This is a homework question in a SCV course, and I'm struggling to approach the problem:
Suppose $f:\mathbb{C}^2 \rightarrow \mathbb{C}$ is an entire holomorphic funtion whose zero set contains the set $ N = \{(z_1,z_2) \in \mathbb{C}^2: z_1=\overline{z_2}\ \}$, prove that $f$ is identically equal to zero.
Now, I am familiar with the identity theorem, stating, handwavingly, that if the zero set of $f$ contains some open set $N\subset \mathbb C^2$ then $f$ is identically equal to $0$.
However, I think I could show that the given set is not open by considering that: $z_1 = \overline{z_2}$ is in some sense the hypersurface $\{ (x_1,y_1,x_1,-y_1) \in \mathbb R^4: x_1,y_1 \in \mathbb R \}$.
Next I thought I would try to use Hurwit'z theorem:
Suppose $\{f_k\}$ is a sequence of nowhere zero holomorphic functions on a domain $U \subset \mathbb C^n$ converging uniformly on compact subsets to a function $f$ . Show that either $f$ is identically zero, or that $f$ is nowhere zero.
Let $(z_{1,0},z_{2,0}) \in N$. If we define $f_k(z) = f(z,z_{2,0}+\frac{1}{k})$, then $f_k$ converge uniformly to $g(z) = f(z, z_{2,0})$ and $g(z_{1,0})=0$ so $g(z)$ must be identically equal to zero (?).
Can I from this assert that there must be zeros of $f$ in some small polydisc about $(z_{1,0}, z_{2,0})$?, hence be able to apply the identity theorem and conclude that indeed $f$ is identically equal to zero?
Next the same question is asked when the zero set of $f$ contains $N' = \{(z_1,z_2)\in\mathbb T^2:z_1=\overline{z_2}$}, I don't see directly why the same argument would not apply in this case. That is, can I still draw the conclusion that $f$ must be identically zero?
We have covered the first chapter of: https://www.jirka.org/scv/
Any direction or hints tied to the above literature would be most helpful!
We want to show that the real two-dimensional plane $\{ z_1 = \overline{z}_2 \}$ is a set of uniqueness.
There exists a complex linear transformation $A : \mathbb{C}^2 \longrightarrow \mathbb{C}^2$ which takes $\{ z_1 = \overline{z}_2 \}$ to $\{ \Im z_1 = \Im z_2 = 0 \}$. Hence, we can map our space linearly to a totally-real space. By the uniqueness theorem on p. 21 of Shabat's Complex Analysis -- Several Complex Variables, if $f$ vanishes on a totally-real space, then $f$ vanishes identically.