Is there a nontrivial analytic function which takes the same values on disjoint circles?

Question

Given two disjoint circles $C_1$ and $C_2$ in the complex plane, is there a nontrivial analytic function $f(z)$ whose values on $C_1$ are the same as its values on $C_2$ up to rotation and dilation?

What little intuition I have about this suggests not. If we had a function $f = u+i~v$ and think of the harmonic functions $u,v$ as potential distributions, we could consider the special case of two nearly osculating circles of very different size. Informally (and far too vaguely) the mutual patterns of influence of the two circles seem qualitatively different, so it seems unreasonable to suppose the functions could be the same. This argument does not work for circles of the same size.

As a guess we would have that $f(z)=u(x,y)+i~v(x,y) = f(w(z))$ for some Mobius transformation $w(z)$ and maybe the Cauchy-Riemann equations cannot be satisfied? I was hoping for an obvious geometric argument but maybe it's not possible?

Edit: It seems that this may be possible for same-radius circles (I didn't want to rule these out) and I would upvote/accept an answer along these lines, but would add 100 points to an answer that proves it's impossible [re-edit: and kal v'chomer possible] for different-radius circles.

Answer 1

Since an entire function is completely determined by its value on a circle, your requirement is in fact very, very strong.

Let $\tau : \Bbb C \to \Bbb C$ be a composition of a translation a rotation and a scaling (which means a map $z \mapsto az+b$ for some $a \in \Bbb C^*$ and $b \in \Bbb C$).

If $f \circ \tau = f$ on some circle $C$, then we must have $f \circ \tau = f$ on all of $\Bbb C$.

If $a \neq 1$, then $f$ has a fixpoint $z_0 = b/(1-a)$, and $f$ has Taylor series around $z_0$, $f(z) = \sum c_k (z-z_0)^k$.
The equation now says $f(z) = f(\tau(z)) = \sum c_k (\tau(z)-z_0)^k = \sum c_k (a(z-z_0))^k $, and so $c_k = a^kc_k$ forall $k$.

From there we get that if $a$ is not an $n$th root of unity for some integer $n$ then $f$ has to be constant, and if it is an $n$th root of unity, then $f$ is a function of $(z-z_0)^n$. For exemple, $f(z) = z^2$ is symmetric around $0$ so you can find two circles with the same image, and so on.

If $a=1$ then $\tau$ is a translation, and the equation simply says that $f$ is $b$-periodic. Now it is a classic result that the $b$-periodic entire functions are exactly the functions of $\exp(2i\pi z/b)$.

Answer 2

[It turns out that this solution was implicit in the version I posted earlier.]

By "circle" below, I'll mean a closed disk, including the circumference. $C(c,r)$ is the circle with center $c$ and radius $r.$

Let $C_1=C(c_1,r_1)$ and $C_2=C(c_2,r_2)$ be two circles in the complex plane. Write $A=\frac{r_2}{r_1}$ and $B=c_2-A c_1,$ so that $k(z)=A z+B$ is the function which maps the circumference of $C_1$ to the circumference of $C_2$ in the natural way (without rotating).

We're interested in finding an analytic function $f$ defined on a region containing $C_1\cup C_2$ such that $f(z)=f(k(z))$ for all $z$ on the circumference of $C_1.$ (This is a stricter requirement than in the question in that we don't allow for rotations, but that's OK because we can find functions $f$ satisfying the stricter requirement.)

