radius of convergence of $\sqrt{\cos z}$ Let $f(z)=\sqrt{\cos z}$ and pick the branch such that $f(0)=1$. Consider the power series of $f$. Find the radius of convergence of power series. 
I claim that the radius of convergence is at least $\pi/2$ then I would like to show that $f(z)$ is not analytic at $z=\pi/2$. So how would I show that? I got stuck on the calculation of $\int_\gamma f'/f$ where $\gamma$ is a closed curve around $\pi/2$. Are there any other easier methods? Thanks
 A: There is no use of the logarithmic indicator $\int_\gamma \frac{f'(z)}{f(z)}\,dz$. Simply, the function $f(x)=\sqrt{\cos x}$ does not have a bounded derivative in a left neighbourhood of $x=\frac{\pi}{2}$, so $f(z)$ cannot be a holomorphic function in a neighbourhood of $z=\frac{\pi}{2}$.

It is also interesting to compute the Fourier series of $f(x)=\sqrt{\cos x}$ over $I=\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.
It is not the most straightforward way to compute the radius of convergence of the Taylor series of $f(x)$ in $x=0$, but it may turn useful as a reference for future problems.
We have:

$$ c_n=\int_{0}^{\pi/2}\sqrt{\cos x}\cos(2nx)\,dx =(-1)^{n+1}\frac{\Gamma\left(-\frac{1}{4}\right)^2}{40\sqrt{2\pi}}\prod_{k=0}^{n}\frac{4k-5}{4k+1}\tag{1}$$

through integration by parts. The Euler product for the $\Gamma$ function hence gives that $\left|c_n\right|$ decays like $n^{-3/2}$. Then we have the following representation for $\sqrt{\cos x}$ over $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$:

$$ \sqrt{\cos x}=\frac{2\sqrt{2}\,\Gamma\left(\frac{3}{4}\right)^2}{\pi^{3/2}}-\frac{1}{2\sqrt{\pi}}\sum_{n\geq 1}(-1)^n\frac{\Gamma\left(n-\frac{1}{4}\right)}{\Gamma\left(n+\frac{5}{4}\right)}\,\cos(2n x).\tag{2}$$

