Convergence radius of $\sqrt{\cos(z)}$ Compute the first 3 non zero terms of the Taylor expansion of $\sqrt{\cos(z)}$ at $z=0$ and determine its convergence radius, considering only the principal branch of the square root.
I've computed the first 3 non zero terms which the problem asks for:
$$f(z)=1-\frac{1}{4}z^2-\frac{1}{96}z^4$$


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*But, how do I determine the radius of convergence if I don't have the general expression for the Taylor series?

*And, what does it mean by the principal branch of the square root? I understand what the principal branch of the logarithm is, but I didn't know it existed for the square root.

The answers were obtained in comments, with the help from Daniel Fischer: 


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*radius of convergence is $\pi/2$, because $\sqrt{\cos z}$ is holomorphic for $|z|<\pi/2$, but is not holomorphic in any neighborhood of $\pi/2$, where cosine is zero. (One way to show this is to notice that the derivative of $\sqrt{\cos z}$ is unbounded there.)

*the principal branch of square root (or of any other fractional power) is defined so that the power of $1$  is equal to $1$. This convention removes the ambiguity present in the choice between two possibilities for the square root.
