line in $mathbb R^2$ and $mathbb R^3$ I am not sure which of these has an analog in $mathbb C^n$ –  novice Oct 28 '10 at 6:05

Since $\mathbb{C}$ is $2$-dimensional, as a real vector space, a one-dimensional subspace of $\mathbb{C}^n$ is actually a complex plane, and is therefore not an ordinary line.

However, such subspaces/"lines" can be expressed as in the real case, as a solution to $n-1$ equations of the form $\lambda_1 z_1+\lambda_2 z_2+ \cdots+ \lambda_n z_n = c$ where c is $0$ if you want a subspace, any complex number if you want any set that "looks" like $\mathbb{C}$.

Edit: Fixed error.

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Thank you, the set that looks like $\mathbb{C}$ is what I was looking for. Does it have a name ? –  novice Oct 28 '10 at 8:49
Unless I'm misinterpreting what you wrote, it looks like you have a single equation there. That would be a "hyperplane" and hence an $(n-1)$-dimensional subspace, not a 1 dimensional one. –  Matt Oct 28 '10 at 16:27
Calling one dimensional complex subspaces of $\mathbb C^n$ "planes" will only bring pain. Don't! –  Mariano Suárez-Alvarez Oct 29 '10 at 14:40

How much do you know about vectors? You can take two points and find the difference between them, as a vector; this is the analogue of the slope of a line in $\mathbb{R}^2$. If the vector is $\mathbf{v}$ and one point is $a$, then points on the line are of the form $a+\lambda\mathbf{v}$, where $\lambda$ is real. (Unless you're talking about a "complex line," which is really a plane.)

Also, if you're just uncomfortable with $\mathbb{C}^n$, you can just rewrite all your points as points in $\mathbb{R}^{2n}$, like $(Re(z_1),Im(z_1),\dotsc,Re(z_n),Im(z_n))$. This is an isomorphism of the real vector space structure of $\mathbb{C}^n$, so in particular it preserves lines and addition. The only thing you lose is complex multiplication.

EDIT: If you're looking for a complex line, just let $\lambda$ be complex in the above vector expression.

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I am confused about the dimension, by line I mean something which is a translate of a one dimensional subspace of the complex vector space $\mathbb{C}^n$ –  novice Oct 28 '10 at 6:49
Thank you, it was very helpful. –  novice Oct 28 '10 at 8:50

It doesn't matter if you work with complex, real numbers, or elements of any field $\mathbb{K}$: if you have a point $p = (z_1, \dots , z_n) \in \mathbb{K}^n$, or any $\mathbb{K}$-vector space $V$, an equation for a straight line in $\mathbb{K}^n$ (or in $V$) passing through $p$ may always be written, for instance, as

$$p + \lambda v \ ,$$

with $v= (v_1, \dots , v_n) \in \mathbb{K}^n$ (or $v\in V$) and $\lambda \in \mathbb{K}$.

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