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.

 

[mathwonk]

If the z2- partial derivative of f at p is non -zero, then near p, on f=0, z2 is indeed a single valued function of z1; but if both partials of f are zero at p, then in general z2 is locally near p only a multiple valued function of z1. I.e., in appropriate coordinates, with p = (0,0), there is in general a parametrization t-->t^n = z1(t), and t-->z2(t), of (each branch of) the curve f=0 in a neighborhood of p. Thus for each value of t^n near t=0, the curve f=0 determines one value of z1, and n values of z2.

I.e. in appropriate coordinates, a punctured neighborhood of p on the curve f=0 always projects as an n to 1 covering space of a punctured neighborhood of z1=0, but when the partials both vanish at p, we will have n > 1.

E.g. y is a 2-valued function of x on every small nbhd of (0,0) for the curve y^2 - x^3 = 0, (with x, y complex).

https://en.wikipedia.org/wiki/Cusp_(singularity)

 

[hax]

the function f , if it is analytic in both z and zbar

then it has poles

f1
f2

which means that the contour cannot be shifted in f1

so f2 degenerates and the essential singularity is the answer to the integral

proof

int f1 f2 = const. f2

so f = f(x)

and I = int f = f2

integrate around f2 and you get the residue

I = f/f2

 

[FactChecker]

the function f , if it is analytic in both z and zbar

This is quite a different problem than the OP asks about.
I would have to think if this is even possible for a non-constant function.
For a function $f(z)=u(x+iy)+i*v(x+iy)$, doesn't that imply the simultanious Cauchy-Riemann equations:
$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$
and
$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial -y} = -\frac{\partial v}{\partial y}$
So $\frac{\partial u}{\partial x}= - \frac{\partial u}{\partial x} \equiv 0$

 

[WWGD]

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.

Can $z_1$ always be expressed in terms of $z_2$? This isn't always possible for Real variables?

 

[mathwonk]

It depends what you mean by "expressed". Do you mean for every z2 near the given one, there is a unique z1 near the given one such that f(z1,z2) = 0, (i.e. z1 is a well defined set theoretic function of z2 determined locally by f=0), and that z1 is given by a power series in z2, (i.e. z1 is even an analytic function of z2)? For this, you need the hypotheses of the implicit function theorem to hold, i.e. you need the partial of f wrt z1 to be ≠0, both in the real and complex cases. Or is the first statement enough, i.e. f=0 determines z1 as a well defined function of z2, but not necessarily analytic? (In the real case this can occur for z2 = z1^3, but apparently not in the complex case.)

Or is it enough that for every z2 near the given one, there is at least one z1 such that f(z1,z2) = 0, i.e. that f=0 determines z1 as a multi - valued function of z2, not necessarily analytic. This does not always hold in the real case since then f=0 can have the given point as an isolated zero (say f(x,y) = x^2+y^2) at (0,0)), but this cannot happen in the complex case unless f=0 is (parallel to) the line z2=0. I.e. other than in this one very special case, the projection from the locus f=0 onto the z2 axis always contains a nbhd of the given point.

I.e. the main difference between the real and complex cases is that, in the complex case the locus where f(z,w) = 0 is of pure (complex) dimension one, i.e. there are no isolated points and this zero locus is a complex curve. Thus unless this curve is parallel to the line z2=0, it always projects to a nbhd of the given value of z2, and hence locally near that value, the curve f=0 is the graph of a multi-valued function of z1. The curve f=0 even decomposes into a finite number of irreducible components locally near the given point, say (0,0), and each of those components looks, in some coordinates, like the graph of z2 = z1^n, or z1 = z2^n. I tried to say this briefly in post #7.

edit: apparently another difference is, (as partially shown by z2 = z1^3), in the real, but not complex case, z1 can be determined by f=0, locally as a set theoretic, but non analytic, function of z1. i.e. the real cube root function is well defined but not analytic at 0.

 

[WWGD]

It depends what you mean by "expressed". Do you mean for every z2 near the given one, there is a unique z1 near the given one such that f(z1,z2) = 0, (i.e. z1 is a well defined set theoretic function of z2 determined locally by f=0), and that z1 is given by a power series in z2, (i.e. z1 is even an analytic function of z2)? For this, you need the hypotheses of the implicit function theorem to hold, i.e. you need the partial of f wrt z1 to be ≠0, both in the real and complex cases. Or is the first statement enough, i.e. f=0 determines z1 as a well defined function of z2, but not necessarily analytic? (In the real case this can occur for z2 = z1^3, but apparently not in the complex case.)

Or is it enough that for every z2 near the given one, there is at least one z1 such that f(z1,z2) = 0, i.e. that f=0 determines z1 as a multi - valued function of z2, not necessarily analytic. This does not always hold in the real case since then f=0 can have the given point as an isolated zero (say f(x,y) = x^2+y^2) at (0,0)), but this cannot happen in the complex case unless f=0 is (parallel to) the line z2=0. I.e. other than in this one very special case, the projection from the locus f=0 onto the z2 axis always contains a nbhd of the given point.

I.e. the main difference between the real and complex cases is that, in the complex case the locus where f(z,w) = 0 is of pure (complex) dimension one, i.e. there are no isolated points and this zero locus is a complex curve. Thus unless this curve is parallel to the line z2=0, it always projects to a nbhd of the given value of z2, and hence locally near that value, the curve f=0 is the graph of a multi-valued function of z1. The curve f=0 even decomposes into a finite number of irreducible components locally near the given point, say (0,0), and each of those components looks, in some coordinates, like the graph of z2 = z1^n, or z1 = z2^n. I tried to say this briefly in post #7.

edit: apparently another difference is, (as partially shown by z2 = z1^3), in the real, but not complex case, z1 can be determined by f=0, locally as a set theoretic, but non analytic, function of z1. i.e. the real cube root function is well defined but not analytic at 0.

Some thing like the results of the inverse/implicit function theorems.

 

[mathwonk]

Yes, the appropriate generalization of the implicit function theorem for this situation is the "Weierstrass preparation theorem". This may indeed be the best way to think about it but I always forget it. The theorem is this:

Th: Given f(z,w) analytic near (0,0), with f(0,0) = 0, if the analytic function of one variable defined by f(0,w) has a zero of finite order n in w at w=0, then f has near (0,0) the same zero locus as a polynomial of form
P(z,w) = w^n + a1(z)w^(n-1) + ...+ an(z), where the aj(z) are analytic functions in z, with aj(0) = 0 for all j.

More precisely, there is an analytic function h(z,w), with h(0,0) ≠ 0, such that near (0,0), f = P.h, so that f and P have the same zero locus there.

The implicit function theorem is the case n=1, so that then f=0 locally equals the graph of the analytic function w = -a1(z).