Does $\mathbb{C}\setminus\{0\}$ with those operations constitute a vector space over $\mathbb{C}$? If given set $V = \mathbb{C}\setminus\{0\}$ and field $\mathcal{F} = \mathbb{C}$ with operations defined for all $\vec{x}, \vec{y} \in V, \vec{x} = x \in \mathbb{C}\setminus\{0\}, \vec{y} = y \in \mathbb{C}\setminus\{0\}$ and $\alpha \in \mathbb{C}$ as:
$$\vec{x} \oplus \vec{y} = x\cdot y$$
$$\alpha \odot \vec{x} = \alpha\cdot x$$
Is it really enough to take $\alpha = 0$, which gives $\alpha\cdot \vec{x} \notin V$ to prove that this is not a vector space over $\mathbb{C}$? I.e. just one counterexample is enough, right?

The answer is yes. The set $\mathbb{C}\backslash \{0\}$ with usual multiplication is an abelian group, so $V$ with $\oplus$ is an abelian group. The problems arise with "scalar multiplication", and although one example is enough, it's also true that distributive laws fail for (almost) any $\alpha,x,y$
$$\alpha\odot (x\oplus y) \neq \alpha \odot x \oplus \alpha\odot y$$
