Prove the existence of an entire function $f$ such that $f(z)=\sin^2(\sqrt{z})$ 
Prove there is an entire function $f$ such that for any branch $g$ of $\sqrt{z}$, $$\sin^2(g(z))=f(z)$$
  for all $z$ in the domain of definition of $g$.

I don't know how to overcome the fact that $g$ is not defined everywhere on the complex plane. Thanks for any help.

We have
$$\sin^2(z) = \frac{1-\cos(2z)}{2}$$
$$\cos(z) = \sum_{k=0}^\infty \frac{z^{2k}(-1)^k}{(2k)!}$$
which gives
$$\sin^2(g(z)) = \frac{1}{2}\sum_{k=1}^\infty \frac{(g^2(z))^k 4^k(-1)^{k+1}}{(2k)!} = \frac{1}{2}\sum_{k=1}^\infty \frac{z^k 4^k(-1)^{k+1}}{(2k)!}$$
except for points on the branch-cut $z\in (-\infty,0]$ where $g$ is not defined, since for any branch $g(z)$ of $\sqrt{z}$ we have $g^2(z) = z$. 
Now it's easy to check that the power-series 
$$f(z) \equiv \frac{1}{2}\sum_{k=1}^\infty \frac{z^k 4^k(-1)^{k+1}}{(2k)!}$$
converges everywhere in the complex plane any therefore representes an entire function which equals $\sin^2(g(z))$ for all points where the latter is defined.
