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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?

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It depends on what you mean by $(\phantom z)^{1/2}$. –  Dylan Moreland Feb 19 '12 at 19:05
    
@DylanMoreland: What do you mean? Suppose I take the principal branch? –  hans Feb 19 '12 at 19:14

2 Answers 2

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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You ask two questions which are not equivalent. Well, it depends what you mean by the function's series representation: If you take the Maclaurin series for the function based on the principal branch of the square root, the series will converge in the circle of radius $\pi/2$ and center at the origin.

However, if you take the complex plane and remove two slits along the real line, namely $(-\infty,-\pi/2)$ and $(\pi/2,\infty)$, what is left is simply connected, and $\cos z$ has no zero there, and so your function will have a unique square root with value $1$ at $z=0$. It will have another square root being the negative of the first, of course, and together the two will define a the Riemann surface of the function.

Think of it this way: You take two copies of the plane, slitted as indicated above, and call them sheets. Now, at every interval $((4k+1)\pi/2,(4k+3)\pi/2)$ along the real axis, where $k$ is any integer, glue the two sheets crosswise, so the top sheet in the upper half plane is glued to the bottom sheet in the lower half plane, and vice versa. Meanwhile, in each interval $((4k-1)\pi/2,(4k+1)\pi/2)$, glue each sheet to itself (upper to lower halfplane). (For the case $k=0$ there is nothing to be done, as the sheets are already connected across the interval $(-\pi/2,\pi/2)$.)

Finally, your function $f$ can be defined to be the principal branch, as defined further up, on the top sheet and its negative on the bottom sheet. The glueing operation as defined above ensure the continuity of $f$ across the real line, except for the points $(2k+1)\pi/2$, which will be branch points of the function.

In hindsight, we see that we could have avoided half the gluing operation by starting with the sheets being copies of the complex plane with the intervals $((4k+1)\pi/2,(4k+3)\pi/2)$ removed. Then we just glue the two sheets crosswise across the remaining intervals.

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