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.


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