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If $f$ has a double zero at a point $z$ then not only $z=\sin z$ there but also the derivatives must be the same, so that $1=\cos z$ there. Letting $w=e^{iz}$ we obtain the equation $w+\frac1w=2i$ from which it follows $w=i(1\pm\sqrt2)$ is pure imaginary. But then $z=\frac{w-\frac1w}{2i}$ is real, a contradiction. [The computational error is corrected in the comments and the other answer.] If $f$ had no simple zeros, then the only zero would be the double zero at $z=0$, so that we can write $f(z)=z^2e^{h(z)}$. Since $\sin z$ grows at most exponentially, the growth of $Re\; h(z)$ cannot be quadratic or higher in $|z|$. Hence $h$ must be a linear polynomial. But the function $f$ is clearly not of that form. [I am grateful to Emil for providing this argument.]

Hence the square root cannot be an entire function. The function $g$ does exist as a real analytic function; its radius of convergence at $0$ is probably the distance to the nearest zero of $f$, but I haven't checked this.

If $f$ has a double zero at a point $z$ then not only $z=\sin z$ there but also the derivatives must be the same, so that $1=\cos z$ there. Then $\sin^2z = 1-\cos^2 z =0$ and therefore $z=\sin z=0$, so that the only double zero is at the origin. If $f$ had no simple zeros, then the only zero would be the double zero at $z=0$, so that we can write $f(z)=z^2e^{h(z)}$. Since $\sin z$ grows at most exponentially, the growth of $Re\; h(z)$ cannot be quadratic or higher in $|z|$. Hence $h$ must be a linear polynomial. But the function $f$ is clearly not of that form. [I am grateful to Emil for providing this argument.] See Borel-Caratheodory.

Hence the square root cannot be an entire function. The function $g$ does exist as a real analytic function; its radius of convergence at $0$ is probably the distance to the nearest zero of $f$, but I haven't checked this.

If $f$ has a double zero at a point $z$ then not only $z=\sin z$ there but also the derivatives must be the same, so that $1=\cos z$ there. Letting $w=e^{iz}$ we obtain the equation $w+\frac1w=2i$ from which it follows $w=i(1\pm\sqrt2)$ is pure imaginary. But then $z=\frac{w-\frac1w}{2i}$ is real, a contradiction. [The computational error is corrected in the comments and the other answer.] If $f$ had no simple zeros, then the only zero would be the double zero at $z=0$, so that we can write $f(z)=z^2e^{g(z)}$. Since $\sin z$ grows at most exponentially, the growth of $Re\; g(z)$ cannot be quadratic or higher in $|z|$. Hence $g$ must be a linear polynomial. But the function $f$ is clearly not of that form. [I am grateful to Emil for providing this argument.]

Hence the square root cannot be an entire function. The function $g$ does exist as a real analytic function; its radius of convergence at $0$ is probably the distance to the nearest zero of $f$, but I haven't checked this.

If $f$ has a double zero at a point $z$ then not only $z=\sin z$ there but also the derivatives must be the same, so that $1=\cos z$ there. Letting $w=e^{iz}$ we obtain the equation $w+\frac1w=2i$ from which it follows $w=i(1\pm\sqrt2)$ is pure imaginary. But then $z=\frac{w-\frac1w}{2i}$ is real, a contradiction. [The computational error is corrected in the comments and the other answer.] If $f$ had no simple zeros, then the only zero would be the double zero at $z=0$, so that we can write $f(z)=z^2e^{h(z)}$. Since $\sin z$ grows at most exponentially, the growth of $Re\; h(z)$ cannot be quadratic or higher in $|z|$. Hence $h$ must be a linear polynomial. But the function $f$ is clearly not of that form. [I am grateful to Emil for providing this argument.]

Hence the square root cannot be an entire function. The function $g$ does exist as a real analytic function; its radius of convergence at $0$ is probably the distance to the nearest zero of $f$, but I haven't checked this.

If $f$ has a double zero at a point $z$ then not only $z=\sin z$ there but also the derivatives must be the same, so that $1=\cos z$ there. Letting $w=e^{iz}$ we obtain the equation $w+\frac1w=2i$ from which it follows $w=i(1\pm\sqrt2)$ is pure imaginary. But then $z=\frac{w-\frac1w}{2i}$ is real, a contradiction. [The computational error is corrected in the comments and the other answer.] Since $f$ is a transcendental function, it must hit every value it attains infinitely many times, including the value $0$. Since the zeros other than $z=0$ are simple, the square root cannot be an entire function. The function $g$ does exist as a real analytic function; its radius of convergence at $0$ is probably the distance to the nearest zero of $f$, but I haven't checked this.

If $f$ has a double zero at a point $z$ then not only $z=\sin z$ there but also the derivatives must be the same, so that $1=\cos z$ there. Letting $w=e^{iz}$ we obtain the equation $w+\frac1w=2i$ from which it follows $w=i(1\pm\sqrt2)$ is pure imaginary. But then $z=\frac{w-\frac1w}{2i}$ is real, a contradiction. [The computational error is corrected in the comments and the other answer.] If $f$ had no simple zeros, then the only zero would be the double zero at $z=0$, so that we can write $f(z)=z^2e^{g(z)}$. Since $\sin z$ grows at most exponentially, the growth of $Re\; g(z)$ cannot be quadratic or higher in $|z|$. Hence $g$ must be a linear polynomial. But the function $f$ is clearly not of that form. [I am grateful to Emil for providing this argument.]

Hence the square root cannot be an entire function. The function $g$ does exist as a real analytic function; its radius of convergence at $0$ is probably the distance to the nearest zero of $f$, but I haven't checked this.