Let $f: \mathbb{C} \to \mathbb{C}$ be an entire function which is not constant. $g: \mathbb{C} \to \mathbb{C}$ is defined as $$ g(z) =\begin{cases} 1, & f(z)=0 \\[2ex] \dfrac{|f(z)|+1}{|f(z)|} \cdot f(z) ,& f(z) \neq 0 \end{cases} $$
And $h: B_1(0) \to \mathbb{C}$ is defined as $ h(z)= \displaystyle\int_{C_{1,0}}^{} g(z-w) \,dz$.
Is the function $h$ bounded function in $B_1(0)$?
I was thinking of going in the direction of showing that $G$ is incomplete, so if incomplete $g$ as a derivative of $h$ the lead to lead does not hold, is that the right direction? I would be happy for a direction or a solution because I am desperate for the question.
Note that $$ |g(z)|=\begin{cases}1 & \text{if } f(z)=0, \\ |f(z)|+1 & \text{if } f(z)\neq0.\end{cases} $$ In particular, $|g|$ is continuous and hence bounded on the compact disk $\overline{ B_2(0)}$. Now, if $z\in \overline{B_1(0)}$ and $w\in\partial B_1(0)$, then $z-w\in\overline{B_2(0)}$, and so $g(z-w)$ is bounded. But then also $h$ is bounded on $B_1(0)$.