TITLE: Real-analytic function of two complex variables, holomorphic in first and anti-holo in second, which vanishes on the diagonal is identically zero.
QUESTION [8 upvotes]: The following theorem is stated as being a well-known result of the theory of several complex variables in a book I am reading (on a more or less unrelated subject):

Let $f:\mathbb C^2\to\mathbb C$ be a real-analytic function such that
  $f$ is holomorphic in the first variable and anti-holomorphic in the
  second variable. If $f(z,z) = 0$ for all $z\in\mathbb C$, then
  $f=0$ identically.

1) Can somebody point me to a reference where this is proved, or provide a proof that does not use extensive machinery of several complex variables?
2) What is the necessity of specifying real-analyticity? Is it false that holomorphicity and anti-holomorphicitiy respectively in the two variables implies real-analyticity?

REPLY [2 votes]: Assuming $f$ is real-analytic leads to a very simple proof; I wasn't aware of this result and know more or less nothing about several complex variables, but I found a proof more or less automatically:
At least in some neighborhood of the origin we have $$f(z,w)=\sum_{n,m\ge0}a_{n,m}z^n\overline{w}^m.$$ So for $r>0$ small enough and $t\in\Bbb R$ we have $$0=\sum_{n,m\ge0}a_{n,m}r^{n+m}e^{i(n-m)t}.$$Uniqueness for Fourier series shows that for small enough $r>0$ and every $k\in\Bbb Z$ we have $$\sum_{n-m=k}a_{n,m}r^{n+m}=0.$$
BUT given $N\ge0$ there exists at most one pair $(n,m)$ with $n-m=k$ and $n+m=N$. Which is to say that the coefficient of $r^{n+m}$ in that last series really is $a_{n,m}$, as it would appear. So one complex variable shows $$a_{n,m}=0.$$