I was reading Gamelin's Complex Analysis and at some point he considers the complex function $f(z)=\frac{1}{\sqrt{z(1-z)}}$ and he says that it's analytic on $\mathbb{C}$\[0,1]. I am having trouble understanding why that is true. I thought that in order for $\sqrt{f(z)}$ to be analytic you would need to exclude the domain $Re(f)\leq 0$ and $Im(f)=0$ which in this case is not what he seems to have done. Doing the calculation I get that the function would be analytic on $\mathbb{C} \setminus ((-\infty,0] \cup [1,\infty))$. I also thought that maybe he defined the argument on $(0,2\pi)$ in which case I get part of his result but also in that case I have to exclude the line $\frac{1}{2}+iy$ which he doesn't do. To give some extra context, he is using the function to calculate $\int_0^1 \frac{dx}{\sqrt{x(1-x)}}$ using exterior domains, so the problem is first to determine where the associated complex function is analytic in order to use the (exterior) residue theorem. Thanks in advance to those willing to help.
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The answer is the complement of $(-\infty, 0]\cup [1,\infty)$, i.e. $(0,1)$ together with all points outside the real axis. Just see when $z(1-z)=t$ has a solution for $t \le 0$. – Kavi Rama Murthy Nov 02 '24 at 12:21
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The square-root is consistent and continuous unless there is a path that starts at $z=z_0$, goes around $z=0$ and returns to $z_0$. Then $\sqrt{z}$ changes to the other square-root, and so does $f(z)$. Same with $z=1$. But this cut means you can't go around $z=0$ without going around $z=1$ as well. Both factors flip their sign, and $f(z)$ is the same sign it always was. – Empy2 Nov 02 '24 at 13:17
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You can examine whether this function has an analytic branch at $z=\infty$. It does. $\sqrt{z(1-z)} = i(z-1/2+1/(8z)+\dots)$ as $z \to \infty$. – GEdgar Nov 02 '24 at 15:01
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THIS ANSWER might be of interest to you. – Mark Viola Nov 05 '24 at 18:27