When I was in grad school, I noticed a funny fact about the zero sets of analytic and harmonic functions. I showed it to several people, and they all shrugged it off as uninteresting.

I guess I’m weird like that, so I’m going to share it with you all in hopes someone else finds it intriguing.

Analytic and Harmonic Functions

I don’t want to get too bogged down in definitions for this. If you’re familiar with these terms, then you already know the definitions. If you’re not, then the definitions don’t really help you understand these types of functions.

Let’s write complex numbers as z=x+iy where x and y are real numbers. An analytic function is just a one-variable function, f(z), that has a derivative. It turns out this is a much stronger condition than for real-valued functions.

Once there is one derivative, there are actually infinitely many. Analytic functions also have power series expansions locally (usually this is the definition).

All the “normal” functions you think of are analytic: polynomials, exponentials, trigonometric functions, etc. Composing, adding, subtracting, multiplying these gives you even more.

A two-variable, real function u: ℝ²→ ℝ is a harmonic function if it satisfies Laplace’s equation (you may have seen this in a first course on differential equations).

Again, don’t dwell too much on this if you haven’t seen it. It’s just a special type of function. I’ll tell you the properties you need to know as they come up.

The key thing for right now is that that analytic and harmonic functions are related very closely. We can write a complex function as f(z)=u(x,y)+i v(x,y) by breaking up the real and imaginary parts.

If f(z) is analytic, then the real part Re(f)=u(x,y) and the imaginary part Im(f)=v(x,y) are both harmonic functions.

We’ll look at some examples in a moment.

The converse is also true. If you have a harmonic function u(x,y), then you can find another function v(x,y) so that f(z)=u(x,y) + i v(x,y) is analytic.

The details aren’t important. The fact is that harmonic functions are just real and imaginary parts of analytic functions. I really want you to let that soak in. The one is always just a part of the other.

It’s what makes the result of this article so surprising.

Examples

Let’s take the easiest examples we can think of and look at what the zero sets look like since that’s what we’ll be considering later.

The function f(z)=z is about as simple an analytic function as it gets. If we solve f(z)=0, we get that z=0 is the only point. The analytic function has a single place where it is 0.

Now if we write it as f(z)=z=x+iy, then it’s easy to write down the real and imaginary parts, Re(f)=u(x,y)=x and Im(f)=v(x,y)=y.

Well, Re(f)=0 when x = 0. This is the whole y-axis. It’s a line. Likewise, Im(f)=0 when y=0. This is the whole x-axis.

The place where Re(f)=0 and Im(f)=0 intersect is (0,0), the place where f(z)=0.

Let’s do one more: f(z)=z²+1. This has exactly two places where it is 0 at z=i and z=-i.

This one takes a slight bit of algebra:

f(z)=(x+iy)²+1
=x²+2ixy+(iy)²+1
=(x² -y²+1)+i(2xy)

So Re(f)=u(x,y)=x²-y²+1 and Im(f)=v(x,y)=2xy. Well, Im(f)=0 when x=0 OR y=0. This means that the zero set consists of both axes.

Re(f)=0 is a hyperbola.

So, again we have u(x,y)=0 is a curve, v(x,y)=0 is a curve, and f(z)=0 consists of two isolated points.

An Aside on These Functions

Before I point out and prove the curious observation, I want to reiterate that these functions can be weird. I used polynomials above, and that can lure you into thinking the below facts are “obvious.”

But if we go back to our definition, an analytic function is just a complex function that has a derivative. It could be something like f(z)=cos(z)-eᶻsin(z²) or even much crazier.

The zero set is not obvious, and the real and imaginary parts aren’t easy to write down.

Second, harmonic functions are just functions that satisfy a certain differential equation. It’s a theorem that they are real and imaginary parts of analytic functions.

This should be a surprising fact if you haven’t seen the proof.

Analytic Functions have Isolated Zeros

If you’ve taken complex analysis, then you’ll probably recognize the result that (non-zero) analytic functions have isolated zeros. It’s one of the most common early things to prove once you have some of the standard tools.

I won’t try to prove it without these tools, but I’ll give you an idea of what it means.

It’s basically the same idea as the two examples above. When we had an analytic function f(z), we saw that f(z)=0 was just a single point in one example and two separate points in another.

Isolated means that if f(ζ)=0 for some ζ, then we can find a radius small enough so that no other point inside the circle of radius r centered at ζ has the property that f(z)=0.

Another way to say this is that the zero set of an analytic function consists of a bunch of separated points.

Zeros of Harmonic Functions are Never Isolated

Harmonic functions have the opposite property. The zeros can never be isolated. This means that whenever u(a,b)=0, no matter how small a radius you pick, you’ll always be able to find another place the function is zero inside a circle centered at (a,b).

Think of it like in the example; the zero sets were curves.

This actually tends to not be proved in standard classes for some reason, but it’s a really cool fact that comes right out of one of the fundamental properties of harmonic functions.

Harmonic functions satisfy something called the mean value property. This says that if you pick a point (a,b), the value of the function at that point, u(a,b), is equal to the average of the function on a circle centered at (a,b).

I won’t prove that, but once you believe it, it’s neat to prove that the zeros of harmonic functions are never isolated.

Suppose u(a,b)=0. Now consider any radius r>0, no matter how small. I’ll show you how to produce a zero on the circle of radius r centered at (a,b). If we do that, then we’ve proven the zeros are never isolated (there’s always another place it’s zero close by).

Well, pick any point on the circle, say (c,d). There are three cases: u(c,d) is positive, zero, or negative.

If u(c,d)=0, then we’re done. We’ve found a place it’s zero.

If it’s positive, then there must be some other place where it’s negative. This is because the average of u on the circle is equal to u(a,b)=0 (the mean value property), and you can’t take an average of all positive numbers and still get zero.

So, we know there’s some point (e,f) on the circle for which u(e,f) is negative. By definition, u(x,y) is a continuous function on the circle, and so by the Intermediate Value Theorem of Calculus, there must be a point between (c,d) and (e,f) where u is 0.

You can run the same argument if u(c,d) was negative. That’s all three cases, and so we’re done! Zeros of harmonic functions can never be isolated.

The Curious Observation

Now we can put it all together to see what’s weird. Analytic functions always have isolated zeros. Harmonic functions are just pieces of analytic functions, and yet harmonic functions never have isolated zeros.

Isn’t that a curious observation?