During reading a solution to a question asking to find a map between to domains, after the author transformed $A=\{|z|<1\}\setminus\{x\in \Bbb{R}:a\le x \le 1\}$ ($-1
How to precisely show that $\sqrt{z}$ can be defined on an obtuse sector?
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complex-analysis
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0Can’t you describe $B$ as $\{re^{i\theta}: 0
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0Why, then, can't it be described in all $\Bbb{D}$? – 2017-02-09
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0What I am trying to understand is where a problem might arise if one doesn't use a branch... – 2017-02-09
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1Doesn’t your $\sqrt z$ jump when you go from $\frac14+i\varepsilon$ to $\frac14-i\varepsilon$? – 2017-02-09
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0You can define the square root function on any open set where you can define the logarithm (by $\sqrt{z} := \exp ( \frac{1}{2} Log z)$ and the most general domain where you can define the Logarithm is any simply connected open which does not contain the origin (and $B$ is an example of such a domain). – 2017-02-09