Consider $f : \{z \in \mathbb{C} : |z| > 1\} \longrightarrow \mathbb{C},\ f(z) = \frac{1}{z}$. Choose the correct statement(s): (A) There are infinitely many entire functions $g$ such that $g(z) = f(z)$ for every $z \in \mathbb{C}$ with $|z| > 1$. (B) There does not exist an entire function $g$ such that $g(z) = f(z)$ for every $z \in \mathbb{C}$ with $|z| > 1$. (C) $g : \mathbb{C} \longrightarrow \mathbb{C}$ with $$g(z) = \begin{cases} 1 - \frac{1}{2}z^{2}, & |z| \leq 1 \\ \frac{1}{z}, & |z| > 1 \end{cases}$$ is an entire function such that $g(z) = f(z)$ for every $z \in \mathbb{C}$ with $|z| > 1$.
Consider $f : \{z \in \mathbb{C} : |z| > 1\} \longrightarrow \mathbb{C},\ f(z) = \frac{1}{z}$. Choose the correct statement(s):\\
(A) There are infinitely many entire functions $g$ such that $g(z) = f(z)$ for every $z \in \mathbb{C}$ with $|z| > 1$.\\
(B) There does not exist an entire function $g$ such that $g(z) = f(z)$ for every $z \in \mathbb{C}$ with $|z| > 1$.\\
(C) $g : \mathbb{C} \longrightarrow \mathbb{C}$ with
$$g(z) = \begin{cases} 1 - \frac{1}{2}z^{2}, & |z| \leq 1 \\ \frac{1}{z}, & |z| > 1 \end{cases}$$
is an entire function such that $g(z) = f(z)$ for every $z \in \mathbb{C}$ with $|z| > 1$.