Show two analytic functions are related by a constant

Question

The problem I'm working on says this: Suppose $f, g$ are two analytic functions on an open set containing $\overline{D}$, where $D = \{ z \in \mathbb{C} : |z| < 1 \}$ is the open disc in $\mathbb{C}$ and $\overline{D}$ is its closure. Suppose $|f(z)| = |g(z)|$ for $z \in \partial D$, and neither $f$ nor $g$ has a zero in $D$. Show that $f = \alpha g$ for some constant coefficient $\alpha \in \partial D$.

What I want to do here is take $h = \frac{f}{g}$ and apply the strong part of Schwarz's Lemma to $h$. If I can do that, then I'm done. The challenge I have is that while I see $h$ is well-defined on $D$, because neither $f$ nor $g$ vanish on $D$, I don't know how to show there's an analytic continuation to (an open neighborhood of) $\overline{D}$. If I do that, then I think I can use a Maximum Modulus argument to say that $h$ takes $D$ to $D$ (i.e. $|h(z)| < 1$ for $z \in D$), putting me in Schwarz's Lemma territory. But as said, I don't see how to get the analytic continuation to $\partial D$, i.e. I don't know how to guarantee there won't be singularities there. I know that a singularity would correspond to a point where $h$ is locally unbounded, and that can't happen along the disk, but I don't know how to be sure it won't be unbounded at nearby points off the disk.

I feel like I see my goal, just don't know how to cover the step I need. I would appreciate any help with this problem.

Thanks!

EDIT: @semiclassical pointed out that I could be clearer about my goal with $h$. I want to show that $h$ is constant of magnitude $1$.

Answer 1

First, note that both $h(z)$ and $\frac{1}{h(z)}$ are well-defined since neither of $g(z)$ or $f(z)$ have zeros in $D$. Then for all $z\in\overline{D}$, we have by the maximum modulus principle

$$|h(z)|\leq 1$$

$$\frac{1}{|h(z)|}\leq 1$$

$$\Rightarrow 1\leq |h(z)|\leq 1$$

$$\Rightarrow |h(z)|=1$$

But this implies that $h(z)$ is constant (again by the maximum modulus principle). We conclude $f(z)=e^{i\theta} g(z)$.

Answer 2

Consider $\overline D\in B(0,1+\delta)$ for some $\delta \gt 0$
(Use compactness of $\overline D$ to show this open ball exists in the domain on which $f,g$ are defined. Technical nit: the OP states the two functions are defined on a common open set but if that open set is not connected then the conclusion in the final paragraph doesn't follow, so I assume we working on a domain).

Then $h=\frac{f}{g}$ may be written as $h(z)=s(z)\cdot r(z)$ for an analytic function $s$ that is non-zero in $\overline B(0,1)$ and a rational function $r$ (why?). Now cancel duplicated roots in the numerator and denominator of $r$. The product, i.e. $h$ with any removable singularities eliminated, is meromorphic on $B(0,1+\delta)$ so it cannot have essential singularities, only poles, and we examine this in the next paragraph.

Suppose there is a pole at $z'\in \partial D$. Consider connected set $S=\partial D \cap B(z',\delta')$ for $\delta'\gt 0$ small enough such that the punctured ball $B(z',\delta')$ does not contain any zeros of $f$. The function $\big\vert\frac{1}{s(z)\cdot r(z)}\big\vert$ is continuous and integer valued on $S$, hence constant i.e. it is $=1$ on $S$ hence $z'$ was not a pole afterall. Similarly $\big \vert s(z)\cdot r(z)\big\vert$ is continuous and may only take finitely many values on $\partial D$ (equal to $1$ except perhaps at zeros of $f$) but $\partial D$ is connected hence its image under $\big \vert s(z)\cdot r(z)\big\vert$ is connected, hence constant. Conclude the behavior of $h$ implied by continuity is to map the unit circle into the unit circle.

Finish: since $h$ is non-zero in $D$, maximum modulus theorem and minimum modulus theorem tell us that for $z\in D$ we have $h(z)=s(z)\cdot r(z)\in \partial D$ and and since a maximum modulus was achieved by a point in $D$, the maximum modulus theorem tells us that $s(z)\cdot r(z)$ must be constant, i.e. $h(z) =s(z)\cdot r(z) = e^{i\theta}$ in $D$. [Alternatively the Argument Principle tells us that $s(z)\cdot r(z)$ maps all values in $D$ into $\mathbb C-D$ and applying Parseval tells us the power series expansion at $0$ is constant, i.e. $s(z)\cdot r(z)$ is constant.]
$\implies f(z)=e^{i\theta}\cdot g(z)$ in $D$ and hence everywhere in our domain.