Show that there is a function $g:\mathbb{C} \backslash[-1,1]\rightarrow\mathbb{C}$ such that $$g(z)^2=z^2-1$$ for all $z \in \mathbb{C}\backslash[-1,1]$
My attempt
Consider $h(z)=\dfrac{z-1}{z+1}$ for all $z\in \mathbb{C}\backslash\{-1\}$
If $\dfrac{z-1}{z+1}=c\in(-\infty,0]$, then $z=\dfrac{c+1}{1-c}\in \mathbb{R}$ then z would contain points in (-1,1] so $h$ maps $\mathbb{C}\backslash[-1,1]\rightarrow \mathbb{C}\backslash(-\infty,0]$
Since $Logz$ is holomorphic on $\mathbb{C}\backslash(-\infty,0]$ it follows that $f(z)=Log\bigg(\dfrac{z-1}{z+1}\bigg)$ is holomorphic on $\mathbb{C}\backslash[-1,1]$
Moreover $J(z)=exp(\dfrac{z}{2})$ is entire we have that $J(f(z))=exp\bigg(\dfrac{1}{2}Log\bigg(\dfrac{z-1}{z+1}\bigg)\bigg)$ is holomorphic on $\mathbb{C}\backslash[-1,1]$
Define $g:\mathbb{C} \backslash[-1,1]\rightarrow\mathbb{C}$ by $$g(z)=(z+1)J(f(z))$$ So g(z) is holomorphic on $\mathbb{C}\backslash[-1,1]$
And $g(z)^2=(z+1)^2J(f(z))^2=(z+1)^2exp\bigg(Log\bigg(\dfrac{z-1}{z+1}\bigg)\bigg)=(z+1)^2\dfrac{z-1}{z+1}=z^2-1$
Is this correct? Thank you