# Continuous holomorphic square root proofs

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. – Thomas Andrews Nov 7 '11 at 18:32
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 $G$. – Thomas Andrews Nov 7 '11 at 18:36
I assume the definition of "region" means it is an open set? – Thomas Andrews Nov 7 '11 at 18:46
Hint: What $z \in \mathbb C$ satisfy $e^z = 0$? – Alexander Thumm Nov 7 '11 at 18:48
@VVV: Wrong, next guess please :D – Alexander Thumm Nov 7 '11 at 19:02

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 $G$.]