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Is $f(z)=(5-2z-2/z)^{1/2}$ regular in the sector A? $$A=\{z\in \mathbb{C};-\pi/2\leq \arg z\leq \pi/2; z\not = 0; z\not = \infty \}$$

Is $f(z)$ regular in $B$? Where $B$ is $A$ with extra restriction $1<Re(z)<2$.

mike
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Let $g(z):=5-2z-2/z$, so $f(z)=\sqrt{g(z)}$. Note that $g$ has a simple zero at $z=2$, hence a small circle around $2$ is mapped by $g$ to a small circle of index $1$ around $0$. It follows that $f$ cannot be defined continuously along $A$.

$B$, however, is simply connected and contains no zeros of $g$, thus every loop in $B$ is mapped by $g$ to a loop of index $0$, and $f$ can be defined continuously (and holomorphically in fact) along $B$.

Amitai Yuval
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  • Thanks a lot for the answer! What about $D$, where $D$ is a restriction of $A$ with $1/2<|z|<2$. And $z=1/2$ is the other zero of $g(z)$. – mike Dec 02 '14 at 21:12
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    @mike All that was said about $B$ still holds for $D$. – Amitai Yuval Dec 02 '14 at 21:45
  • Thanks you very much! I really appreciate your detailed explanation. Best regards – mike Dec 02 '14 at 22:03
  • Is $f(z)$ regular in $B'$? Where $B'$ is $A$ with extra restriction $1 \leq Re(z) \leq 2$? Thanks – mike Dec 20 '14 at 11:10
  • @mike Depends on your definition of "regular". You can set $f(2)=0$, and make $f$ continuous. It won't be holomorphic at $2$, of course. But then again, $B'$ is not open in $\mathbb{C}$, and $2$ belongs to the boundary, so I don't know what holomorphic even means in this context. – Amitai Yuval Dec 20 '14 at 12:41
  • http://math.stackexchange.com/questions/815082/a-question-about-theorem-2-in-de-bruins-1950-paper-the-roots-of-trigonometric Please take a look at another question of my mine in the hypelink above. I think that the regular inside of $B'$ means holomophic and regular on the boundary means continuous. Could you please check the EDIT part in my question cited in the hyperlink above to make sure that they are correct? Best – mike Dec 22 '14 at 04:08