Complex valued function $ \cos\sqrt z$ I'm looking for the right argument why the function $ \cos\sqrt z$ is analytic on the whole complex plane. As far as I understand, a holomorphic branch of $\sqrt z$ can only be found on the cut plane (without negative numbers) since the Argument function isn't continuous everywhere. Hence $ \cos\sqrt z$ is at least holomorphic on the same domain, but how to justify that it is actually holomorphic everywhere?

$w = \cos \sqrt{z}$
$w = \cos z^\frac{1}{2}$
$w = \cos \left (e^{\frac{1}{2} ( \ln |z| + i \, Argz + i 2n\pi)} \right )$ for all $n \in \mathbb{N}$
$w = \cos \left (e^{\frac{1}{2} (\ln |z| + i \, Argz)} e^{ i n \pi} \right )$ for all $n \in \mathbb{N}$
Recall that $e^{i n \pi} = \cos n \pi + i \sin n \pi$.
For $n = 2k$, $k \in \mathbb{N}$, we have $w = \cos \left ( e^{\frac{1}{2} (\ln |z| + i \, Argz)} \right )$
For $n = 2k+1$, $k \in \mathbb{N}$, we have $w = \cos \left ( - e^{\frac{1}{2} (\ln |z| + i \, Argz)} \right ) = \cos \left ( e^{\frac{1}{2} (\ln |z| + i \, Argz)} \right )$
Since $\cos$ is an even function, $\cos (-p) = \cos p$. For example, if $\sqrt{|z|} e^{\frac{1}{2} i \, Argz} = \pi$, then $w = \cos \pi = -1$, and $w = \cos \left (- \pi \right ) = -1$.
Thus, there is no issue of discontinuity like there is with $\sin \sqrt{z}$. There is no need to restrict the domain. Hence, there is not need for branch cuts. Therefore, you get analyticity.
