Assume we have a vector space $V$ over the field $F$. A linear combination of elements $v_1, \dots, v_k$ of $V$ is an expression of the form $$ c_1 v_1 + \dots + c_k v_k, \quad \text{where } c_i \in F. $$ My question is: what's the point of $c_i$ being in $F$?
My guess is to be able to represent all elements of $V$ using linear combinations of $v_i$. However, this guess got me confused when $F$ is the complex field.
For the sake of argument, assume $V=\mathbb{R}^2$. It is clear to me that if we consider the linear combination (with linearly independent $v_1, v_2 \in V$) $$ c_1 v_1+ c_2 v_2, \quad \text{with } c_i \in F = \mathbb{N} \text{ (natural numbers)} $$ then there would be elements of $V=\mathbb{R}^2$, which we cannot represent using this linear combination, since all possible combinations can't fill all of $V=\mathbb{R}^2$. Hence, we resort to $c_i$ in $F = \mathbb{R}$ that does the job.
I fail, however, to have a similar understanding when $V=\mathbb{C}^2$ for why should $F = \mathbb{C}$ and not $\mathbb{R}$. That is, I can't see which elements of $V = \mathbb{C}^2$ can't be represented using $$ c_1 v_1+ c_2 v_2, \quad \text{with } c_i \in F = \mathbb{R} \text{ (real numbers)} $$ I guess my imagination fails me when it comes to complex numbers... or maybe they're not called complex for nothing after all :-)
Edit
Thinking about a simpler case where $V = \mathbb{C}$, linear combinations (actually scaling a single element) $$ c_1 v_1, \quad \text{with } c_i \in F = \mathbb{R} \text{ (real numbers)} $$ would not allow to fill up all the complex plane. So this might be a reason for taking $c_i \in F = \mathbb{C}$.
Any further comments, corrections are welcome.
Disclaimer: I wanted this to be a comment, but it is too long for a comment.
We discard $\mathbb N$ because it is not a field to begin with.
Given an abelian group $V$, there are virtually many fields over which we can consider $V$ a vector space. Formally speaking, a vector space is a 2-tuple $(V,F)$, where $V$ is an abelian group and $F$ is a field acting on $V$ in such a way that etc. So, in some sense, the field over which you are considering your object to be a vector space is not intrinsically related to your object; you choose it (out of all the possible candidates).
For the case where $V=\mathbb C^2$, since both $\mathbb R$ and $\mathbb C$ are fields that act on $V$ appropriately (i.e. according to the axioms of a vector space), they are both candidates for a field over which $V$ can be considered a vector space. However, $\mathbb C^2$ as a vector space over $\mathbb R$ is not the same thing as $\mathbb C^2$ as a vector space over $\mathbb C$.
I don’t think I understand your question, but I thought I’d point these out because I sensed that you might have a misunderstanding of some sort. I hope this helped.