Find the orders of the entire functions \[ \sin^2(\sqrt z) \qquad\text{and}\qquad \exp(\sin z). \]
Solution
1. The function\[ F(z)=\sin^2(\sqrt z) \]is entire.
Use the identity \[ \sin^2w=\frac{1-\cos(2w)}2. \] Since the power series for \(\cos(2\sqrt z)\) contains only even powers of \(\sqrt z\), it is a power series in \(z\): \[ F(z) =\frac12\sum_{n=1}^{\infty} (-1)^{n+1}\frac{(2\sqrt z)^{2n}}{(2n)!}. \] Hence \(F\) is entire.
2. The order of \(F\) is at most \(1/2\).
For every \(w\in\mathbb C\), \[ |\sin w|\le e^{|w|}. \] Thus for \(|z|=r\), \[ |F(z)| =|\sin^2(\sqrt z)| \le e^{2\sqrt r}. \] Therefore \[ \log M_F(r)\le2\sqrt r, \] which implies \[ \limsup_{r\to\infty} \frac{\log\log M_F(r)}{\log r}\le\frac12. \]
3. The order of \(F\) is at least \(1/2\).
On the negative real axis, take \(z=-r\) and \(\sqrt z=i\sqrt r\). Then \[ |F(-r)| =|\sin^2(i\sqrt r)| =\sinh^2(\sqrt r). \] For large \(r\), \[ \sinh^2(\sqrt r)\ge \frac1{16}e^{2\sqrt r}. \] Hence \[ \log M_F(r)\ge2\sqrt r-O(1), \] and so the order is at least \(1/2\).
4. Consequently\[ \boxed{\rho\!\left(\sin^2(\sqrt z)\right)=\frac12.} \]
Combine 2 and 3.
5. Let\[ G(z)=e^{\sin z}. \]Then \(G\) has infinite order.
Take \[ z_r=\frac\pi2+ir. \] Since \[ \sin\left(\frac\pi2+ir\right)=\cosh r, \] one has \[ |G(z_r)|=e^{\cosh r}. \] Put \(R_r=|z_r|\). Then \(R_r\sim r\), while \[ \log M_G(R_r)\ge\cosh r\ge\frac12e^r. \] Thus \[ \log\log M_G(R_r)\ge r-O(1). \] Since \(\log R_r\sim\log r\), \[ \frac{\log\log M_G(R_r)}{\log R_r} \longrightarrow\infty. \] Therefore \[ \boxed{\rho(e^{\sin z})=\infty.} \]