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Find a holomorphic function on $\mathbb{C}$ \ $( (-\infty,-1]\cup [1,\infty) ) $ such that $f(0)=i$ and $(f(z))^2=z^2-1$.

I really don't have idea how to start it. The first thing I tried to do is consider $f(z)=(z^2-1)^{1/2}$ and choose good branch to satisfy the conditions, but it didn't work to me.

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Hint: where on the complex plane is $z^2-1$ when $z$ is in $(-\infty,-1] \cup [1,\infty)$?

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    $z^2-1 \in [0,\infty) \implies z^2 \in [1,\infty)\implies z \in [1,\infty) \cup (-\infty,-1]$ – 2017-02-06
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    To Robert, it will be $(-\infty,-2] \cup [0,\infty).$ – 2017-02-06
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    $z^2-1 \in [0,\infty)$ is correct. So what branch of the square root do you want to use? – 2017-02-06
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    choose the branch $\mathbb{C}$ \ $[0,\infty)$. Is it correct? – 2017-02-06