Proof of The Complex Coordinate Space is a Complex Vector Space

Let $u,v,w\in\mathbb{C}^{n}$ and $\lambda,\mu\in\mathbb{C}$, with components and componentwise operations as in The Complex Coordinate Space. Two $n$-tuples are equal exactly when their corresponding components are equal, so each condition of Vector Space over a Field follows from the corresponding identity for each component, and those identities hold in the field of complex numbers by its field axioms. Explicitly, for each $k$ with $1\le k\le n$:

Condition 1. The $k$-th component of $(u+v)+w$ is $(u_{k}+v_{k})+w_{k}=u_{k}+(v_{k}+w_{k})$, the $k$-th component of $u+(v+w)$, by associativity of addition in $\mathbb{C}$.

Condition 2. The $k$-th components of $u+v$ and $v+u$ are $u_{k}+v_{k}=v_{k}+u_{k}$, by commutativity of addition in $\mathbb{C}$.

Condition 3. Let $0_{\mathbb{C}^{n}}=(0,\dots,0)$. The $k$-th component of $u+0_{\mathbb{C}^{n}}$ is $u_{k}+0=u_{k}$, so $u+0_{\mathbb{C}^{n}}=u$ for every $u\in\mathbb{C}^{n}$.

Condition 4. Let $w=(-u_{1},\dots,-u_{n})$, where $-u_{k}$ is the additive inverse of $u_{k}$ in $\mathbb{C}$. The $k$-th component of $u+w$ is $u_{k}+(-u_{k})=0$, so $u+w=0_{\mathbb{C}^{n}}$.

Condition 5. The $k$-th component of $\lambda(\mu u)$ is $\lambda(\mu u_{k})=(\lambda\mu)u_{k}$, the $k$-th component of $(\lambda\mu)u$, by associativity of multiplication in $\mathbb{C}$.

Condition 6. The $k$-th component of $1u$ is $1u_{k}=u_{k}$.

Condition 7. The $k$-th component of $\lambda(u+v)$ is $\lambda(u_{k}+v_{k})=\lambda u_{k}+\lambda v_{k}$, the $k$-th component of $\lambda u+\lambda v$, by the distributive law.

Condition 8. The $k$-th component of $(\lambda+\mu)u$ is $(\lambda+\mu)u_{k}=\lambda u_{k}+\mu u_{k}$, the $k$-th component of $\lambda u+\mu u$, again by the distributive law together with commutativity of multiplication.

Hence $\mathbb{C}^{n}$ with these operations is a complex vector space. By claims 1 and 2 of Elementary Identities in a Vector Space, its zero vector and the additive inverse of each vector are unique, so they are exactly the elements $0_{\mathbb{C}^{n}}=(0,\dots,0)$ and $-u=(-u_{1},\dots,-u_{n})$ exhibited in conditions 3 and 4.