Let $g: U \to \mathbb{C}$ be an analytic function in its open and simply connected domain $U \subseteq \mathbb{C}$.
Question 1: Every time I read about complex analytic function, it is always explained that exist $g^{(n)}(z)$ analytic in $U$, for natural numbers $n$. However, my question is: Does $g$ possesses infinite anti-derivatives analytic in $U$, i.e. $g^{(n)}(z), \,\,n \in \mathbb{Z^-}$ aswell?
Answer 1: I guess so. Since $g$ is simply connected there are no "holes", therefore any closed path will be equal to zero.
However, the following one is a little trickier.
Question 2: Let's suppose $g$ has a removable singularity $z_0$ somewhere on the border of $U$. Will $z_0$ be also considered a removable singularity of the anti-derivative of $g$, $G$? Broadly speaking, will $z_0$ be a removable singularity of every $g^{(n)}, \,\, n \in \mathbb{Z^-}$?
At once, there is a counterexample in which the last question of Q2 is false: $\ln(z)$. However, I still have no idea about the first of Q2.
Any references would be appreciated.
EDIT I tried to answer Q2 by myself. However, I'm stuck right here:
$$\displaystyle\lim_{z \to a} (z-a) \cdot G(z)- \displaystyle\lim_{z \to a} [\,\,(z-a) \cdot \displaystyle\lim_{z \to a}G(z)\,\,] = 0$$
Here is how I achieved it:
Firstly, we say $a$ is a removable singularity of $g$. Then,
$$\displaystyle\lim_{z \to a} (z-a) \cdot g(z)= 0$$
Secondly, we integrate from $a$ to $z$, for any $z \in U$. Thus, after integration by parts, we achieve
$$\displaystyle\lim_{z \to a} (z-a) \cdot G(z)- \displaystyle\lim_{z \to a} [\,\,(z-a) \cdot \displaystyle\lim_{z \to a}G(z)\,\,] = 0$$
Thanks
You can refer to this Wikipedia article: https://en.wikipedia.org/wiki/Antiderivative_(complex_analysis)#Existence