## Some uniqueness sets of holomorphic function

Suppose ${f\in \mathcal{O}(\mathbb{C}^2)}$, if ${f}$ is zero on the following set
(a) A real hyperplane in ${\mathbb{C}^2}$
(b) The real two dimensional plane ${\{z_1=\bar{z}_2\}}$
(c) the arc ${\{z_1=\bar{z}_2,y_1=x_1\sin(1/x_1)\}}$
Show that ${f}$ must be zero on ${\mathbb{C}^2}$.

Proof: (a) Suppose the hyperplane has normal vector ${(1,0)\in \mathbb{C}^2}$ and passing through origin, then ${(iy,z_2)}$ will be in this hyperplane for any ${y\in \mathbb{R}}$.

Fix ${z_2}$, ${f(\cdot,z_2)}$ is a holomorphic function in ${\mathbb{C}}$. Because zero point of holomorphic function is isolated unless a constant, ${f(iy,z_2)=0\quad\forall\,y\in \mathbb{R}}$ will imply ${f(z_1,z_2)\equiv 0}$.

(b) Let ${w_1=\frac{z_1+z_2}{2}}$, ${w_2=\frac{z_1-z_2}{2i}}$, then

$\displaystyle \{z_1=\bar{z}_2\}=\{\text{Im}\,w_1=\text{Im}\,w_2=0\}$

${g(w_1,w_2)=f(z_1,z_2)}$ is also a holomorphic function in ${w_1,w_2}$. As we did in part (a), fix ${w_1=x\in \mathbb{R}}$, ${g(x,y)=0}$ imply ${g(x,w_2)=0}$ for any ${w_2\in\mathbb{C}}$. Since ${x}$ is arbitrary, then ${g(w_1,w_2)\equiv 0}$.

(c) Use the transformation of part (b), ${\{z_1=\bar{z}_2, y_1=x_1\sin \frac{1}{x_1}\}}$ is equivalent

$\displaystyle \gamma=\{\text{Im}\,w_1=\text{Im}\,w_2=0,w_2=w_1\sin\frac{1}{w_1}\}$

Consider ${g(w_1,w_2)}$, then ${g(\cdot,0)=0}$ has a sequence of roots ${w_1}$ converging to 0, then ${g(w_1,0)=0}$ for any ${w_1\in \mathbb{C}}$. From the Taylor expansion, there exists ${g_1\in \mathcal{O}(\mathbb{C}^2)}$ such that

$\displaystyle g(w_1,w_2)=w_2g_1(w_1,w_2)$

Since ${g(w_1,w_2)}$ is zero on ${\gamma}$, then ${g_1(w_1,w_2)}$ is zero on ${\gamma}$ when ${w_2\neq 0}$. By continuity, ${g_1(w_1,w_2)}$ is zero on ${\gamma}$.

Then we can repeat the above procedure, ${\exists\, g_2}$ such that

$\displaystyle g_1(w_1,w_2)=w_2g_2(w_1,w_2),\quad g=w_2^2g_2$

Then we must have ${g\equiv 0}$, otherwise we can continue this process infinitely. $\Box$

Remark: The idea of part (b) is due to Lun Zhang.