Non-constant holomorphic function on an annulus with constant modulus on boundary has at least two zeros on the annulus

Question

Let $A$ be the annulus $A:=\{z\in\Bbb C:1<|z|<2\}$. Let $f$ be a non-constant holomorphic function in a neighborhood of $A$, and suppose that $\big|f(z)\big|=1$ on $\partial A$.

Prove that $f$ has at least two zeros in $A$.

$\textbf{Attempt 1}$ Now, $f$ has no zero in $A$ implies $1/f$ is continuous on $\overline A$ and non-constant holomorphic in $A$. So, the maxima of $1/f$ occurs in $\partial A$. Therefore, both maxima and minima occur in $\partial A$. That is $f:A\to \Bbb S^1$, contradicts to open mapping theorem. So, $f$ has at least one zero in $A$. We can't talk about the multiplicity of this zero. So, what about second zero?

$\textbf{Attempt 2}$ Note that $\partial A=\{|z|=1\}\sqcup \{|z|=2\}$. If $f\big|_{|z|=1}$ and $f\big|_{|z|=2}$ both are Jordan curves with counter-clockwise orientation, then number of zeros of $f$ in $A$ equals to $\int_{\partial A}\frac{f'}{f}=\int_{|z|=1}\frac{f'}{f}+\int_{|z|=2}\frac{f'}{f}\geq 1+1=2$. But, in the question there is no assumption on injectivity of $f$ on $\partial A$.

Any help will be appreciated. Thanks in advance.

Answer 1

To solve this in a more elementary way, first let's change variables $z \to z/\sqrt 2$ to symmetrize the annulus to be $A=1/\sqrt 2 < |z| < \sqrt 2$ and assume there is an $f$ as required with some simple root $f(c)=0, c \in A, |f|_{\partial A}=1$. Using the rotation, $f_1(z) =f(e^{i\arg c}z)$ we can assume $1/\sqrt 2 <c < \sqrt 2$

We claim that there is analytic $F$ on $\bar A$ and meromorphic on $\mathbb C^*$ with the following properties:

$F(c)=0, F(z) \ne 0, z \ne c, z \in A, |F(z)|=1, |z|\sqrt 2=1$ and $|F(z)|=\sqrt 2/c, |z|=\sqrt 2$ and show how this leads to a contradiction, while noting that for $c=1, G(z)=\frac{F^2}{z\sqrt 2}$ satisfies $|G(z)|=1, z\in \partial A$ and $1$ is the only (double) root of $G$ so the conclusion is actually sharp.

First from the existence of $F$ we get that $g=f/F$ is analytic and non-zero on $\bar A$ and $|g(z)|=1, |z|\sqrt 2=1$ while $|g(z)|=\frac{c}{\sqrt 2}<1, |z|=\sqrt 2$ since $1/\sqrt 2 <c < \sqrt 2$

Since $\log |g|$ is harmonic and equal to $b\log |z\sqrt 2|$ on the boundary of $A$, where $b=\frac{\log (c/\sqrt 2)}{\log 2}$, it follows that $\log |g|=b\log |z\sqrt 2|$ on the full annulus hence by the usual manipulations we would get that "$g=b_1z^{b}, b_1 \ne 0$" and that is not analytic on the annulus unless $b$ is integral; however $|\log (c/\sqrt 2)|<\log 2$ so $|b| <1, b \ne 0$ and that is a contradiction.

(for a rigorous proof of the last claim, see my answer to: Help with exercise in complex analysis on the existence of a mapping)

To construct $F$ we define it by Schwarz reflection essentially, so:

$F(z)=(1-z/c)\frac{\Pi_{k=1}^{\infty}(1-z/4^kc)(1-c/4^kz)}{\Pi_{k=1}^{\infty}(1-cz/2^{2k-1})(1-1/(cz)2^{2k-1})}, z \ne 0$

so the zeroes of $F$ are at $c,4^{\pm k}c, k \ge 1$ integral, and the poles at $1/2^{\pm (2k-1)}c, k \ge 1$ integral

Normal convergence on compact subsets of $\mathbb C^*$ (away from the poles of course) is immediate by taking logarithms so $F$ is indeed analytic on $\bar A$ and meromorphic on $\mathbb C^*$ with a simple zero at $c \in A$ (and no other zeroes in $A$ of course), while easy substitutions show that $F(z)F(1/2z)=1, F(z)F(2/z)=2/c^2$ which combined with $F$ "real" (or if you prefer conjugate invariant, $F(\bar z)=\bar F(z)$) gives the required modulus equalities noting that on $|z|\sqrt 2=1, \bar z=1/2z$ etc

Answer 2

By the reflection principle, $f$ extends as a meromorphic function to $\mathbb C^{\ast}$ and is invariant under the multiplicative group $\langle 4 \rangle \subset \mathbb C^{\ast}$. Now $\mathbb C^{\ast}/\langle 4\rangle$ is an elliptic curve and a non-constant meromorphic function on an elliptic curve has covering degree at least $2$. Therefore, $f$ has at least two zeroes (counting multiplicity) in the annulus $\tfrac12 < \lvert z\rvert<2$. By construction, these zeroes must be situated in the original annulus (otherwise $f$ would have a pole there).

Answer 3

There's a very short topological oriented proof of this. You have
$f\big(\overline A\big)\subseteq \overline B(0,1)$ (the closed unit ball/disc). Maximum modulus theorem tells us that $f\big( A\big)\subseteq B(0,1)$. (In fact these are both equalities.)

Let $\Gamma :=(\gamma,\sigma)$ be the cycle comprised of two simple closed circular curves with $\gamma$ positively oriented (counter-clockwise) around modulus 2 and $\sigma$ being negatively oriented around modulus 1. Then Artin's (homologous) criterion applies and by the Argument Principle if $n\big(f\circ\Gamma,w\big)=1$ for every $w\in B(0,1)\implies f$ is injective on $A$. Note that the first paragraph implies $n\big(f\circ\Gamma,w\big)\geq 1$ for at least one $w\in B(0,1)$.

Observe $n\big(f\circ\Gamma,0\big)=n\big(f\circ\Gamma,w\big)$ for any $w\in B(0,1)$ since $0$ and $w$ are in the same component of $\mathbb C-f\circ\Gamma$. This implies $f(A)=B(0,1)$.

Finally suppose $n\big(f\circ\Gamma,0\big)=1$:
Then $f:A\longrightarrow B(0,1)$ would be a homeomorphism (open mapping theorem) but the latter is simply connected and the former is not, which is contradiction.

Conclude: $n\big(f\circ\Gamma,0\big)\geq 2$ i.e. by Argument Principle $f$ has (at least) 2 zeros in $A$, with multiplicity.