Does there exist an open connected set in the complex plane on which the identity function has an analytic square root but not an analytic logarithm?
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If I have an analytic logarithm I can define an analytic square root from it. I am asking for the converse. An analytic square root does not give me an analytic logarithm in any simple way. – Kavi Rama Murthy Jan 22 '16 at 08:13
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The question has been answered here https://math.stackexchange.com/q/2624819/42969, is that correct? – Martin R Sep 09 '21 at 11:25
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@MartinR You are right. David Ullrich's answer is correct. – Kavi Rama Murthy Sep 09 '21 at 11:32
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Suppose $U \subseteq \mathbb{C}$ is a connected open set on which an analytic square root can be defined. Then it follows easily that $U$ cannot contain the origin, and every closed curve in $U$ must have even winding number around the origin. But any closed curve in the plane with nonzero winding number around the origin contains in its image a simple closed curve with winding number one around the origin, so $U$ cannot have any closed curves with nonzero winding number around the origin, and hence an analytic logarithm exists on $U$.
See also this question.
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"... any closed curve in the plane with nonzero winding number around the origin contains in its image a simple closed curve with winding number one around the origin ..." – I think that is intuitively clear to me, but could you give a hint how to prove that rigorously? – Martin R Jan 23 '16 at 17:34
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I just noticed that the solution to my question is easy: it is a consequence of the proof of Riemann Mapping Theorem. For example we can take w_0=0 in the proof of that theorem in Rudin's Real and Complex Analysis. Thus the region is necessarily simply connected. – Kavi Rama Murthy Jan 25 '16 at 07:50
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I withdraw my last comment. Riemann Mapping Theorem requires existence of square roots for all holomorphic functions that don't have zeros. I Don't understand Jim Belk's argument. Apart from the question raised by Martin, I also don't understand the last step in Jim's Argument. – Kavi Rama Murthy Jan 25 '16 at 08:14
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I agree that my argument skips over some important details. Unfortunately, I don't have time to work on filling these in, so I've placed a bounty on the question in the hopes of attracting someone else to give a more complete answer. – Jim Belk Jan 26 '16 at 17:15