We'll find solutions if neither circle is entirely contained in the other. (Of course, the case where $C_1=C_2$ also has a solution, namely $f(z)=z\text{.)}$

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Claim 1: Neither circle is entirely contained in the other iff $\lvert c_1-c_2 \rvert \gt \lvert r_1-r_2 \rvert.$

$(\Longrightarrow)\;$ If neither circle is entirely contained in the other, there exists a point $z_1\in C_1\setminus C_2$ and a point $z_2\in C_2\setminus C_1.$ So:

\begin{align} \lvert z_1-c_1 \rvert &\le r_1, \\ \lvert z_1-c_2 \rvert &\gt r_2, \\ \lvert z_2-c_1 \rvert &\gt r_1, \\ \text{and }\lvert z_2-c_2 \rvert &\le r_2. \end{align}

By the triangle inequality,

\begin{align} \lvert z_2-c_1 \rvert &\le \lvert c_1-c_2 \rvert + \lvert z_2-c_2\rvert \\ \text{and }\lvert z_1-c_2 \rvert &\le \lvert c_1-c_2 \rvert + \lvert z_1-c_1\rvert. \end{align}

It follows that

\begin{align} \lvert c_1-c_2 \rvert &\ge \lvert z_2-c_1\rvert -\lvert z_2-c_2\rvert \gt r_1-r_2 \\ \text{and } \lvert c_1-c_2 \rvert &\ge \lvert z_1-c_2\rvert -\lvert z_1-c_1\rvert \gt r_2-r_1, \end{align}

so

$$ \lvert c_1-c_2\rvert \gt \lvert r_1-r_2\rvert. $$

$(\Longleftarrow)\;$ Suppose one circle is entirely contained in the other; without loss of generality, $C_1 \subseteq C_2.$ We need to show that $\lvert c_1-c_2 \rvert \le \lvert r_1-r_2 \rvert.$

If $c_1=c_2,$ then $\lvert c_1-c_2 \rvert \le \lvert r_1-r_2 \rvert$ holds trivially.

So assume $c_1\ne c_2.$ Every line through $c_1$ intersects the circumference of $C_1$ in two points. Let $z$ be the point on the circumference of $C_1$ that lies on the line containing $c_1$ and $c_2,$ and which is on the opposite side of $c_1$ from $c_2.$ We have $\lvert z-c_1\rvert=r_1$ and, since $C_1\subseteq C_2,$ $\lvert z-c_2\rvert \le r_2.$ Since $c_1$ is on the line segment connecting $z$ and $c_2,$ we have $$\lvert z - c_1\rvert + \lvert c_1-c_2\rvert=\lvert z-c_2\rvert,$$

so

$$r_1 + \lvert c_1-c_2\rvert \le r_2.$$

Since $C_1\subseteq C_2,$ the diameter of $C_1$ must be less than or equal to the diameter of $C_2,$ so $r_1\le r_2,$ and it follows that

$$\lvert c_1-c_2\rvert \le r_2-r_1 = \lvert r_1-r_2 \rvert,$$

completing the proof of Claim 1.

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We'll assume that neither circle is contained within the other, so that $\lvert c_1-c_2 \rvert \gt \lvert r_1-r_2 \rvert,$ and we'll then construct a function $f$ with the desired property. Note that our hypothesis here implies that $c_1 \ne c_2.$

First, if $r_1=r_2,$ the solution is easy: just take any $f$ that is periodic with period $c_2-c_1.$ To be more specific, if $g$ is any entire function, then $f(z)=g\left(\exp\left({\frac{2\pi iz}{c_2-c_1}}\right)\right)$ will work.

Now look at the case where $r_1\ne r_2.$ Recall the function $k(z)=Az+B$ from above. $A$ is a real number not equal to $1.$ The point $v=\frac{B}{1-A}$ is a fixed point of $k,$ as you can verify by calculating $k(v).$

Claim 2: The fixed point $v$ does not lie in $C_1\cup C_2.$

Here's how to see this:

\begin{align} \lvert v-c_1\rvert &= \Big\lvert \frac{B}{1-A}-c_1 \Big\rvert \\&=\Big\lvert \frac{c_2-\frac{r_2}{r_1}c_1}{1-\frac{r_2}{r_1}}-c_1 \Big\rvert \\&=\Big\lvert \frac{r_1c_2-r_2c_1}{r_1-r_2}-c_1 \Big\rvert \\&=\Big\lvert \frac{r_1c_2-r_2c_1-r_1c_1+r_2c_1}{r_1-r_2} \Big\rvert \\&=r_1\Big\lvert \frac{c_2-c_1}{r_1-r_2}\Big\rvert \\&\gt r_1 \end{align}

