$f$ is an entire function of exponential type (that is, $|f(z)|\lesssim e^{C|z|}$), and this implies that there are other zeros, in addition to $z=0$: if not, then its Hadmard factorization would be of the form $f(z)=Az^2e^{az}$, but clearly $f$ does not have such a representation.

However, as already explained by Mikhail in his answer (when the computational error is corrected), these additional zeros will be simple: if $f(z_0)=f'(z_0)=0$, then $\sin z_0=z_0$, $\cos z_0=1$, and taking squares and then the sum shows that $1=1+z_0^2$, so $z_0=0$.

Thus there is no entire $g$ with $g^2=f$.

$f$ is an entire function of exponential type (that is, $|f(z)|\lesssim e^{C|z|}$), and this implies that there are other zeros, in addition to $z=0$: if not, then its Hadmard factorization would be of the form $f(z)=Az^2e^{az}$, but clearly $f$ does not have such a representation.

However, as already explained by Mikhail in his answer (when the computational error is corrected), these additional zeros will be simple: if $f(z_0)=f'(z_0)=0$, then $\sin z_0=z_0$, $\cos z_0=1$, and taking squares and then the sum shows that $1=1+z_0^2$, so $z_0=0$.

Thus there is no entire $g$ with $g^2=f$.

$f$ is an entire function of exponential type (that is, $|f(z)|\lesssim e^{C|z|}$), and this implies that there are other zeros, in addition to $z=0$: if not, then its Hadmard factorization would be of the form $f(z)=Az^2e^{az}$, but clearly $f$ does not have such a representation.

However, as already explained by Mikhail in his answer (when the computational error is corrected), these additional zeros will be simple: if $f(z_0)=f'(z_0)=0$, then $\sin z_0=z_0$, $\cos z_0=1$, and taking squares and then the sum shows that $1=1+z_0^2$, so $z_0=0$.

Thus there is no entire $g$ with $g=f^2$.

$f$ is an entire function of exponential type (that is, $|f(z)|\lesssim e^{C|z|}$), and this implies that there are other zeros, in addition to $z=0$: if not, then its Hadmard factorization would be of the form $f(z)=Az^2e^{az}$, but clearly $f$ does not have such a representation.

However, as already explained by Mikhail in his answer (when the computational error is corrected), these additional zeros will be simple: if $f(z_0)=f'(z_0)=0$, then $\sin z_0=z_0$, $\cos z_0=1$, and taking squares and then the sum shows that $1=1+z_0^2$, so $z_0=0$.

Thus there is no entire $g$ with $g^2=f$.