Show that $$f(z)=\frac{\sin\sqrt z}{\sqrt z}$$ is an entire function of finite order $\rho$ and determine $\rho$.
I observed that the two determinations of the square root differ only for the signum. Since $\sin(-z)=-\sin z$, we have that $f(z)$ is well defined, and entire because it's the ratio of two entire functions with denominator never vanishing. For the order i use the Taylor expansion $$\sin z=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n+1}}{(2n+1)!}$$ which for $z=\sqrt z$ gives $$\sin\sqrt z=\sum_{n=0}^{\infty}(-1)^n\frac{z^{n}\sqrt z}{(2n+1)!}$$ Thus $$\frac{\sin\sqrt z}{\sqrt z}=\sum_{n=0}^{\infty}(-1)^n\frac{z^{n}}{(2n+1)!}$$ Then we have $$(2n+1)!\geq2^n n!$$ hence $$\bigg|\frac{\sin\sqrt z}{\sqrt z}\bigg|\leq\large\sum_{n=0}^{\infty}\left(\frac{|z|}{2}\right)^{n}\cdot\frac{1}{n!}=e^{|z|/2}$$
This (if correct) shows that $\rho\leq\frac{1}{2}$. How can be shown the identity?
To show $\rho \geq 1/2$, write
$$ f(z) = \frac{\sin \sqrt{z}}{\sqrt{z}} = \frac{e^{i\sqrt{z}}-e^{-i\sqrt{z}}}{2i\sqrt{z}} $$
and show that
$$ f(-x) \sim \frac{e^{\sqrt{x}}}{2\sqrt{x}} $$
as $x \to \infty$ with $x > 0$.
Clarification: If $f$ is of order $\rho'$ then for any $\rho>\rho'$ there is $C$ such that $$|f(z)| \leq C \exp(|z|^\rho)$$ so that $|f(z)| \exp(-|z|^\rho)$ is bounded. In particular, $$|f(-x)| \exp(-x^\rho)$$ is bounded as $x \to \infty$. Now suppose $\rho'< 1/2$, pick $\rho\in (\rho',1/2)$ and use the asymptotic estimate above.