and

\begin{align} \lvert v-c_2\rvert &= \Big\lvert \frac{r_1c_2-r_2c_1}{r_1-r_2}-c_2 \Big\rvert \\&=\Big\lvert \frac{r_1c_2-r_2c_1-r_1c_2+r_2c_2}{r_1-r_2} \Big\rvert \\&=r_2\Big\lvert \frac{c_2-c_1}{r_1-r_2}\Big\rvert \\&\gt r_2, \end{align}

which proves Claim 2.

$$ $$

For any angle $\theta,$ let $R_\theta=\lbrace v+te^{i\theta} \mid t\text{ is a non-negative real}\rbrace$ be the ray with endpoint $v$ at angle $\theta.$ Note that the function $k$ maps each $R_\theta$ to itself; in fact, $k$ maps $R_\theta$ one-one onto $R_\theta.$

Since $v$ does not lie in $C_1,$ there exist $\theta_1$ and $\theta_2$ such that $0 \le \theta_1 \lt 2\pi,$ $\,\theta_1 \lt \theta_2 \lt \theta_1+\pi,$ and the rays $R_\theta$ with $\theta_1 \le \theta \le \theta_2$ are precisely the rays with endpoint $v$ that intersect $C_1.$ It follows that $X_1=\lbrace \theta \mid 0 \le \theta \lt 2\pi \text{ and }R_\theta\cap C_1 \ne \emptyset\rbrace$ is either the closed interval $ [\theta_1, \theta_2]$ or the union $[\theta_2,2\pi)\cup [0,\theta_1]$ of two intervals; in either case, it has measure less than $\pi.$

Similarly, $X_2=\lbrace \theta \mid 0 \le \theta \lt 2\pi \text{ and }R_\theta\cap C_2 \ne \emptyset\rbrace$ has measure less than $\pi.$

It follows that $\operatorname{m}(X_1\cup X_2) \le \operatorname{m}(X_1) + \operatorname{m}(X_2) \lt 2\pi.$ (Here, $\operatorname{m}(X)$ denotes the measure of the set $X.$ By the way, there's no need to use anything complicated like Lebesgue measure here, although one could; we just have the union of a finite number of intervals.)

So $\lbrace \theta \mid 0 \le \theta \lt 2\pi \text{ and }R_\theta\cap (C_1\cup C_2) = \emptyset\rbrace$ is a non-empty set and, in fact, contains an open interval.

Fix any $\theta_0$ such that $R_{\theta_0}\cap (C_1\cup C_2) = \emptyset$ and $\theta_0 \ne 0.$

Let $R$ be the ray $\lbrace te^{i\theta_0}\mid t\text{ is a non-negative real number}\rbrace;$ this is just $R_{\theta_0}$ translated so that it starts at the origin instead of at $v.$

Let $L$ be a branch of the logarithm function that is analytic on $\mathbb{C}\setminus R.$

Note that $\exp\left(L(z^2)-2L(z)\right)=1,$ so $L(z^2)-L(z)$ is always an integral multiple of $2 \pi i.$ Since $\mathbb{C}\setminus R$ is connected, it follows that there is an integer constant $k$ such that $L(z^2)-2L(z)=2\pi ki$ for all $z$ such that neither $z$ nor $z^2$ lies on $R.$

