Simply connected set and continuous extension of an analytic function

Question

$\forall x\in\mathbb{R}$ let $\Delta_x=\{x+iy,\ y\le 0\}$. Let the open connected set $G=\mathbb{C}\setminus (\Delta_{-1}\cup\Delta_0\cup\Delta_1)$.
I proved that there is a unique analytic function $f:G\rightarrow \mathbb{C}$ statisfying $f(z)^2=\frac{1}{z(1-z^2)}\ \forall z\in G$ and $f\left (\frac{1}{2}\right )=\left (\frac{8}{3}\right )^{\frac{2}{3}}$.
I then proved that there is a unique analytic function $g:G\rightarrow \mathbb{C}$ such that $g'=f$ and $g\left (\frac{1}{2}\right )=0$.
Now I have to prove that $g$ can be extended continuously at the points $-1,\ 0$ and $1$.
I know that an expression of $g$ can be given by $g(z)=\int_{\gamma}f(w)dw$ where the path $\gamma$ is any path in $G$ from $\frac{1}{2}$ to $z$. But I don't know how to prove that $g(z)$ has finite limits as $z$ goes to $-1$, $0$ or $1$.
And I'm curious as to whether or not $g$ can be extended analytically at these three points (meaning whether or not there exists $r>0$ and $h:G\cup D(-1,r)\cup D(0,r)\cup D(1,r)\rightarrow \mathbb{C}$ analytic such that $h(z)=g(z)\ \forall z\in G$).
I'd be grateful if anyone can help me with these two questions.
Thank you!

Answer 1

The second question has a negative answer, since if $g$ extends analytically around $0$ say, $g'$ obviously does too and since $g'(z)=f(z)=k(z)z^{-1/2}$ with $k$ analytic on a small disc around $0$, $k(1)=\pm 1$ (see below) we get a contradiction as $f$ is unbounded there; similar proofs apply to $-1,1$

For the first point, using $f(z)=k(z)z^{-1/2}$, $k$ defined and analytic on a small disc centered at $0$, with $k^2(z)=\frac{1}{1-z^2}$ we get $k(0)=\pm 1, k'(0)=0$. But then $\frac{2k(z)}{z^2}=\pm \frac{2}{z^2}+k_1(z)$ has a meromorphic primitive at $0$ (there is no $\frac{1}{z}$ term) so we can find $K$ holomorphic at $0$ with $(\frac{K}{z^2})'=\frac{2k(z)}{z^2}$ and $K(0)=0$ or $zK'-2K=2zk$ near $0$ and $K(0)=0$

But then $g(z)=\frac{K(z)}{\sqrt z}+C$ on a small punctured disc at zero since $g'=f$ by computation and we can choose $C$ accordingly to match and then $g$ is continuos at $0$ since $K(z)=zK_1(z)$, $K_1$ analytic hence continuos at $0$.

Changing variables we can repeat this at $-1,1$ so we are done!