Suppose we have a vector space $ \langle V, \oplus, \otimes \rangle$ where $V = \mathbb{C}$. Define $z \oplus w := |zw| \quad \forall z,w \in V$ and $\lambda \otimes z := \lambda \cdot z \quad \forall \lambda \in \mathbb{C} \quad \forall z \in V$. Verify whether each axiom for vector spaces holds.
I can verify whether most of the axioms hold/fail except I run into the following logical dilemma:
To verify whether there exists a $\overrightarrow{\mathbf{0}}$ for all $z \in V$ such that $z \oplus \overrightarrow{\mathbf{0}} = z \quad \forall z \in V$, note that $|zw|$ is always a nonnegative real number (by definition of modulus of a complex number). We take $z = 1 + i$ as a counterexample, and take an arbitrary $x$ where $x$ is our choice of the zero element (identity element under the operation $\oplus$). Then $z \oplus x \neq z$ since we get that $z \oplus x = |(1+i)x| \geq 0$ i.e. $z \oplus x \in \mathbb{R}$ for any $x$ we choose to be the zero vector. Hence such $\mathbf{\overrightarrow{0}}$ does not exist.
Next, to verify whether $\forall z \in V$, $\exists z^* \in V$ such that $z \oplus z^* = \mathbf{\overrightarrow{0}}$ (inverse element under the operation $\oplus$): We have proven that $\mathbf{\overrightarrow{0}}$ does not exist in $V$.We can take $z = 1+i$ as a counterexample, and clearly, no $z^*$ exists such that $z \oplus z^* = \mathbf{\overrightarrow{0}}$ as $\mathbf{\overrightarrow{0}}$ does not even exist in $V$ and since the zero vector of any vector space $\langle V, \oplus, \otimes \rangle$ must be unique, we conclude that the axiom fails i.e. for all $z$, there cannot exist a $z^*$ such that $z + z^* = \mathbf{\overrightarrow{0}}$. No inverse element under $\oplus$ exists.
However, since $\mathbf{\overrightarrow{0}}$ does not even exist in $V$, shouldn't I be able to conclude that the second aforementioned axiom holds by vacuous truth? Then we would have a logical contradiction. Is my approach given above correct?