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Prove the multifunction $[z^{1/2}]$ has no holomorphic branch on $B(0,r)$.

Here is my attempt:

Let $f: B(0,r) /(-\infty,0] \to \mathbb{C}$ be a holomorphic branch of $[z^{1/2}]$. Note $-f$ defines another holomorphic branch on the same cutplane.

For sake of contradiction assume $h: B(0,r) \to \mathbb{C}$ is a holomorphic branch of $[z^{1/2}]$. We want $h(z) \in [z^{1/2}]$, which requires $|z|^{1/2} = |h(z)|$ and $\arg(h(z)) \in [\arg(z^{1/2})]$, so $\arg(h(z)) = \frac{\theta}{2} + 2\pi n(z)$, where $\theta \in [\arg(z)]$ is the principal argument and $n: \mathbb{C} \to \mathbb{Z}$ is some integer valued function. Therefore $h(z)=|z^{1/2}|e^{\frac{i( \theta + 2 \pi n(z))}{2}}$.

Consider $g(z) :=\frac{h(z)}{f(z)}$, where we assume $f(z) \neq 0$. Then, $g(z)=e^{i\pi n(z)} \in \{-1,1\}$. Since $g$ is continuous on a connected domain $g$, is constant, so that $h= \pm f $, which is impossible because $ \pm f$ cannot be holomorphically extended to $B(0,r)$*. That completes my argument, as we have hopefully arrived at a contradiction.

*[$ \lim_{z \to -1} h(z)= \pm f(z)$ approaching above the cut $\neq \lim_{z \to -1} h(z)= \pm f(z)$ approaching below the cut.]

If correct, does this method generalize?

Is my working correct? Thanks!

  • Continue analytically along a circle to find that $f(e^{2i\pi} z)\ne f(z) \implies f$ isn't analytic on $0<|z|<r$. Note that $z^{1/2},|z-1|< 1$ is "locally analytic on $\Bbb{C}^$" its analytic continuation exists along every curve in $\Bbb{C}^$ starting at $1$. – reuns Dec 21 '19 at 02:45
  • Thanks! I so far haven't looked at analytic continuation, so could you perhaps verify my approach? –  Dec 21 '19 at 10:19
  • My idea was to mimic the answer provided here https://math.stackexchange.com/questions/2662779/show-holomorphic-branch-of-log-z-on-omega-mathbbc-backslash-z-text-re?rq=1 –  Dec 21 '19 at 11:54
  • See also my answer to https://math.stackexchange.com/q/3358049 . – Paul Frost Dec 25 '19 at 11:03

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