Let $f(z) = \frac{\sin(z^{\alpha})}{z^{\alpha}}$ for $\alpha > 1$ and $z \in \mathbb{C}\setminus\{0\}$. Then $f$ is holomorphic (if a recall correctly ...) My question is: is $f$ holomorphic at $z = 0$ ? Can we use Riemann continuation theorem (because $\lim_{z\to0} f(z) = 1$) to say that $f$ can be analytically continued in $0$ hence forming an entire function?
I have calculated $$f'(z) = \frac{\cos(z^{\alpha})\alpha z^{\alpha-1}}{z^{\alpha}} -\alpha \frac{\sin(z^{\alpha})}{z^{\alpha+1}} = \alpha \left( \frac{\cos(z^\alpha)- \frac{\sin(z^\alpha)}{z^{\alpha}}}{z}\right)$$ We are interested in $\lim_{z\to 0} f'(z)$. Using L'Hospital rule we obtain, that if $\exists \lim_{z\to 0}f'(z)$ then $$ \lim_{z\to 0} f'(z) = \lim_{z\to 0}\alpha \left(\sin(z^{\alpha})\alpha z^{\alpha-1} - f'(z)\right) = \lim_{z\to 0} -\alpha f'(z)$$ hence, if exists then $f'(0) = 0$. Is it correct? How to calculate $f''(0)$ ?
If $\alpha$ is an integer $\ge 1$, then yes, $$f(z) = \begin{cases} \frac{\sin(z^\alpha)}{z^\alpha} & z\ne 0 \\ 1 & z=0 \end{cases} $$ is entire.
You can simply plug $z^\alpha$ into the power series for $\sin$ and divide out $z^\alpha$ term by term -- this gives a power series that must converge at every $z$, because the series for $\sin w$ converges for every $w$, in particular when $w$ happens to be $z^\alpha$.
If $\alpha$ is not an integer, then you need to make a branch cut to define $z^\alpha$. And it will generally not be the case that when you approach that branch cut along two different branches, the values of $\sin(w)/w$ will match up between them. So the function can't even be continuous on $\mathbb C\setminus 0$ then.
(Except, as @Wojowu notes, when $\alpha$ is a half-integer, because then the two branches of $z^{\alpha}$ are each other's negatives, and $\sin(w)/w$ happens to be an even function. Alternatively, in that case, plugging $z^\alpha$ into the power series and dividing, the choice of branch cancels out term by term.)