Why is $\cos \sqrt z$ entire but $\sin \sqrt{z}$ isn't? I've been trying to formulate a way of comparing these two functions, in order to find out why the function $\sin \sqrt z$ is not entire, but I couldn't find a good way of doing that. What I tried so far:


*

*I wrote the series of both functions:
\begin{align}\cos(\sqrt{z})&=\sum_{n=0}^\infty \frac{(-1)^nz^n}{(2n)!}\\
\sin(\sqrt{z})&=\sum_{n=0}^\infty \frac{(-1)^nz^n\cdot\sqrt{z}}{(2n+1)!}\end{align}

*The $e$ version:
\begin{align}\cos(\sqrt{z})&=\frac{e^{i\sqrt{z}}+e^{-i\sqrt{z}}}{2}\\
\sin(\sqrt{z})&=\frac{e^{i\sqrt{z}}-e^{-i\sqrt{z}}}{2i} ,\end{align}
but they didn't help me on solving the problem. How could I progress from here? I could not find a way of writing Cauchy-Riemann equations for $\cos$ or $\sin$. 
 A: Hint Suppose $F(z) := f(\sqrt{z})$ is analytic at $z = 0$ for some choice of branch cut of $\sqrt{\cdot}$, say, $F(w)$ has power series $$F(w) \sim a_0 + a_1 w + a_2 w^2 + \cdots$$ at $w = 0$. What is the power series of $f(w) = F(w^2)$ at $w = 0$?

Additional hint We have $$f(w) \sim a_0 + a_1 w^2 + a_2 w^4 + \cdots .$$

This shows that not only is $\sin \sqrt{z}$ not entire, it's not analytic in any neighborhood of $z = 0$ (for any choice of branch cut).
A: Given any argument $z$, there are exactly two values for the square root, $w$ and $-w$.  If the sine of $w$ equals the sine of $-w$ for all $z$, then the two different possible values for the square root of $z$ lead to only a single value of the sine and thus no branch cut is required.  But, if $w$ and $-w$ give different sines for any $z$, then there must be a branch cut to separate the two sine values therefore you can't have an entire function.
It turns out this doesn't work because $\sin (-w)=-\sin w$ and thus $\ne \sin w$ except where the sine value happens to be zero.  We are forced to accept branch cuts between the zeroes of $\sin w$, meaning $w=\sqrt{z}$ is a multiple of $\pi$ and $z$ itself us a multiple of $\pi^2$.  Can you diagram a relatively simple choice for the needed branch cuts In the complex plane?
Now try it with the cosine function using the same reasoning above, but there is one little difference.  Cosine is an even function, not odd, so $\cos (-w)=+\cos w$ instead of $-\cos w$.  How does that change the rest of your conclusions above for the cosine function, as opposed to the sine?
Always watch your signs.
