Suppose $w = f(z), \space z \in \mathbb{C} \space$ is an entire function which fails to take any values belonging to some curve $\gamma$ in the $w$-plane. Show that $f(z)$ is constant.
This problem appears in my textbook "Complex Analysis with Applications" by Silverman. I'm having a very tough time interpreting this question, let alone beginning to prove it.
What do they mean when they say "$f$ fails to take any values belonging to $\gamma$ in the $w$-plane?" This sentence does not make sense to me. It sounds like they are saying the domain of $f$ does not include points in $\gamma$, but $f$ is entire, so that cannot be. Do they mean that the image of $f$ does not contain the curve $\gamma$?
They gave a hint to consider the function $g(z) = \phi(f(z)), \space z \in \mathbb{C}$, where $\phi: \mathbb{C} \mathbb{P} \setminus \gamma \to \mathbb{D}_1(0)$ maps the extended complex plane with boundary $\gamma$ to the unit disk.
However, I don't know where to begin using this hint. Riemann Mapping Theorem tells us that for any simply connected domain $G$ in the extended plane whose boundary contains more than one point in $G$, there exists a unique univalent function $w = f(z)$ which maps $G$ conformally onto the unit disk.
As far as I can tell, $\mathbb{C} \mathbb{P} \setminus \gamma$ is not simply connected, so I see no reason why the $\phi$ defined above need be univalent (or even exist). Nor do I see how such a fact would help me if it was. Clearly, the fact that $f$ is entire needs to enter somewhere, but where?
Indeed, $\Bbb C\Bbb P\setminus \gamma$ need not be simply connected - or even connected. But only one of its connected components, $U_0$ say, contains the image of $f$. The complement of $U_0$ is connected because each of the connected components of $\Bbb C\Bbb P\setminus \gamma$ touches $\gamma$. We conclude that $U_0$ is simply connected and can be Riemann-mapped to the unit disk.