How to show existence of a hyperplane in $\mathbb{C}^n$ which misses finitely many points Let $E\subset\mathbb{C}^n$ be a finite subset and $0\notin E$. Then show that there exists a hyperplane of $\mathbb{C}^n$ which passes through the origin and does not contain any point of $E$.
I just have a heuristic argument: If a hyperplane intersects any of the point in $E$ then we can rotate it by small angle such that it misses every point of $E$, since E is discrete. 
But I am not able to make this argument precise.
Is there any easy way to see this result?
 A: Every Hyperplane through the origin can be written as $\{x \in \mathbb{C}^n \mid \langle v, x \rangle = 0\}$, where $v \ne 0$ is a normal vector of the hyperplane. We can use this fact to prove the assertion via induction.
Let $E = \{e_1, \ldots, e_m\}$. If $m = 1$, the assertion holds trivially. For $m > 1$ we can find (by our induction hypothesis) a vector $v \ne 0$ so that $\langle v, e_i\rangle \ne 0$ holds for all $1 \le i < m$. Now note that for any $t \in \mathbb{R}$ we have
$$\begin{align*}
\langle v + t e_m, e_m \rangle &= \langle v, e_m \rangle + t \langle e_m, e_m
\rangle \\
\langle v + t e_m, e_i \rangle &= \langle v, e_i \rangle + t \langle e_m, e_i\rangle \qquad \text{for}\ 1 \le i < m
\end{align*}$$
It is easy to see that there is a $t = t_0$ so that all of these terms are nonzero (just note that each term is zero for at most one value of $t$). But this means the Hyperplane $\{x \in \mathbb{C}^n \mid \langle v + t_0e_m, x\rangle = 0\}$ does not contain any element from $e$.
