My motivation for this question is that I'm wondering if the subspace Zariski topology of $\mathbb{R}^n\subseteq\mathbb{C}^n$ is the same as the Zariski topology over $\mathbb{R}^n$, something that would be true if $\mathbb{R}^n$ was a variety of $\mathbb{C}^n$.
Since for one variable, $\Im(z)=0$ gives me the real axis I can easily generalise this to $n$ variables. I can write $\Im(z)=\frac{z-\overline{z}}{2i}$, so my question reduces to:
Is there a way to express $\overline{z}$ as a complex polynomial on the variable $z$?
No. If $\bar z = f(z)$ was a polynomial in $z$, then $\|z\|=z\cdot\bar z=z\cdot f(z)=:g(z)$ would be a complex polynomial whose only zero is $0$, therefore you'd have $g(z)=\alpha \cdot z^{k+1}$ for some $k\in\Bbb N$ and $\alpha\in\Bbb C^\times$, hence $f(z)=\alpha\cdot z^k$. Substituting $2$ and $3$ for $z$ quickly leads to the conclusion that $f(z)=z$ or $2^{-k}=\alpha=3^{-k}$, the latter being absurd. This yields the contradiction $-i=f(i)=i$.