Does constant modulus on boundary of annulus imply constant function? Suppose I have a function $f:\mathbb{C}\rightarrow \mathbb{C}$, holomorphic on some neighborhood of an annulus $r\le|z|\le R$, $r<R$. If, for $z\in\{|z|=r\text{ or }|z|=R\}$, $|f(z)|=C$ for some constant C, does it follow that $f(z)$ is a constant function?
 A: No, it does not follow. Look at $$f(z)=z+1/z$$
As $z\rightarrow 0$ the function $|f|$ tends to $\infty$, similary when $z\rightarrow \infty$ On the other hand, $|f| |_{S^1}$ is bounded by some constant $C$ (e.g. $C =2 $ :-) 
Choose a regular value $R$ of $|f|$ bigger than $C=|f| |_{S^1}$. $f^{-1}(R)$ will be a smooth $1$- dimensional submanifold of $\mathbb{C}$ with at least one component interior to the unit ball and one to the exterior of the unit ball. To see that (one of) the curves outside the unit ball will actually surround the unit ball at least once just look at the definition of $f$, which for large values of $|z|$ is just a distortion of $z$ (choose $R$ bigger if necessary). The same reasoning shows that the curve in the interior may be chosen so it surrounds the origin (i.e. the origin is contained in the disc bounded by the curve), because $f$ is basically a distortion of $1/z$ near $z=0$. 
That is, you can choose two components of $f^{-1}(R)$ which form the boundary of a (topological) annulus $G$ containing $S^1$ as a homotopically nontrivial curve, on which $f$ is defined and holomorphic. 
Any topological annulus is conformally equivalent to a standard annulus $A(r) := \{z: 1 < |z| < r\}$ - this is not trivial, but well known. Let $\phi: A(r) \rightarrow G$ denote a conformal equilvalence. Now look at $f\circ\phi$. This is holomorphic and it's norm  equals $R$ at the boundary.
(This, admittedly does not answer the question completely, since it is not clear whether $f\circ\phi$ can be extended holomorphically beyond the annulus. But it's continuous up to the boundary, so I think it's close enough to an answer to at least state the result.)
A: The answer is no. An example of such a function is the Ahlfors function, which in the case of the annulus is a 2-to-1 mapping of the annulus onto the unit disk, which extends to be holomorphic on a neighborhood of the annulus and which maps each boundary circle to the unit circle. 
More generally, let $\Omega$ be a finitely connected domain , whose boundary consists of $n$ disjoint analytic Jordan curves. Then there is a function $F$, called the Ahlfors function for $\Omega$, which has the following properties :


*

*$F$ is a $n$-to-$1$ holomorphic mapping of $\Omega$ onto the unit disk.

*$F$ extends analytically across each boundary curve of $\Omega$, and maps each of theses curves homeomorphically onto the unit circle.


The Ahlfors function is the solution to an extremal problem involving a quantity called analytic capacity. When $\Omega$ is simply connected, the Ahlfors function is simply the Riemann map. A good reference is the book "analytic capacity and measure" by Garnett. See also "Geometric function theory: explorations in complex analysis", by Krantz, theorem 4.5.9.
