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Let $G\subset \mathbb{C}$ be a domain and let $f$ be a logarithmic function on $G$. Then it will be now shown that:

i) $ \displaystyle{ w(z)= \exp(\frac{1}{2}f(z))}$ is a holomorphic square root on $G$, e.g. $(w(z))^{2} = z \ \forall z \in G.$

ii) Every continuous function $w:G\rightarrow \mathbb{C}$ with $(w(z))^{2}=z \ \forall z$ is of the form $w(z)=\pm \exp(\frac{1}{2}f(z))$.

iii) If $0\in G$ then there would be no holomorphic square root on G.

VVV's work

i) The holomorphy of $w(z)$ follows directly from the theorem for composition of holomorphic functions. Every logarithmic function is of the form $f(z) \log(|z|) + i\phi + i2\pi\mathbb{Z} $

so: $w(z) = \exp(\frac{1}{2}(\log(|z|)+i\phi)) = |z|^{1/2}e^{i\frac{\phi}{2}} \Rightarrow (w(z))^{2} = |z|e^{i\phi} = z$

ii) in i) it is shown that $w(z)=\exp(\frac{1}{2}f(z))$ is a solution of $(w(z))^{2}=z \ \forall z\in G$. Since $(-w(z))^{2} = (-1)^{2}(w(z))^{2} = (w(z))^{2}$ also $-w(z)$ must fulfill this criteria. How does one show that these are all solutions?

iii) Assume $0\in G$, then look at $w(0) = \exp(\frac{1}{2}f(0))$ since $\log(0)$ isn't defined so neither can the holomorphic square root exist.

Are these proofs correct ? Does anybody see how to show that in $ii)$ $w(z)$ and $-w(z)$ are the only solutions? Please do tell me

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    You can't use the first part to prove (iii). Th first part shows that *if* you have a log function on $G$, then you have a square root. It does not show that if you have a square root, you necessarily have a log function.2011-11-07
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    For (ii): Let $w_0(z)$ be a square root function on $G$. Then $(w_0(z)-w(z))(w_0(z)+w(z)) = w_0(z)^2-w(z)^2 = 0$. So $w_0$ must agree with one of $w(z)$ or $-w(z)$ for a large enough set of values in $G$ that they must agree everywhere on $G$.2011-11-07
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    so my iii) is wrong, what else can I do ?2011-11-07
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    I assume the definition of "region" means it is an open set?2011-11-07
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    Hint: What $z \in \mathbb C$ satisfy $e^z = 0$?2011-11-07
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    @ThomasAndrews: open and simply connected2011-11-07
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    I edited …………..2011-11-07
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    What $z\in \mathbb{C}$ satisfy $e^{z}=0$ all z with $|z| \rightarrow \infty$2011-11-07
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    @VVV: Wrong, next guess please :D2011-11-07
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    @AlexanderThumm: The lack of a value for $log(0)$ doesn't imply the lack of a square root function, so that's hardly a useful hint.2011-11-07
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    z with $z\rightarrow -\infty$ . But how does this help!2011-11-07
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    @ThomasAndrews: after having proven ii), it is.2011-11-07
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    No, (ii) is still only true if you have a logarithm on $G$ - if there is a log on $G$, then the only square roots on $G$ are of that form. It doesn't say that if there is a square root, then there is a log.2011-11-07
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    To clarify: if $0\in G$ and $(w(z))^2 = z$ for all $z \in G$ we have $(w(0))^2 = 0$, hence $w(0) = 0$. Now by ii) we know, that $0 = w(0) = \pm e^x$ for some $x \in \mathbb C$, which is impossible.2011-11-07
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    No, (ii) is predicated on having a logarithm on $G$, not a general theorem.2011-11-07
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    but if you put $x = -\infty$ then $0=e^{x}$????2011-11-07
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    @VVV: There is no useful sense in which $-\infty$ is a complex number. You can add a point at infinity to the complex plane, but it is not "directional." In particular, then, $\lim_{z\to\infty} e^z$ is undefined on $\mathbb C$2011-11-07

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For (iii), use that if $w(z)^2 = z$, then $2w(z)w'(z)=1$. So $w'(0)$ cannot be defined.

For (ii): As mentioned in comments, in general, if $a,b$ are holomorphic on $G$, and $a(z)b(z)=0$ for all $z\in G$, then one of $a$ or $b$ is identically zero. So, if $w$ is a holomorphic square root function, and $w_0$ is another, then, since $0=w(z)^2-w_0(z)^2 = (w(z)-w_0(z))(w(z)+w_0(z))$, then one of $w(z)-w_0(z)$ or $w(z)+w_0(z)$ must be identically zero on $G$.

[The harder thing to prove, but not part of the problem, is that, if you can define a holomorphic square root on $G$, then you can define a logarithm on $G$.]