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How does one determine the number of branches of $f(z)$ = $\sqrt{z(1-z)}$ on the set $\Omega$ = $\mathbb{C}$ \[0,1]?

And then prove that $f(z) = \sqrt{z} + \sqrt{1-z}$ on the same set, has no branches?

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The function $z \mapsto z(1-z)$ is non-zero on $\Omega$, so there are two single-valued analytic branches on the simply connected domain $\Omega \setminus (1,\infty) = \mathbb{C} \setminus [0,\infty)$. Now if you continue the function along a circle of radius $R>1$ from the upper boundary of the branch cut $(1,\infty)$ to its lower boundary, both $\sqrt{z}$ and $\sqrt{1-z}$ pick up a phase factor of $-1$, so $f$ as their product changes by a phase factor of $(-1)^2 = 1$, i.e., not at all. This shows that both branches of $f$ are analytic in $\Omega$.

You can follow almost the same argument for the second function, except that in this case the function as a sum of two square roots, which both pick up a phase factor of $-1$, also changes by a phase factor of $-1$. So in this case there is no continuous, and thus no analytic branch of the function in $\Omega$.

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    For any branch of $\sqrt{z}$ and $\sqrt{1-z}$, their product is a branch of $\sqrt{z(1-z)}$. And you can get all the branches in this way. This is the reason that this does not cause a problem here.2012-12-22