Let $f(z_1,z_2)=z_2$ where $(z_1,z_2)\in \mathbb{C}^2$.
I have a few questions regarding this function:
- This function is typically used as a canonical example of a function that is zero on $\mathbb{R}^2$ but not on $\mathbb{C}^2$. This is done by saying that the set of zeros is given by \begin{align} N(f)= \{ \mathbb{C} \cdot e_1\} \end{align} where $e_1$ is one of the standard bases and identifying $\mathbb{C} $ with $\mathbb{R}^2$ and, hence, $N(f)$ with $\mathbb{R}^2$. However, this seems so unnatural. Why is this the example that is used? I guess I am missing something very fundamental in my understanding and would like for this point to be clarified.
- Is there a better example of a function that is zero on $\mathbb{R}^2$?
I suspect the answer to (2) illuminates the answer to (1). In particular, the inclusion of $\mathbb R$ into $\mathbb C$ gives a natural inclusion of $\mathbb R^2$ into $\mathbb C^2$ - consisting, namely, of the pairs $(z_1,z_2)$ where both $z_1$ and $z_2$ are real.
So, a natural example of such an $f$ would be a function $f(z_1,z_2)$ such that $f$ vanished when both $z_1$ and $z_2$ were real. This, however, does imply that $f=0$.
To see this, define $g_z(z_2)=f(z,z_2)$ - that is, a function obtained by fixing an argument. Note that $g_z$ is also complex analytic. However, when $z$ is real, $g_z$ is zero on $\mathbb R$ which implies that $g_z=0$ everywhere, using facts about complex functions of one variable. Thus $f(z_1,z_2)=0$ whenever $z_1$ is real. However, we can repeat the argument for the other variable, letting $h_z(z_1)=f(z_1,z_2)$. Then, for every $z$ we get that $h_z$ vanishes on the real axis, thus is zero, thus $f(z_1,z_2)=0$ everywhere. So, there are no really good examples of this occurring - they all have to be unnatural like the one you encountered.
I would guess the point that the projection map $f(z_1,z_2)=z_2$ is there to make is that the vanishing set of a complex-analytic function of two variables behaves quite differently from the vanishing set of a one variable function (which, for instance, is always discrete). I think it's rather clunky to think of the vanishing set as $\mathbb R^2$ - it's really $\mathbb C \times \{0\}$ in set notation - but some people think of $\mathbb C$ as $\mathbb R^2$, so would think of $\mathbb C^2$ as $\mathbb R^2\times \mathbb R^2$, which justifies the claim.