TL;DR

Let, ##f(z_1,z_2)##, be entitre analytic in both ##z_1## and ##z_2##. If ##f(z_1,z_2)=0## has a solution, is it true that ##f## is zero along some curve in ##z_1,z_2##?

This question crops up in solving electromagnetic boundary value problems. For problems with rotational symmetry, if ##f## has a node at ##(\theta,\phi)## then ##(\theta,\phi')=0## for all other ##\phi'##. This (I think) implies that, $$f(\theta,\phi)=F(\theta)G(\phi)$$ which, for the problems I'm considering, isn't going to happen except under very special circumstances.

My thinking is, if a node exists at ##(z_1,z_2)##, then some curve in ##(z_1,z_2)## must exist along which ##f## is zero.

Ah, problem solved. In the form asked, one simply solves ##f(z_1,z_2)=0## for ##z_2## in terms of ##z_1##. The real problem that was troubling me is addressed by realizing real polar coordinates, ##(\theta,\phi)##, are in fact just a single complex coordinate. The functions, ##f##, are really entire functions of a single complex coordinate. The zeros of analytic functions that are not identically zero, are all isolated points.

are z1, z2 complex variables? if so, the zeroes of an analytic function of 2 complex variables are never isolated.

mathwonk said:

are z1, z2 complex variables? if so, the zeroes of an analytic function of 2 complex variables are never isolated.

I was slowly tumbling to this fact. ##z_1## and ##z_2## correspond to analytically continued polar coordinate angles. In my problem, each zero corresponds to a new vector in a separable Hilbert space. Non-isolated zeros are problematic. Fortunately, the sphere is conformally mapped to the complex plane using, $$\xi=e^{i\phi}\cot(\theta/2)$$. So looking for isolated zeros makes more sense.

I don't know if the argument for my statement is of interest, but it is nice and fairly short; namely, Riemann showed for analytic functions of one complex variable, if they are bounded in a punctured neighborhood of an isolated point, then they extend analytically also to that point.

Hartogs proved the analogous result for analytic functions of 2 complex variables, without assuming boundedness; i.e. a function of two complex variables which is analytic in a punctured neighborhood of an isolated point, always extends analytically also to that point.

Taking logs gives the desired result: i.e. if f is an analytic function of 2 complex variables, and f is non zero on a punctured neighborhood of an isolated point p, then since such a punctured neighborhood is simply connected, f has an analytic logarithm g = ln(f), on that punctured neighborhood. But then by Hartogs that logarithm g extends analytically to p, so f extends also to p as f = exp(g). Since the exponential function is never zero, f(p) = exp(g(p)) ≠ 0.

Cool. For an analytic function, ##f##, in two complex variables one may solve ##f(z_1,z_2)=0## for ##z_2## as a function of ##z_1##. Zeros, as you pointed out, are not isolated points. For analytic functions of a single variable, the zeros are isolated if the function is not identically 0.