I claim the set of three linearly independent vectors in $\mathbb{C}^2$ to be empty.
But I'm not sure if the following proof is correct.
Let $x$ and $y$ be linearly independent vectors in $\mathbb{C}^2$ over the field $\mathbb{C}$. Assume that $z \neq 0$ in $\mathbb{C}^2$ is not a linear combination of $x$ and $y$. Then, for all $\alpha_1, \alpha_2$ in $\mathbb{C}$, we have that
$$ \begin{eqnarray} z & = & z_1 (1, 0) + z_2 (0, 1) \\ & \neq & \alpha_1 x + \alpha_2 y \\ & = & (\alpha_1 x_1 + \alpha_2 y_1)(1, 0) + (\alpha_1 x_2 + \alpha_2 x_2)(0, 1). \end{eqnarray} $$
This requires that either $z_1 \neq (\alpha_1x_1 + \alpha_2y_1)$, or $z_2 \neq (\alpha_1x_2 + \alpha_2y_2)$, or both.
But if $z_1 \neq (\alpha_1x_1 + \alpha_2y_1)$, for all $\alpha_1, \alpha_2$, this can only be the case if $x_1 = y_1 = 0$. This, in turn, causes $\{x, y\}$ to be linearly dependent, contradicting the assumption. A similar argument applies for $z_2$. Hence, either $z = 0$ or $z$ is a linear combination of $x$ and $y$.