Using Rouche's for function constant on a circle Let $c\in\mathbb{R}$. A non-constant function $f(z)$ is holomorphic in $|z|<2$. Suppose $|f(z)|=c$ for all $|z|=1$. Show that  $f(z)$ must have a root in $|z|<1$.
Here there is an answer using the maximum principle. Since the question deals with showing the existing of a root inside a circle, I wonder if it can be solved using Rouche's theorem.

We can prove it using Rouché's theorem, but the way that I see smells artificial. Nevertheless:
Since $f$ is non-constant, we have $c > 0$. Let $g(z) = \frac12 f(z)$. Let $h(z) = g(z) + g(0)$. Since $f$, and hence $g$ is non-constant, by the maximum principle (sorry, can't do it entirely without that), we have $\lvert g(0)\rvert < \frac{c}{2}$, whence
$$\lvert h(z)\rvert \leqslant \lvert g(z)\rvert + \lvert g(0)\rvert = \frac{c}{2} + \lvert g(0)\rvert < c$$
on the unit circle. Thus, by Rouché's theorem, $f$ has as many zeros in the unit disk as
$$f(z) - h(z) = f(z) - \frac12\left(f(z) + f(0)\right) = \frac12\left(f(z) - f(0)\right).$$
The latter function evidently has at least one zero in the unit disk.
