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it's true that if I have $\Omega\subset\mathbb{C}$ a connected open non-empty set, then there exists a holomorphic function $f:\Omega\to\mathbb{C}$ such that $[f(z)]^2=z$, for all $z\in\Omega$ and $f(\Omega)$ is in a half plane of $\mathbb{C}$?

I think that it's necesary to "redefine" a branch of the square root function.

Can anyone give me some approach of the answer?

Thanks in advance.

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  • $\begingroup$ If it's the open right half plane, then the principal square root applied to f will work. You're now just a translation and a rotation away from this for the general case. $\endgroup$
    – zhw.
    Apr 13, 2016 at 4:18
  • $\begingroup$ @user21820 I may have misunderstood the problem. $\endgroup$
    – zhw.
    Apr 13, 2016 at 4:28
  • $\begingroup$ To the OP: First, f cannot have a zero in $\Omega.$ Second, are you assuming $f(\Omega)$ is contained in a half plane? $\endgroup$
    – zhw.
    Apr 13, 2016 at 4:30
  • $\begingroup$ Yes, I'm assuming that $f(\Omega)$ is contained in a half plane, but I don't know which one. $\endgroup$
    – jnaf
    Apr 13, 2016 at 4:32
  • $\begingroup$ @zhw.: It's I who didn't see the extra condition. And I made a mistake in my answer; missing $2πi$. But my answer works in general. $\endgroup$
    – user21820
    Apr 13, 2016 at 4:36

1 Answer 1

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Let $Ω = \{ z : z \in \mathbb{C} \land 0<|z|<2 \}$, which is basically an open annulus surrounding $0$, and let $g(t) = \exp(it)$ for any $t \in [0,2π]$, so that $g$ is basically a circle within $Ω$ with centre $0$. Now let $h = \arg \circ f \circ g$, and so $h$ is continuous and $\frac{1}{2π}(2h(t)-t) \in \mathbb{Z}$ for any $t \in [0,2π]$. (Note that every integer-valued continuous function is constant.) Thus $\frac{1}{2π}(2h(t)-t)$ is constant over all $t \in [0,2π]$, which gives a contradiction at $t = 0$ and $t = 2π$ (which are basically the endpoints of $g$).

Closely related question

$\def\less{\smallsetminus}$If $Ω$ is a simply-connected subset of $\mathbb{C} \less \{0\}$, then there is actually a branch of the square-root on $Ω$, though this branch may not lie within a half-plane. Below is a sketch of the proof, which reveals exactly when such holomorphic branches are possible. (The above example has $Ω$ that cannot be expanded to any simply-connected subset of $\mathbb{C} \less \{0\}$, because of that curve in $Ω$ with non-zero winding number around $0$.)

First prove that you can define a branch of $\ln$ that is holomorphic on your domain. Let $c$ be a point in your domain, and $a$ be such that $e^a = c$. Then let $g(z) = a + \frac{1}{2πi} \int_γ \frac{1}{z}$ where $γ$ is some path from $c$ to $z$. Use simply-connectedness to prove that $g$ is independent of the path and hence prove that it is holomorphic, and also that it satisfies $\exp(g(z)) = z$ for any $z$ in the domain. Then define $f(z) = \exp(\frac{1}{2}g(z))$. Prove that $f$ is a branch of $\sqrt{\ }$ as required.

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  • $\begingroup$ I understand the reasoning, but it's fails if my domain is connected? $\endgroup$
    – jnaf
    Apr 13, 2016 at 4:43
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    $\begingroup$ @Kas123: Sorry I don't know why I'm so careless today. My answer just now only works if $1$ is in the domain. To make it work in all cases, you need to start from some arbitrary point, so you need to add the appropriate constant to make it have the property of $\ln$. If the domain is just connected but not simply-connected set, it may be impossible to define the square-root, such as for an annulus around zero. $\endgroup$
    – user21820
    Apr 13, 2016 at 4:54
  • $\begingroup$ You shouldn't use the symbol f above, as that already has meaning. $\endgroup$
    – zhw.
    Apr 13, 2016 at 4:55
  • $\begingroup$ @zhw.: Okay I switched $f,g$. Anyway as I just commented, my solution may be for a different question... $\endgroup$
    – user21820
    Apr 13, 2016 at 5:01

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