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?


2 Answers 2


The two branches of $\sqrt{z}$ differ only by a sign, while the cosine function is even. Thus the ambiguity in the square root is undone by the application of the cosine.

Another way to see it is to use the power series $$\cos w=\sum_{n=0}^\infty \frac{(-1)^n w^{2n}}{(2n)!},$$ insert $w=\sqrt{z}$, and to get $$\cos \sqrt{z}=\sum_{n=0}^\infty \frac{(-1)^n z^{n}}{(2n)!}.$$


$w = \cos \sqrt{z}$

$w = \cos z^\frac{1}{2}$

$w = \cos \left (e^{\frac{1}{2} ( \ln |z| + i \, Argz + i 2n\pi)} \right )$ 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.

  • $\begingroup$ What is derivative at zero? $\endgroup$
    – Sigur
    Commented Jun 25, 2018 at 22:36

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