Similarly, for each $z_0$ and each $z$ in $\mathbb{C}\setminus R$ such that $z_0z$ also belongs to $\mathbb{C}\setminus R,$ we have $\exp\left(L(z_0 z)-L(z)-L(z_0)\right)=1,$ so there is an integer constant $k_{z_0},$ apparently depending on $z_0,$ such that $L(z_0 z)-L(z)-L(z_0)=2 \pi k_{z_0} i$ for all $z\in\mathbb{C}\setminus R$ such that $z_0z$ also belongs to $\mathbb{C}\setminus R.$ But you can see that each $k_{z_0}$ must actually equal $k\!:$ If $z_0^{\,2}$ is also not in $R,$ then $2 \pi k_{z_0} = L(z_0 z_0)-L(z_0)-L(z_0)=L(z_0^2)-2L(z_0) = 2 \pi k.$ If $z_0^{,2}\in R,$ then let $w$ be any complex number whose argument is not $\theta_0,$ $\theta_0/2,$ $\theta_0/2+ \pi,$ or $\arg(z)-\theta_0$ $\pmod{2\pi}.$ Then we have $w\not\in R,$ $w^2\not\in R,$ and $wz\not \in R.$ So we already know that $k_w=k,$ and it follows that $2 \pi k_z=L(z w)-L(z)-L(w)=2\pi k_w=2\pi k.$

So we have an integer constant $k$ such that $L(z_1 z_2) = L(z_1)+L(z_2) + 2 \pi k i$ for all $z_1, z_2 \in \mathbb{C}\setminus R$ such that $z_1z_2$ is also not in $R.$

This lets us switch to a more convenient branch of the logarithm: $\log^*(z)=L(z)+2\pi ki.$ This branch has the advantage that $\log^*(z_1 z_2)=\log^*(z_1)+\log^*(z_2)$ for all $z_1, z_2 \in \mathbb{C}\setminus R$ such that $z_1z_2$ is also not in $R,$ as one can verify by calculating.

Since $\theta_0\ne 0,$ the branch cut $R$ includes no positive real numbers. In particular, $A\not\in R.$ It's also easy to verify that $\log^*(x)=\ln(x)$ for every positive real $x,$ using the fact that $\log^*(1^2)=2\log^*(1).$

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We'll now construct our $f.$

Let $g$ be any entire function (for any $g,$ we'll get a corresponding $f).$ Recall that $A$ is a positive real not equal to $1.$ So we can set $$f(z)=g\left(\exp\left(2\pi i {\frac{\log^* (z - v)}{\ln(A)}}\right)\right).$$ The function $f$ is analytic on $\mathbb{C}-R_{\theta_0}.$

Note that $k(z)=A(z-v)+v,$ so, for any $z\in \mathbb{C}-R_{\theta_0},$ we have \begin{align}f(k(z)) &=g\left(\exp\left(2\pi i \frac{\log^* (A(z-v))}{\ln(A)}\right)\right) \\&=g\left(\exp\left(2\pi i \frac{\log^*(A) + \log^*(z-v))}{\log^*(A)}\right)\right) \\&=g\left(\exp\left(2\pi i + \frac{\log^*(z-v))}{\log^*(A)}\right)\right) \\&=g\left(\exp\left(\frac{\log^*(z-v))}{\log^*(A)}\right)\right) \\&=f(z). \end{align}

Answer 3

Here's an even simpler proof that this is impossible for nonconstant entire functions $f$ when the circles $C_1$ and $C_2$ are of different sizes:

Suppose there is a rotation and dilation $\tau : \mathbb{C} \to \mathbb{C}$, $\tau(x) = ax+c$, $\tau(C_1) = C_2$, such that $(f \circ \tau)|_{C_1} = f|_{C_1}$.

Since by the principle of permanence an analytic function is entirely determined by its values on $C_1$, we get $f \circ \tau = f$, hence $f \circ \tau^n = f$ for all $n \in \mathbb{Z}$.

Now choose any $x \in C_1$ that is not a fixed point of $\tau$, and let $A = \{\tau^n(x) : n \in \mathbb{Z}\}$. We know that $f$ is constant on $A$. If $A$ has an accumulation point, then $f$ is constant on $\mathbb{C}$, again by the principle of permanence. The only way for $A$ not to have an accumulation point is if $a$ is a root of unity. This of course implies that $C_1$ and $C_2$ have the same size.