Let $f\rightarrow \mathbb{C} \setminus \{ 0 \} \rightarrow \mathbb{C}$ be holomorphic. Show that the following are equivalent:
There exists $g : \mathbb{C} \rightarrow \mathbb{C}$ holomorphic, such that $g(z) = f (z)$ for all $z \neq 0$.
There exists $g : \mathbb{C} \rightarrow \mathbb{C}$ continuous, such that $g(z) = f (z)$ for all $z \neq 0$.
There exists $\varepsilon > 0 $ such that $f$ is bounded in $\dot{B}_\varepsilon = \{ z \in \mathbb{C} \mid \lvert z \rvert < 0 \} \setminus \{ 0 \}$.
$\lim_{z \rightarrow 0} z f(z) = 0$.
Proof so far:
$\Longrightarrow $2) is obvious b.c. holomorphic functions are continuous
$\Longrightarrow$ 3) Let $\varepsilon > 0$, since g is continuous, g is bounded in $B_\varepsilon(0)$, thus also in $\dot{B}_\varepsilon$. Now, since $f(z) = g(z)$ for $z \in \dot{B}_\varepsilon$, $f$ is also bounded in $\dot{B}_\varepsilon$
$\Longrightarrow$ 4) Let $M \geq 0$ s.t. $\lvert f(z) \rvert \leq M, \forall z \in \dot{B}_\varepsilon$. Then, since $\lvert \cdot \rvert$ is continuous on $\mathbb{C}$: $\lvert \lim_{z \rightarrow 0} z f(z) \rvert \leq \lim_{z \rightarrow 0} \lvert z \rvert M = 0 \Longrightarrow \lim_{z \rightarrow 0} z f(z) = 0$
My argument in 2)=>3) seems wrong, is there something I'm overlooking?
Now I'm struggling on 4)=>1), any tips?