Properties of a non-constant analytic map from the annulus $A(0,1,2)$ to the unit disk such that $\lvert f(z)\rvert = 1$ on $\partial A(0,1,2)$ Let $f$ be an analytic map that sends the annulus $A(0,1,2)$ to the unit disk such that $|z|=1,|z|=2$ get mapped to the points $|f(z)| = 1$. Furthermore f is not constant. 
Prove:
1) $f$ has at least one zero point.
2) For $|w|<1$, the number of $z$ with $f(z) = w$ does not depend on $w$
3) Show that $f$ has at least $2$ zero points!
1) I already got stuck at one, using the maximum-modulus theorem, I know that $\lvert f(z)\rvert$ has to have its maximum on the boundaries. With the minimum-modulus theorem I know that if $f$ has no zero in the annulus, then the minimum of $\lvert f\rvert$ will be attained on the boundary. This means that $\lvert f(z)\rvert$ has its minimum and maximum on the boundary. but then we have that $f$ must be a constant function (because it has a local maximum everywhere), which is a contradiction to our assumption.
2) I have no clue on what to do here, mainly because I don't know what $f$ looks like.
3) I have a feeling I need to combine 1) and 2) here, but I don't seem to see how.
Any tips would be appreciated :)
Kees
 A: First, for completeness, let's recapitulate the proof of existence of a zero - although that will also follow from part 2).
Since $f$ is non-constant by assumption, and $\lvert f(z)\rvert \equiv 1$ on the boundary of the annulus, $f(A(0,1,2))$ is an open subset of the closed unit disk, hence there is a $z_0$ with $\lvert f(z_0)\rvert < 1$. By the minimum modulus principle, $f$ has a zero in $A(0,1,2)$.
For part 2), the crucial fact is the

Proposition: Let $V\subset \mathbb{C}$ be a bounded open set with piecewise smooth boundary, and $h$ a meromorphic function on a neighbourhood of $\overline{V}$ such that $h$ has neither zeros nor poles on $\partial V$. Then $$N(0) - N(\infty) = \frac{1}{2\pi i} \int_{\partial V} \frac{h'(z)}{h(z)}\,dz,\tag{1}$$ where $N(w)$ is the number of times the value $w\in \widehat{\mathbb{C}}$ is attained in $V$, counting multiplicities.

This application of the residue theorem is also known as the argument principle. Since $n(\partial V, z) = 1$ for $z\in V$ and $n(\partial V, z) = 0$ for $z \notin \overline{V}$, the residue theorem tells us
$$\frac{1}{2\pi i} \int_{\partial V} \frac{h'(z)}{h(z)}\,dz = \sum_{z\in V} \operatorname{Res}\biggl(\frac{h'}{h}; z\biggr).$$
If $h(z) = (z-z_0)^k\cdot g(z)$ in a neighbourhood of $z_0$ where $g$ is holomorphic with $g(z_0) \neq 0$, then we can write
$$\frac{h'(z)}{h(z)} = \frac{k(z-z_0)^{k-1}g(z) + (z-z_0)^kg'(z)}{(z-z_0)^kg(z)} = \frac{k}{z-z_0} + \frac{g'(z)}{g(z)}$$
in that neighbourhood of $z_0$, which shows $\operatorname{Res} \Bigl(\frac{h'}{h}; z_0\Bigr) = k$. So the residue is $0$, except in zeros or poles of $h$, where it is the multiplicity of the zero resp. minus the order of the pole, so the sum of the residues is the number of zeros minus the number of poles, both counted with multiplicity.
Since $f$ has no poles and $\lvert f(z)\rvert \equiv 1$ on $\partial A(0,1,2)$, if we already knew that $f$ is holomorphic in a neighbourhood of $\overline{A(0,1,2)}$, or at least that $f'$ extends continuously to the boundary, we could thus directly compute
$$N(w) = \frac{1}{2\pi i} \int_{\lvert z\rvert = 2} \frac{f'(z)}{f(z)-w}\,dz - \frac{1}{2\pi i} \int_{\lvert z\rvert = 1} \frac{f'(z)}{f(z)-w}\,dz$$
for $w\in \mathbb{D}$, and since the right hand side is a continuous function of $w$ conclude that $N(w)$ is constant, since $N(w)$ only attains integer values. The existence of an analytic continuation of $f$ to a neighbourhood of $\overline{A(0,1,2)}$ - even a meromorphic extension to $\mathbb{C}\setminus \{0\}$ - follows by Schwarz' reflection principle.
Without that knowledge, we have to use circles of radius $1+\varepsilon$ and $2-\varepsilon$ for small $\varepsilon > 0$, noting that given $0 < r < 1$, we have
$$f^{-1}(\overline{D(0,r)}) \subset A(0,1+\varepsilon, 2-\varepsilon)$$
for all sufficiently small $\varepsilon > 0$ since $\lvert f(z)\rvert \equiv 1$ on $\partial A(0,1,2)$. Thus for $\lvert w\rvert \leqslant r$, we have
$$N(w) = \frac{1}{2\pi i} \int_{\lvert z\rvert = 2-\varepsilon} \frac{f'(z)}{f(z)-w}\,dz - \frac{1}{2\pi i} \int_{\lvert z\rvert = 1+\varepsilon} \frac{f'(z)}{f(z)-w}\,dz,$$
and conclude as above that $N(w)$ is constant on $\overline{D(0,r)}$. Since $r < 1$ was arbitrary, it follows that $N(w)$ is constant on $\mathbb{D}$.
Finally, to show $N(0) \geqslant 2$, we proceed by contradiction. If we had $N(0) = 1$, then since $N(w)$ is constant on $\mathbb{D}$, $f$ would be a bijection between $A(0,1,2)$ and $\mathbb{D}$. But bijective holomorphic functions have a holomorphic, in particular continuous, inverse, so $f$ would be a homeomorphism. But the unit disk is simply connected, and the annulus $A(0,1,2)$ is not simply connected, hence the two are not homeomorphic, and therefore $N(w) \neq 1$ for all $w\in\mathbb{D}$.
