# Existence of square roots and logarithms

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?

• Sorry, I am unable to use your hint. Kindly elaborate. – Kabo Murphy Jan 22 '16 at 4:47
• 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. – Kabo Murphy Jan 22 '16 at 8:13

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 $U$.

• "... 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
• 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. – Kabo Murphy Jan 25 '16 at 7:50
• 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. – Kabo Murphy Jan 25 '16 at 8:14
• 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

Here's a topological proof, along the lines of what Jim Belk proposed. I have tried to make it as precise as possible. Let's denote the punctured plane as $$\Bbb C^* := \Bbb C \setminus \{0\}$$, and throughout argument of a complex number will be $$[0, 2\pi)$$-valued.

I claim that the analytic square root is defined over $$U$$ if and only if $$U \subset \Bbb C^*$$ is a connected open subset such that $$U$$ does not intersect image of some proper ray $$\gamma : [0, \infty) \to \Bbb C^*$$ with $$\gamma(0) = 0$$.

First note that the map $$f : \Bbb C^* \to \Bbb C^*$$ given by $$f(z) = z^2$$ is a holomorphic double cover. For the forward direction, as $$\Omega = \Bbb C^* \setminus \gamma([0, \infty))$$ is a simply connected domain, $$f|_{\Omega} : f^{-1}(\Omega) \to \Omega$$ must be the trivial cover as every covering space of a simply connected domain must be trivial. This implies existence of a biholomorphism $$g : \Omega \to f^{-1}(\Omega)$$ which defines a section of the cover, i.e., $$f(g(z)) = z$$ for all $$z \in \Omega$$. $$g$$ defines the required analytic square root over $$\Omega$$.

For the other direction, suppose that $$U$$ intersects every proper ray emanating from the origin. In particular, it intersects the radial lines $$\theta = \theta_0$$. For any point $$p = r_0\exp(\theta_0) \in U$$ consider the open sector of the plane $$S_\delta(p) = \{r_0\exp(\theta_0) \in \Bbb C : r_0 - \delta < r < r + \delta, \theta_0 - \delta < \theta < \theta_0 + \delta\}$$, where $$\delta$$ is very small, so that $$S_\delta(p_0) \cap U$$ is a connected open neighborhood of $$p_0$$. Starting from a point $$z_0 \in U \cap [0, \infty)$$, iteratively mark down a sequence of points $$\{z_k\} \in U$$ such that $$z_{n+1} \in S_\delta({z_n})\cap U$$ and $$\text{arg}(z_{n+1}) > \text{arg}(z_n)$$ by moving a constant angular distance counterclockwise from $$z_n$$ to $$z_{n+1}$$ while staying in the $$\delta$$-sector. Eventually stop the algorithm for some $$N$$ when $$\text{arg}(z_0) - \text{arg}(z_N)$$ is less than the constant we're moving by, so $$z_0 \in S_\delta(z_N)$$. Since for each pair $$z_i, z_{i+1}$$ (where indices are modulo $$N+1$$) lies in $$S_\delta(z_i) \cap U$$, a connected open subset of $$\Bbb C$$ and therefore path-connected, there is a path $$\gamma_i$$ joining $$z_i$$ and $$z_{i+1}$$ contained in $$S_\delta(z_i) \cap U$$, hence in $$U$$.

Take the path $$\gamma = \gamma_1 * \gamma_2 * \cdots * \gamma_N$$ obtained by concatenating these individual paths. This is a loop $$\gamma : [0, 1] \to U$$ with basepoint $$z_0$$ which has winding number $$1$$ about the origin, therefore represents the generator of $$\pi_1(\Bbb C^*$$). As $$f$$ is a double cover, the map $$f_* : \pi_1(\Bbb C^*) \to \pi_1(\Bbb C^*)$$ is multiplication by $$2$$, once the identification $$\pi_1(\Bbb C^*) \cong \Bbb Z$$ has been made. $$[\gamma] \in \pi_1(\Bbb C^*)$$ does not belong to image of $$f_*$$, therefore $$\gamma$$ must lift to a path which is not a loop by the covering map $$f$$, but by path lifting lemma it must join the two distinct points in the fiber $$f^{-1}(z_0)$$. This implies the covering space $$f : f^{-1}(U) \to U$$ is connected, and therefore cannot even have a topological section, let alone a holomorphic one. Phrased otherwise, we have proved there is no analytic square root defined over $$U$$.

This implies if $$U \subset \Bbb C^*$$ has an analytic square root defined over it, it must miss a proper ray emanating from the origin. Since analytic logarithm is simply a (locally defined) section of the holomorphic covering map $$\exp : \Bbb C^* \to \Bbb C^*$$ given by the exponential, we can play out the same argument in the first half of the proof of the previous claim: if $$U \subset \Omega := \Bbb C^* \setminus \gamma([0, \infty))$$ for some proper ray $$\gamma$$, then $$\exp : \exp^{-1}(\Omega) \to \Omega$$ must be the trivial cover by simple-connectedness of $$\Omega$$. Therefore there is a holomorphic section $$g : \Omega \to \Bbb C^*$$ such that $$\exp(g(z)) = z$$ - this $$g$$ is precisely (some branch of a) holomorphic logarithm defined over $$\Omega$$. Therefore the analytic logarithm is defined over all of $$\Omega$$, hence in particular on the subset $$U \subset \Omega$$, as desired.