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I am studying complex analysis out of Gamelin. I am reading section 1.4 on roots and principal values, but there's not a lot of worked examples.

Question: For the function $F(z) = \sqrt{z(z-1)(z-2)(z-3)}$, does there exist a continuous branch $f$ for the domain $\mathbb{C} \setminus [0,2] \cup [3,\infty)$ for which $\text{Re}(f(i)) > 0$?

This question is really simple to me if the domain were different (say if the excluded part was $[0,1] \cup [2,3]$ because these domain boundaries reflect the sign changes of $z(z-1)(z-2)(z-3)$). Basically, if I assumed the inside of the square root was real, for which $z$ would the inside be negative, then I just use the branch for $\sqrt{z}$ over the domain excluding the negative real axis. But I do not see how to apply this intuition for the above domain. Any help would be appreciated!

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    No, that doesn't work. – 2017-01-25
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    That was my guess, but I only conclude that because I don't see how to write a branch. Why can't it work? – 2017-01-25
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    The branch cut from $3$ to $\infty$ is fine. Now, we need $3$ more. Suppose we choose to cut the plane from $2$ to $-\infty$ and from $0$ to $-\infty$. The leaves one more branch cut that originates from $z=0$. What shall we do with that? – 2017-01-25
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    Let $\sqrt{\cdot}$ be the principal square root. Then consider $G(z) = F(z) / \sqrt{-z}\sqrt{z-2}\sqrt{3-z}$. With an appropriate modification along $[0, 2] \cup [3, \infty)$, this function can be made continuous on all of $\Bbb{C}$. At the same time, we have $G(z)^2 = z-1$. Certainly this is impossible. – 2017-01-25

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