Is $e^{\sqrt{z}}$ an entire function

Question

I was wondering whether $e^{\sqrt{z}}$ is an entire function. We know that the composition of two entire functions is entire. But $\sqrt{z}$ is a multivalued function which is analytic in the principal branch. So if we take the principal branch of $\sqrt{z}$, is $e^{\sqrt{z}}$ analytic? If so, what is the order of growth of it. The order of growth of an entire function $f$ is defined as
$\sigma=\displaystyle{\limsup\limits_{r\rightarrow\infty}\frac{\log\log M(r,f)}{\log r}}$, where $M(r,f)=\displaystyle{\sup_{|z|=r}|f(z)|}$.

Answer 1

Let $f(z)=e^{g(z)}$ where $g(z)^2=z.$ Then $f(0)=1,$ so for $z\ne 0$ we have $$\frac {f(z)-f(0)}{z-0}=\frac {e^{g(z)}-1}{g(z)^2}=$$ $$=\frac {g(z)/1!+g(z)^2/2!+g(z)^3/3+...}{g(z)^2}=A+B$$ where $A=\frac {1}{g(z)}$

and $B=1/2!+g(z)/3!+...$

Now $g(z)\to 0$ as $z\to 0.$ So $B\to 1/2$ as $z\to 0.$ But $A$ does not converge as $z\to 0.$ So $f'(0)$ does not exist.