Zeros of functions analytic in 2 variables

[Paul Colby]

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.

 

[Paul Colby]

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.

 

[mathwonk]

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

 

[Paul Colby]

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.

 

[mathwonk]

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.

 

[Paul Colby]

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.