Let $G=\{z\in\mathbb{C}: |z-2|<1\}$ and assume f is holomorphic function on the closed disk $\overline{G}$ expect for a simple pole $z_0\in G$. If $|f(z)|=1$ for every $z\in \partial G$ then show for every complex number with $|a|>1$, $f^{-1}(a)$ contains exactly on point.
My idea was to define a curve for which the function $z\to f(z)-a$ had no zeros inside and then use the argument principle to conclude that $\frac{1}{2\pi i} \int_{\gamma} \frac{f'(z)}{f(z)-a}dz=1$, however, I was not able to define the curve and I was having trouble using some of the assumptions of the problem. Any help would be very appreciate it.
Let $\psi$ a holomorphic bijection mapping the unit disc onto $G$ st the simple pole of $f$ is the image of the origin and consider $g(w)=f\circ \psi(w), |w|< 1$; it is easy to see that $g$ extends holomorphically to the closed unit disc since $f$ does so to $\bar G$ ($\psi$ is Mobius), $g$ has a simple pole at the origin and is holomorphic everywhere else inside the unit disc, while $|g(w)|=1, |w|=1$
Then $h(w)=wg(w)$ is a holomorphic function on the closed unit disc satisfying $h(0) \ne 0, |h(w)|=1, |w|=1$ (we know all such as being finite Blaschke products but that is not necessary here)
But now for $|a|>1$ we have $h(w)-aw, aw$ have the same number of zeroes, namely one, inside the unit disc by Rouche since obviously $|h(w)|=1<|a|=|aw|$ on the unit circle, so the equation $g(w)=a$ has a unique root inside the unit circle, hence so does $f(z)=a$ inside $G$ as $\psi$ is bijection
Note that the proof above (modified with $h=z^kg$) shows that under same hypothesis except that $f$ has a pole of order $k$ inside $G$, then the preimage of any $|a|>1$ has precisely $k$ elements (up to multiplicity) inside $G$