TITLE: Convergence/Analyticity of a complex function
QUESTION [0 upvotes]: Where is the function $(\cos (z))^{1\over 2}$ analytic? In other words, where in the complex plane $\mathbb C$ is the above function's series representation convergent?

REPLY [0 votes]: If an entire function has no zeroes, then it has an entire square root.  Otherwise, you have to make branch cuts between the zeroes in such a way that every simple closed loop in the plane not passing across a branch cut encircles an even number of zeroes, counting multiplicity.  (If there are only finitely many zeroes, some branch cuts may need to go to infinity.)  Harald Hanche-Olsen's answer gives one way of doing this.
More generally, for $n$th roots, replace "an even number of zeroes" with "a number of zeros divisible by $n$".  For logarithms, replace with "no zeroes".
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